Free Math Worksheets

Build your own math worksheets from 21,000 problems for grades 3 to 12, from fractions to calculus. Every problem includes step-by-step solutions.

Mean and median

Click problems to add them to your worksheet.

Difficulty: 1 – 5

5387356

The ordered data set is \(2, 4, 7, x, 13, 16, 21\). Its median is \(11\). Find \(x\).

Hints

- The data set is already ordered. - Which position is exactly in the middle? - Check that the value keeps the list in order.

Solution

1. With seven ordered values, the median is the fourth value. 2. The fourth value is \(x\). 3. Therefore, \(x = 11\).

Answer

\(x = 11\)

5387776

The jersey numbers on a basketball team are \(4, 7, 10, 12, 18, 23, 31\). Someone calculates an average jersey number of \(15\). Decide whether this mean gives useful information about the team.

Hints

- Decide whether the numbers represent amounts or only labels. - Ask whether the difference between two numbers has a meaningful interpretation. - Not every list of numbers should be summarized with a mean.

Solution

1. Jersey numbers are labels, not measured quantities. 2. Differences between jersey numbers do not represent meaningful amounts. 3. Therefore, the calculated mean of \(15\) does not describe a typical characteristic of the team.

Answer

The mean is not useful because jersey numbers are labels rather than measured quantities.

5387836

Ten teenagers have a total of \(16\) pets. The mean is \(1.6\) pets per person. How should this value be interpreted? Is it correct to say that every person has \(1.6\) pets?

Hints

- Distinguish a statement about the whole group from one about each person. - Consider whether the quantity can take non-whole-number values for an individual. - Interpret the mean as an equal share of the total.

Solution

1. The mean represents an equal mathematical distribution of \(16\) pets among \(10\) people. 2. Pets are counted in whole numbers, so \(1.6\) cannot be one person's actual number of pets. 3. Therefore, it is incorrect to say that every person has \(1.6\) pets. 4. The correct interpretation is that the group has an average of \(1.6\) pets per person.

Answer

The group has an average of \(1.6\) pets per person. This does not mean that every person has \(1.6\) pets.

5115696

Six students in a bowling club scored \(105\), \(118\), \(92\), \(126\), \(110\), and \(103\) points. a) Find the mean score. b) A scoring error is later corrected: the first score was \(117\), not \(105\). Find the new mean and the amount by which the mean increased.

Hints

- Add the scores and divide by the number of students. - Determine how much the corrected score changes the total. - Divide the change in the total among all six scores.

Solution

1. The original total is \(105+118+92+126+110+103=654\). 2. The original mean is \(654\div6=109\). 3. The corrected score increases the total by \(117-105=12\), so the new total is \(654+12=666\). 4. The new mean is \(666\div6=111\). 5. The mean increased by \(111-109=2\) points.

Answer

a) \(109\) points b) The new mean is \(111\) points, an increase of \(2\) points.

5115786

The mean height of a group of \(5\) students is \(60\) inches. A new student who is \(66\) inches tall joins the group. a) Find the new mean height for the group of \(6\) students. b) Explain without calculating whether the new mean must be greater or less than the original mean.

Hints

- Use the original mean and group size to find the original total height. - Add the new student’s height, then divide by the new group size. - Compare the new value with the old mean.

Solution

1. The original total height is \(5\times60=300\) inches. 2. The new total is \(300+66=366\) inches. 3. The new mean is \(366\div6=61\) inches. 4. The new value, \(66\) inches, is greater than the original mean of \(60\) inches, so adding it must increase the mean.

Answer

a) \(61\) inches b) The new mean must be greater because the added height is greater than the original mean.

5115816

Six students collected these amounts of blueberries: \(11\), \(16\), \(10\), \(18\), \(14\), and \(15\) ounces. a) Find the mean amount collected. b) How many students collected more than the mean? Is that more than half of the students?

Hints

- Add all six amounts and divide by \(6\). - Compare each amount with the mean. - Find half of \(6\) before answering the final question.

Solution

1. The total amount is \(11+16+10+18+14+15=84\) ounces. 2. The mean is \(84\div6=14\) ounces. 3. The amounts above the mean are \(16\), \(18\), and \(15\), so \(3\) students collected more than the mean. 4. Half of \(6\) is \(3\), so this is exactly half, not more than half.

Answer

a) \(14\) ounces b) \(3\) students; no, that is exactly half of the group.

5117616

During one school week, Maya collects recyclable bottles for a class project. From Monday through Wednesday, she collects exactly \(4\) bottles each day. On Thursday and Friday, she collects \(9\) bottles each day. How many bottles does she collect in all, and what is the mean number collected per day?

Hints

- Find the total for the first three days. - Find the total for the last two days. - Divide the five-day total by \(5\).

Solution

1. From Monday through Wednesday, Maya collects \(3\times4=12\) bottles. 2. On Thursday and Friday, she collects \(2\times9=18\) bottles. 3. The total is \(12+18=30\) bottles. 4. The mean is \(30\div5=6\) bottles per day.

Answer

Maya collects \(30\) bottles in all, for a mean of \(6\) bottles per day.

5121396

Find the number exactly halfway between each pair of values on a number line. Use the mean of the two values. a) \(12.4\) and \(18.6\) b) \(-5\) and \(3\) c) \(-\frac{1}{4}\) and \(\frac{3}{4}\) d) \(-2.5\) and \(-7.5\)

Hints

- How do you find the mean of two values? - Picture the values on a number line. Where is the point that is the same distance from both? - Pay close attention to positive and negative signs.

Solution

1. For a), calculate \(\frac{12.4 + 18.6}{2} = \frac{31}{2} = 15.5\). 2. For b), calculate \(\frac{-5 + 3}{2} = \frac{-2}{2} = -1\). 3. For c), calculate \(\frac{-\frac{1}{4} + \frac{3}{4}}{2} = \frac{\frac{2}{4}}{2} = \frac{1}{4}\). 4. For d), calculate \(\frac{-2.5 + (-7.5)}{2} = \frac{-10}{2} = -5\).

Answer

a) \(15.5\) b) \(-1\) c) \(\frac{1}{4}\) d) \(-5\)

5317486

The bar graph shows the daily high temperatures during a fall school week from Monday through Friday. a) Read the high temperature for each day. b) Find the mean high temperature for the five days.

Figure for problem 531748

Hints

- Read each bar using the y-axis scale. - Add the five temperatures. - Divide the sum by \(5\).

Solution

1. The graph shows Monday, \(54^\circ\text{F}\); Tuesday, \(60^\circ\text{F}\); Wednesday, \(58^\circ\text{F}\); Thursday, \(52^\circ\text{F}\); and Friday, \(66^\circ\text{F}\). 2. The total is \(54+60+58+52+66=290\). 3. The mean is \(290\div5=58^\circ\text{F}\).

Answer

a) Monday: \(54^\circ\text{F}\); Tuesday: \(60^\circ\text{F}\); Wednesday: \(58^\circ\text{F}\); Thursday: \(52^\circ\text{F}\); Friday: \(66^\circ\text{F}\) b) \(58^\circ\text{F}\)

5317866

The bar graph shows the nightly low temperatures, in degrees Celsius, for seven days in late fall. Find the median low temperature.

Figure for problem 531786

Hints

- What does the median represent? - Order the values from least to greatest. - Which value is in the middle of a list with seven values?

Solution

1. Order the seven temperatures: \(-5\,^\circ\text{C}, -4\,^\circ\text{C}, -3\,^\circ\text{C}, -1\,^\circ\text{C}, 1\,^\circ\text{C}, 2\,^\circ\text{C}, 4\,^\circ\text{C}\). 2. With seven values, the median is the fourth value in the ordered list. 3. The median is \(-1\,^\circ\text{C}\).

Answer

The median low temperature is \(-1\,^\circ\text{C}\).

5350476

A school cafeteria records the number of fruit servings sold during the morning break from Monday through Friday. The bar graph shows the results. a) State the number of servings sold on each day. b) Find the mean number of fruit servings sold per day. c) The cafeteria’s goal is a mean of at least \(18\) servings per day. Was the goal met? Explain.

Figure for problem 535047

Hints

- Read each bar using the y-axis scale. - Add all five daily values. - Divide the weekly total by \(5\). - Compare the mean with the goal.

Solution

1. The graph shows Monday, \(14\); Tuesday, \(18\); Wednesday, \(12\); Thursday, \(22\); and Friday, \(14\) servings. 2. The total is \(14+18+12+22+14=80\) servings. 3. The mean is \(80\div5=16\) servings per day. 4. Since \(16<18\), the goal was not met.

Answer

a) Monday: \(14\); Tuesday: \(18\); Wednesday: \(12\); Thursday: \(22\); Friday: \(14\) b) \(16\) servings per day c) No. The mean of \(16\) is less than the goal of \(18\).

5387136

A lookout tower recorded these visitor counts on seven mornings: \(128, 143, 137, 151, 146, 132, 157\). Find the mean number of visitors per morning.

Hints

- Add all seven observations. - Divide by the number of observations. - Check that the result lies between the least and greatest values.

Solution

1. Add the seven visitor counts: \(128 + 143 + 137 + 151 + 146 + 132 + 157 = 994\). 2. Divide by the number of mornings: \(994 \div 7 = 142\).

Answer

The mean number of visitors is \(142\) per morning.

5387196

The bar graph shows how many defect-free wooden parts six inspection stations checked in one hour. Find the mean number of defect-free parts per station.

Figure for problem 538719

Hints

- Read each bar value from the vertical axis. - Record exactly six values. - Check whether the mean is reasonable compared with the bar heights.

Solution

1. Read the six values from the graph: \(14, 18, 13, 17, 16,\) and \(12\). 2. Their sum is \(14 + 18 + 13 + 17 + 16 + 12 = 90\). 3. The mean is \(90 \div 6 = 15\) parts per station.

Answer

The mean is \(15\) defect-free parts per station.

5387256

A cold-storage room has these seven temperature differences from its target temperature, in degrees Celsius: \(-3, -1, 2, 4, -2, 1, 6\). Find the mean difference.

Hints

- Pay attention to signs when adding. - Positive and negative differences can cancel. - Interpret the sign of the result.

Solution

1. Add the signed differences: \(-3 + (-1) + 2 + 4 + (-2) + 1 + 6 = 7\). 2. Divide by the number of measurements: \(7 \div 7 = 1\,^\circ\text{C}\). 3. The positive mean means the temperature averaged \(1\,^\circ\text{C}\) above the target.

Answer

The mean difference is \(1\,^\circ\text{C}\), so the temperature averaged \(1\,^\circ\text{C}\) above the target.

5387266

Six teams assemble the same small display booth. Their times are \(11.5\), \(12.2\), \(10.8\), \(13.1\), \(12.4\), and \(12.6\) minutes. Find the mean setup time, rounded to the nearest tenth of a minute.

Hints

- Keep the decimal points aligned when adding. - Round only after dividing. - Check that the result lies between the least and greatest times.

Solution

1. Add the six times: \(11.5 + 12.2 + 10.8 + 13.1 + 12.4 + 12.6 = 72.6\) minutes. 2. Divide by \(6\): \(72.6 \div 6 = 12.1\) minutes. 3. To the nearest tenth, the mean is \(12.1\) minutes.

Answer

The mean setup time is \(12.1\) minutes.

5387296

Nine teams took these numbers of minutes to set up a light-art installation: \(12, 7, 9, 15, 10, 8, 11, 14, 6\). Find the median setup time.

Hints

- Order the values from least to greatest. - Find the position with the same number of values on each side. - Check that all nine times are included.

Solution

1. Order the times: \(6, 7, 8, 9, 10, 11, 12, 14, 15\). 2. With nine values, the median is the fifth value. 3. The median is \(10\) minutes.

Answer

The median setup time is \(10\) minutes.

5387306

Eight short nature trails have these lengths: \(4.2\,\text{km}, 5.1\,\text{km}, 3.8\,\text{km}, 6\,\text{km}, 4.7\,\text{km}, 5.5\,\text{km}, 4.4\,\text{km}, 5\,\text{km}\). Find the median trail length.

Hints

- Order the lengths from least to greatest. - With an even number of values, two values are in the middle. - Find the mean of those two middle values.

Solution

1. Order the lengths: \(3.8, 4.2, 4.4, 4.7, 5, 5.1, 5.5, 6\) kilometers. 2. The two middle values are \(4.7\) and \(5\). 3. The median is \((4.7 + 5) \div 2 = 4.85\,\text{km}\).

Answer

The median trail length is \(4.85\,\text{km}\).

5387336

The bar graph shows the number of checkpoints on seven orienteering courses. Find the median number of checkpoints.

Figure for problem 538733

Hints

- Read all seven bar values. - The route order does not matter for the median. - Order the values before finding the middle.

Solution

1. Read the values from the graph: \(5, 8, 6, 9, 7, 10,\) and \(6\). 2. Order the values: \(5, 6, 6, 7, 8, 9, 10\). 3. The middle value is \(7\), so the median is \(7\) checkpoints.

Answer

The median is \(7\) checkpoints.

5387386

Exactly one of the numbers \(5\), \(10\), \(14\), or \(25\) is inserted into the ordered list \(4, 7, 9, 12, 16, 20\). Which number must be inserted so that the new median is \(10\)?

Hints

- Place each possible value in the ordered list. - With seven values, one position determines the median. - You do not need to recompute everything for each choice.

Solution

1. After one number is inserted, there are seven values, so the fourth value is the median. 2. Inserting \(10\) gives \(4, 7, 9, 10, 12, 16, 20\). 3. The fourth value is \(10\), so \(10\) is the required number.

Answer

\(10\) must be inserted.

5387396

Noah tries to find the median of \(18, 12, 15, 21, 14, 16, 19\). He chooses the fourth listed value, \(21\), because there are seven numbers. Explain his error and find the correct median.

Hints

- Are the values already ordered from least to greatest? - The original list position is not necessarily the middle value by size. - Order the values and check that all seven are included.

Solution

1. The values are not ordered, so the fourth position in the original list does not determine the median. 2. Order the values: \(12, 14, 15, 16, 18, 19, 21\). 3. The fourth ordered value is \(16\), so the median is \(16\).

Answer

Noah must order the data first. The correct median is \(16\).

5387456

The ordered data set is \(5, 6, 7, 8, 9, 10, 11\). Then the value \(100\) is added. Find the original median and the new median.

Hints

- Find the middle before and after adding the value. - The number of values changes from odd to even. - The new value belongs at the end of the ordered list.

Solution

1. With seven values, the fourth value is the median, so the original median is \(8\). 2. After adding \(100\), the ordered list is \(5, 6, 7, 8, 9, 10, 11, 100\). 3. The middle values are \(8\) and \(9\), so the new median is \((8 + 9) \div 2 = 8.5\).

Answer

Original median: \(8\) New median: \(8.5\)

5387466

A second \(7\) is added to the ordered data set \(2, 5, 7, 9, 12, 14, 18\). Find the new median.

Hints

- Insert the additional value in the correct position. - Repeated values are allowed. - With eight values, use the two middle positions.

Solution

1. The new ordered list is \(2, 5, 7, 7, 9, 12, 14, 18\). 2. With eight values, the middle values are the fourth and fifth values, \(7\) and \(9\). 3. The new median is \((7 + 9) \div 2 = 8\).

Answer

The new median is \(8\).

5387476

Which data set has a median of \(9\)? A: \(2, 4, 6, 8, 12, 18\) B: \(1, 5, 7, 11, 12, 14\) C: \(3, 4, 9, 10, 13, 20\)

Hints

- Each data set is already ordered. - Use only the two middle positions in each set. - Compare each median with the target value.

Solution

1. Each data set has six values, so find the mean of the third and fourth values. 2. For A, the median is \((6 + 8) \div 2 = 7\). 3. For B, the median is \((7 + 11) \div 2 = 9\). 4. For C, the median is \((9 + 10) \div 2 = 9.5\).

Answer

Data set B has a median of \(9\).

5387516

A team had these rankings in eleven regional competitions: \(2, 5, 1, 7, 4, 3, 6, 8, 9, 2, 4\). Find the median ranking.

Hints

- Put the smaller ranking numbers first. - Find the middle position after ordering the values. - Keep repeated rankings in the list.

Solution

1. Order the rankings: \(1, 2, 2, 3, 4, 4, 5, 6, 7, 8, 9\). 2. With eleven values, the median is the sixth value. 3. The sixth value is \(4\).

Answer

The median ranking is \(4\).

5387816

Two dance groups each have \(15\) members. The groups practice an average of \(7.2\) hours and \(8.4\) hours per month, respectively. Can you average the two group means to find the mean for all \(30\) members? Explain and calculate the combined mean.

Hints

- Compare the two group sizes first. - Equal group sizes give the two means equal weight. - You can check the result using the groups' total practice hours.

Solution

1. The two means represent equal numbers of people, so they have equal weight. 2. The combined mean is \((7.2 + 8.4) \div 2 = 7.8\) hours per month. 3. The same result follows by adding the two groups' total practice hours and dividing by \(30\).

Answer

Yes. Because the groups are the same size, the combined mean is \(7.8\) hours per month.

5387826

Four model bridges are \(9\,\text{cm}, 10\,\text{cm}, 10\,\text{cm},\) and \(12\,\text{cm}\) tall. Their mean height is \(10.25\,\text{cm}\), even though no bridge has exactly that height. Does that make the mean unusable? Explain.

Hints

- A mean does not have to appear in the data set. - Think of the mean as an equal share of the total. - Check whether the data represent genuine measurements.

Solution

1. The mean is \((9 + 10 + 10 + 12) \div 4 = 10.25\,\text{cm}\). 2. A mean does not have to be one of the observed values. 3. It represents the total height distributed equally among the four bridges and lies between the minimum and maximum. 4. Therefore, it is a useful summary.

Answer

No. The mean of \(10.25\,\text{cm}\) is a useful summary even though no bridge has exactly that height.

5387896

The pie chart shows how students travel to school. Jordan codes “walk” as \(1\), “bike” as \(2\), “bus” as \(3\), and “train” as \(4\), then plans to calculate a mean. Evaluate Jordan's plan.

Figure for problem 538789

Hints

- Decide whether the numbers would be measurements or only labels. - Consider whether changing the codes would change the result. - Use summaries that do not depend on arbitrary coding.

Solution

1. The numbers \(1\) through \(4\) are arbitrary codes for categories. 2. The distances between the codes do not represent measurable differences between transportation methods. 3. A different coding system would produce a different mean. 4. Therefore, a mean is not meaningful. Percentages or the most common category are appropriate summaries.

Answer

A mean is not meaningful because the numbers are only category codes. The transportation methods should be described using their percentages or the most common category.

5387996

Five teams have \(3, 5, 7, 9, 11\) water containers. All containers will be redistributed so that every team receives the same number. Which measure gives the new number per team?

Hints

- Imagine combining all the containers first. - Divide the total equally among the teams. - The question asks about equal sharing, not only a typical position.

Solution

1. There are \(3 + 5 + 7 + 9 + 11 = 35\) containers altogether. 2. An equal distribution gives \(35 \div 5 = 7\) containers per team. 3. The arithmetic mean represents this equal-share value. 4. The median also happens to be \(7\), but the equal-sharing interpretation comes from the mean.

Answer

Each team receives \(7\) containers. The arithmetic mean represents this equal distribution.

5388066

The ordered data set is \(2, 4, 8, 10\). Its median is \(6\), even though \(6\) is not in the data set. Is the median therefore incorrect or meaningless?

Hints

- Recall how to find the median for an even number of values. - Count how many values lie on each side. - A summary statistic does not have to appear in the data set.

Solution

1. With four values, the median is the mean of the two middle values, \(4\) and \(8\). 2. The median is \((4 + 8) \div 2 = 6\). 3. Two data values lie on each side of \(6\). 4. The median still divides the ordered data into two equal halves even though it is not an observed value.

Answer

The median of \(6\) is correct and meaningful because it divides the ordered data into two equal halves.

5115706

A basketball team scored \(42\), \(38\), \(55\), and \(45\) points in its first four games. The team wants a mean of exactly \(46\) points after five games. How many points must the team score in the fifth game?

Hints

- Use the target mean to find the required total for five games. - Add the first four scores. - Subtract the current total from the required total.

Solution

1. A mean of \(46\) points over \(5\) games requires a total of \(46\times5=230\) points. 2. The first four games total \(42+38+55+45=180\) points. 3. The fifth-game score must be \(230-180=50\) points.

Answer

The team must score \(50\) points in the fifth game.

5115796

Two classes compare their results in a charity run. Class 6A has \(20\) students and ran a mean distance of \(4.5\) miles per student. Class 6B has \(25\) students and ran a mean distance of \(4.0\) miles per student. Determine whether each statement is true or false, and justify your decision mathematically. a) “Class 6A ran a greater total distance than Class 6B.” b) “Because Class 6A has the greater mean, every student in Class 6A ran farther than every student in Class 6B.”

Hints

- Multiply each class size by its mean to find the total distance. - A group mean does not identify individual values. - A larger group can have a greater total even with a smaller mean.

Solution

1. Class 6A’s total distance is \(20\times4.5=90\) miles. 2. Class 6B’s total distance is \(25\times4.0=100\) miles. 3. Statement a is false because \(90<100\). 4. Statement b is false. A mean summarizes all values but does not determine any individual student’s distance. Individual distances in the two classes may overlap.

Answer

a) False. Class 6A ran \(90\) miles in total, while Class 6B ran \(100\) miles. b) False. The means do not show how far each individual student ran.

5115806

A basketball team has \(5\) players. Their mean number of points in one game was exactly \(12\). Four players scored \(8\), \(15\), \(10\), and \(14\) points. a) How many points did the fifth player score? b) If the other four scores stayed the same, how many points would the fifth player have needed for the team mean to be \(15\) points?

Hints

- Multiply the mean by the number of players to find the required total. - Add the four known scores. - Subtract the known total from each required total.

Solution

1. A mean of \(12\) for \(5\) players requires a total of \(5\times12=60\) points. 2. The four known scores total \(8+15+10+14=47\) points. 3. The fifth player scored \(60-47=13\) points. 4. A mean of \(15\) would require a total of \(5\times15=75\) points. 5. The fifth player would need \(75-47=28\) points.

Answer

a) \(13\) points b) \(28\) points

5115826

Jordan earned \(8\), \(12\), \(15\), \(9\), and \(11\) points on five math quizzes. a) Find the mean score. b) Find the distance from the mean to the highest score and from the mean to the lowest score. c) Jordan wants a mean of exactly \(12\) points after a sixth quiz. What score is needed on the sixth quiz?

Hints

- Add the five scores and divide by \(5\). - Identify the greatest and least scores before finding their distances from the mean. - Use the target mean to find the required six-quiz total.

Solution

1. The current total is \(8+12+15+9+11=55\), so the mean is \(55\div5=11\). 2. The highest score is \(15\), so its distance from the mean is \(15-11=4\). 3. The lowest score is \(8\), so its distance from the mean is \(11-8=3\). 4. A mean of \(12\) after \(6\) quizzes requires a total of \(12\times6=72\) points. 5. The required sixth score is \(72-55=17\).

Answer

a) \(11\) b) Distance to the highest score: \(4\); distance to the lowest score: \(3\) c) \(17\) points

5115846

Jordan and Maya are training for a \(100\)-meter race. During one week, each runner records five times. <table> <tr><th>Name</th><th>Run 1</th><th>Run 2</th><th>Run 3</th><th>Run 4</th><th>Run 5</th></tr> <tr><td>Jordan</td><td>\(14.5\,\text{s}\)</td><td>\(13.8\,\text{s}\)</td><td>\(14.2\,\text{s}\)</td><td>\(15.1\,\text{s}\)</td><td>\(14.4\,\text{s}\)</td></tr> <tr><td>Maya</td><td>\(13.9\,\text{s}\)</td><td>\(14.1\,\text{s}\)</td><td>\(14.0\,\text{s}\)</td><td>\(14.3\,\text{s}\)</td><td>\(14.7\,\text{s}\)</td></tr> </table> Find each runner’s mean time. Who was faster on average?

Hints

- Add each runner’s five times separately. - Divide each total by \(5\). - In a race, a smaller time is faster.

Solution

1. Jordan’s total time is \(14.5+13.8+14.2+15.1+14.4=72.0\,\text{s}\). 2. Jordan’s mean is \(72.0\div5=14.4\,\text{s}\). 3. Maya’s total time is \(13.9+14.1+14.0+14.3+14.7=71.0\,\text{s}\). 4. Maya’s mean is \(71.0\div5=14.2\,\text{s}\). 5. Since \(14.2<14.4\), Maya was faster on average.

Answer

Jordan’s mean time was \(14.4\,\text{s}\). Maya’s mean time was \(14.2\,\text{s}\), so Maya was faster on average.

5115876

The weights of \(20\) students’ backpacks are summarized in the table. <table> <tr><td>Weight (pounds)</td><td>\(8\)</td><td>\(9\)</td><td>\(10\)</td><td>\(11\)</td></tr> <tr><td>Frequency</td><td>\(6\)</td><td>\(8\)</td><td>\(4\)</td><td>\(2\)</td></tr> </table> a) Find the relative frequency of each weight as a percent. b) Find the mean backpack weight.

Hints

- Divide each frequency by \(20\) to find its relative frequency. - Multiply each weight by its frequency before adding. - Divide the total backpack weight by \(20\).

Solution

1. The relative frequencies are: \(8\) pounds, \(\frac{6}{20}=30\%\); \(9\) pounds, \(\frac{8}{20}=40\%\); \(10\) pounds, \(\frac{4}{20}=20\%\); \(11\) pounds, \(\frac{2}{20}=10\%\). 2. The total weight is \(6\times8+8\times9+4\times10+2\times11=48+72+40+22=182\) pounds. 3. The mean is \(182\div20=9.1\) pounds.

Answer

a) \(8\) pounds: \(30\%\); \(9\) pounds: \(40\%\); \(10\) pounds: \(20\%\); \(11\) pounds: \(10\%\) b) \(9.1\) pounds

5115886

A \(5\)-point math quiz was given to \(25\) students. The score distribution was: <table> <tr><td>Score</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td></tr> <tr><td>Number of students</td><td>\(3\)</td><td>\(7\)</td><td>\(10\)</td><td>\(4\)</td><td>\(1\)</td></tr> </table> a) Find the mean quiz score. b) A recording error is discovered: one student’s score was entered as \(1\) point instead of \(5\) points. Find the corrected mean and the amount by which it increases.

Hints

- Multiply each score by the number of students who earned it. - Add those products and divide by \(25\). - Determine how the correction changes the total score.

Solution

1. The original total is \(3\times1+7\times2+10\times3+4\times4+1\times5=68\) points. 2. The original mean is \(68\div25=2.72\) points. 3. Correcting a \(1\) to a \(5\) increases the total by \(5-1=4\), so the corrected total is \(72\). 4. The corrected mean is \(72\div25=2.88\) points. 5. The mean increases by \(2.88-2.72=0.16\) point.

Answer

a) \(2.72\) points b) The corrected mean is \(2.88\) points, an increase of \(0.16\) point.

5115896

A survey asked \(25\) students how many books they read last month. - \(4\) students read \(0\) books. - \(10\) students read \(1\) book. - \(6\) students read \(2\) books. - \(5\) students read \(3\) books. Determine whether the mean number of books per student is greater than or less than \(1.5\). Show the calculation.

Hints

- Multiply each number of books by its frequency. - Add the products to find the total number of books. - Divide by the \(25\) students, then compare with \(1.5\).

Solution

1. The total number of books is \(4\times0+10\times1+6\times2+5\times3=37\). 2. The mean is \(37\div25=1.48\) books per student. 3. Since \(1.48<1.5\), the mean is less than \(1.5\).

Answer

The mean is \(1.48\) books per student, which is less than \(1.5\).

5116176

A candy store mixes \(3\) pounds of strawberry candy priced at \(\$14\) per pound with \(2\) pounds of lemon candy priced at \(\$9\) per pound. What is the average cost per pound of the mixture?

Hints

- Find the total cost of each type of candy. - Add the costs and the weights separately. - Divide the total cost by the total weight.

Solution

1. The strawberry candy costs \(3\times\$14=\$42\). 2. The lemon candy costs \(2\times\$9=\$18\). 3. The total cost is \(\$42+\$18=\$60\), and the total weight is \(3+2=5\) pounds. 4. The average cost is \(\$60\div5=\$12\) per pound.

Answer

The mixture costs an average of \(\$12\) per pound.

5116186

Two classes participate in a charity walk. Class 6A has \(24\) participants who collected a mean of \(\$10\) each. Class 6B has \(26\) participants who collected a mean of \(\$15\) each. Find the mean amount collected per participant for both classes combined.

Hints

- The two class sizes are different, so do not simply average \(\$10\) and \(\$15\). - Find the total collected by each class. - Divide the combined total by the combined number of participants.

Solution

1. Class 6A collected \(24\times\$10=\$240\). 2. Class 6B collected \(26\times\$15=\$390\). 3. Together, the classes collected \(\$240+\$390=\$630\) from \(24+26=50\) participants. 4. The combined mean is \(\$630\div50=\$12.60\) per participant.

Answer

The combined mean is \(\$12.60\) per participant.

5116196

A gardener has three rain barrels. The first contains \(30\) gallons of water, and the second contains \(40\) gallons. The gardener wants the three barrels to contain a mean of exactly \(36\) gallons. a) How many gallons must be in the third barrel? b) Without a new calculation, explain how much water each barrel would contain if all the water were distributed equally among the three barrels.

Hints

- Multiply the target mean by \(3\) to find the required total. - Subtract the contents of the first two barrels. - A mean can be interpreted as an equal share.

Solution

1. A mean of \(36\) gallons for \(3\) barrels requires a total of \(3\times36=108\) gallons. 2. The first two barrels contain \(30+40=70\) gallons. 3. The third barrel must contain \(108-70=38\) gallons. 4. The mean is the equal-share amount, so distributing the total evenly would put \(36\) gallons in each barrel.

Answer

a) \(38\) gallons b) \(36\) gallons in each barrel

5117626

Jordan is reading a \(123\)-page book. During the first \(4\) days, Jordan reads exactly \(15\) pages per day. Jordan wants to finish the book in exactly one week, or \(7\) days. What mean number of pages per day must Jordan read during the remaining \(3\) days?

Hints

- Find how many pages have already been read. - Subtract from the total number of pages. - Divide the remaining pages by the remaining days.

Solution

1. Jordan reads \(4\times15=60\) pages during the first four days. 2. The number of pages remaining is \(123-60=63\). 3. There are \(7-4=3\) days remaining. 4. The required mean is \(63\div3=21\) pages per day.

Answer

Jordan must read a mean of \(21\) pages per day during the remaining \(3\) days.

5121406

At noon on five consecutive days, the temperatures were: \(14\,^\circ\text{C}\), \(17\,^\circ\text{C}\), \(13\,^\circ\text{C}\), \(16\,^\circ\text{C}\), and \(15\,^\circ\text{C}\). a) Find the mean temperature for the five days. b) A temperature is recorded on a sixth day. What must that temperature be for the mean of all six days to be exactly \(16\,^\circ\text{C}\)?

Hints

- What total is needed when you know the desired mean and the number of values? - Compare the needed total for six days with the total of the first five temperatures.

Solution

1. For a), add the five temperatures: \(14 + 17 + 13 + 16 + 15 = 75\). Then divide by \(5\): \(75 \div 5 = 15\). The mean temperature is \(15\,^\circ\text{C}\). 2. For b), a mean of \(16\,^\circ\text{C}\) over \(6\) days requires a total of \(6 \times 16 = 96\). 3. The first five temperatures total \(75\), so the sixth temperature must be \(96 - 75 = 21\,^\circ\text{C}\).

Answer

a) \(15\,^\circ\text{C}\) b) \(21\,^\circ\text{C}\)

5121416

A basketball player averaged \(18\) points per game in the first four games of the season. a) How many total points did she score in those four games? b) She scores \(28\) points in the fifth game. Find her new mean after five games. c) How many points must she score in the sixth game for her mean to return to \(18\) points per game? Explain without doing a long calculation.

Hints

- How can you find the total when you know the mean and the number of values? - For part c), compare the fifth-game score with the target mean of \(18\). - What happens to a mean when a new value equals the mean? What happens when it is smaller?

Solution

1. For a), multiply the mean by the number of games: \(4 \times 18 = 72\) points. 2. For b), the new total is \(72 + 28 = 100\) points. The new mean is \(100 \div 5 = 20\) points per game. 3. For c), scoring \(18\) points in the fifth game would have kept the mean at \(18\). Instead, she scored \(10\) extra points because \(28 - 18 = 10\). 4. To balance those extra points, she must score \(10\) fewer than \(18\) in the sixth game: \(18 - 10 = 8\) points. Check: \(100 + 8 = 108\), and \(108 \div 6 = 18\).

Answer

a) \(72\) points b) \(20\) points per game c) \(8\) points. Her fifth-game score was \(10\) points above the target mean, so her sixth-game score must be \(10\) points below it.

5126986

A class was surveyed about how many books each student read during summer vacation. <table> <tr><td>Books read</td><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(12\)</td></tr> <tr><td>Number of students</td><td>\(5\)</td><td>\(8\)</td><td>\(6\)</td><td>\(4\)</td><td>\(1\)</td><td>\(1\)</td></tr> </table> a) Find the mean and median number of books read. b) One student read \(12\) books, much more than the others. Which measure from part a) better represents a typical student in this class? Briefly explain.

Hints

- Use the frequencies to find the total number of books. - Consider how one unusually high value affects the mean. - Locate the middle value using cumulative frequencies. - Think about whether increasing only the largest value would change the median.

Solution

1. There are \(5 + 8 + 6 + 4 + 1 + 1 = 25\) students. 2. The total number of books is \(0 \times 5 + 1 \times 8 + 2 \times 6 + 3 \times 4 + 4 \times 1 + 12 \times 1 = 48\). The mean is \(48 \div 25 = 1.92\) books. 3. With \(25\) values, the median is the value in position \(13\). Positions \(1\) through \(5\) are \(0\), and positions \(6\) through \(13\) are \(1\), so the median is \(1\) book. 4. The median better represents a typical student because the unusually high value of \(12\) pulls the mean upward.

Answer

a) Mean: \(1.92\) books; median: \(1\) book b) The median better represents a typical student because the value \(12\) has a strong effect on the mean.

5227156

Consider the numbers \(8.5\), \(7.2\), \(9.0\), \(8.3\), \(7.5\), and \(8.1\). 1. Find their mean. 2. For each number, find its signed deviation using \(\text{value}-\text{mean}\). 3. Find the sum of all the deviations. What do you notice?

Hints

- Add the values and divide by \(6\). - Subtract the mean from each value, keeping the signs. - Combine positive and negative deviations carefully.

Solution

1. The sum is \(8.5+7.2+9.0+8.3+7.5+8.1=48.6\), so the mean is \(48.6\div6=8.1\). 2. The deviations are \(8.5-8.1=0.4\), \(7.2-8.1=-0.9\), \(9.0-8.1=0.9\), \(8.3-8.1=0.2\), \(7.5-8.1=-0.6\), and \(8.1-8.1=0.0\). 3. Their sum is \(0.4+(-0.9)+0.9+0.2+(-0.6)+0.0=0\). The positive and negative deviations balance.

Answer

1. \(8.1\) 2. \(0.4\), \(-0.9\), \(0.9\), \(0.2\), \(-0.6\), and \(0.0\) 3. The sum is \(0\); deviations from the mean balance.

5227166

Find the mean and each signed deviation from the mean, using \(\text{value}-\text{mean}\), for the data set \(\frac{1}{2}\), \(1.2\), \(\frac{4}{5}\), \(1\frac{1}{2}\), and \(0.7\).

Hints

- Convert the fractions and mixed number to decimals. - Add all five values and divide by \(5\). - Subtract the mean from each original value, keeping the sign.

Solution

1. Write the values as decimals: \(0.5\), \(1.2\), \(0.8\), \(1.5\), and \(0.7\). 2. Their sum is \(0.5+1.2+0.8+1.5+0.7=4.7\), so the mean is \(4.7\div5=0.94\). 3. The signed deviations are \(0.5-0.94=-0.44\), \(1.2-0.94=0.26\), \(0.8-0.94=-0.14\), \(1.5-0.94=0.56\), and \(0.7-0.94=-0.24\).

Answer

Mean: \(0.94\) Signed deviations: \(-0.44\), \(0.26\), \(-0.14\), \(0.56\), and \(-0.24\)

5317056

A weather station recorded the daily high temperature from Monday through Friday. The bar graph shows the results. a) What was the mean daily high temperature for the school week? b) On how many of the five days was the temperature above the mean?

Figure for problem 531705

Hints

- Read each temperature from the graph. - Add the five values and divide by \(5\). - Compare each daily temperature with the mean.

Solution

1. The graph shows Monday, \(58^\circ\text{F}\); Tuesday, \(64^\circ\text{F}\); Wednesday, \(56^\circ\text{F}\); Thursday, \(60^\circ\text{F}\); and Friday, \(62^\circ\text{F}\). 2. The mean is \(\frac{58+64+56+60+62}{5}=\frac{300}{5}=60^\circ\text{F}\). 3. Tuesday and Friday are above \(60^\circ\text{F}\), so \(2\) days were above the mean.

Answer

a) \(60^\circ\text{F}\) b) \(2\) days

5317366

Twenty students were asked how much weekly allowance they receive. The bar graph shows the results. a) Find the mean weekly allowance. b) Find the median weekly allowance.

Figure for problem 531736

Hints

- How can you use the bar heights to confirm the number of students? - How can you find the total amount represented by each bar? - Which positions are in the middle of an ordered data set with \(20\) values? - What do you do with the two middle values?

Solution

1. The bar heights represent \(3 + 5 + 6 + 4 + 2 = 20\) students. 2. Find the total allowance represented by the graph: \((3 \times \$10) + (5 \times \$15) + (6 \times \$20) + (4 \times \$25) + (2 \times \$30) = \$385\). 3. The mean is \(\$385 \div 20 = \$19.25\). 4. With \(20\) values, the median is the mean of the \(10\)th and \(11\)th values in order. Both values are \(\$20\), so the median is \(\$20\).

Answer

a) The mean weekly allowance is \(\$19.25\). b) The median weekly allowance is \(\$20\).

5317406

Jordan recorded the number of steps taken each school day from Monday through Friday. The bar graph shows the results. a) Read the number of steps for each day. b) Find the total number of steps for the school week. c) Find the mean number of steps per day.

Figure for problem 531740

Hints

- Use the y-axis scale to read each bar. - Add all five daily values. - Divide the total by \(5\) to find the mean.

Solution

1. The graph shows Monday, \(8000\); Tuesday, \(11{,}000\); Wednesday, \(7000\); Thursday, \(9000\); and Friday, \(10{,}000\) steps. 2. The total is \(8000+11{,}000+7000+9000+10{,}000=45{,}000\) steps. 3. The mean is \(45{,}000\div5=9000\) steps per day.

Answer

a) Monday: \(8000\); Tuesday: \(11{,}000\); Wednesday: \(7000\); Thursday: \(9000\); Friday: \(10{,}000\) steps b) \(45{,}000\) steps c) \(9000\) steps per day

5317756

A sixth-grade class surveyed students about the number of books they read during summer break. The bar graph shows the results. a) How many students participated in the survey? b) Find the mean number of books read per student. c) Another class also has \(20\) students and read a mean of exactly \(3.5\) books per student. How many books did that class read in all?

Figure for problem 531775

Hints

- Add all bar heights to find the number of students. - Multiply each number of books by its frequency. - Divide the total number of books by the number of students. - For part c), multiply the mean by the class size.

Solution

1. The total number of students is \(2+5+6+5+2=20\). 2. The total number of books is \(1\times2+2\times5+3\times6+4\times5+5\times2=60\). 3. The mean is \(60\div20=3\) books per student. 4. The other class read \(3.5\times20=70\) books in all.

Answer

a) \(20\) students b) \(3\) books per student c) \(70\) books

5317836

The bar graph shows the daily number of visitors to an outdoor adventure park during one summer week. a) Find the mean number of visitors per day. b) Give two possible reasons why attendance varies so much during the week.

Figure for problem 531783

Hints

- Read each daily value from the graph. - Add all seven values and divide by \(7\). - Think about how weekends, weather, and special events can affect attendance.

Solution

1. The graph shows Monday, \(40\); Tuesday, \(60\); Wednesday, \(110\); Thursday, \(50\); Friday, \(140\); Saturday, \(480\); and Sunday, \(520\) visitors. 2. The weekly total is \(40+60+110+50+140+480+520=1400\). 3. The mean is \(1400\div7=200\) visitors per day. 4. Possible reasons include greater free time on weekends, differences in weather, special events, or organized group trips on particular days.

Answer

a) \(200\) visitors per day b) Possible reasons include higher weekend attendance and changes in weather or scheduled group events.

5317856

A bar graph shows the daily difference, in centimeters, between a lake’s water level and its normal level for seven days. Find the mean difference for the week.

Figure for problem 531785

Hints

- How do you find the mean of a set of values? - Add all values shown in the graph, paying close attention to their signs. - By what number should you divide the total?

Solution

1. Add the seven differences shown in the graph: \(12 + 8 + (-4) + (-10) + (-15) + (-3) + 5 = -7\,\text{cm}\). 2. Divide by the number of days: \(-7 \div 7 = -1\,\text{cm}\). 3. The negative mean means that the water level averaged \(1\,\text{cm}\) below its normal level.

Answer

The mean difference is \(-1\,\text{cm}\), so the water level averaged \(1\,\text{cm}\) below normal.

5318046

The graph shows the number of hours of sunshine each day in one city during a week. a) Find the mean number of sunshine hours per day. b) On how many days were the sunshine hours greater than the mean?

Figure for problem 531804

Hints

- Read each daily value from the graph. - Add all seven values and divide by \(7\). - Compare each day with the mean.

Solution

1. The graph shows Monday, \(6\) hours; Tuesday, \(4\); Wednesday, \(8\); Thursday, \(5\); Friday, \(7\); Saturday, \(9\); and Sunday, \(3\). 2. The total is \(6+4+8+5+7+9+3=42\) hours. 3. The mean is \(42\div7=6\) hours per day. 4. Wednesday, Friday, and Saturday are above the mean, so there are \(3\) such days.

Answer

a) \(6\) hours per day b) \(3\) days

5318176

A class measured the number of hours of sunshine on five consecutive days. The results are shown in the bar graph. a) Find the mean number of sunshine hours per day from Monday through Friday. b) On Saturday, the sunshine duration increased the mean for all six days to exactly \(6.5\) hours. How many hours of sunshine were recorded on Saturday?

Figure for problem 531817

Hints

- Read each weekday value carefully from the graph. - How do you find the mean of several values? - What total is needed for six days to have a mean of \(6.5\) hours? - Compare that total with the total for the first five days.

Solution

1. Read the five values from the graph: Monday \(6\) hours, Tuesday \(8\) hours, Wednesday \(4\) hours, Thursday \(9\) hours, and Friday \(3\) hours. 2. Their total is \(6 + 8 + 4 + 9 + 3 = 30\) hours, so the mean is \(30 \div 5 = 6\) hours. 3. A mean of \(6.5\) hours over \(6\) days requires a total of \(6 \times 6.5 = 39\) hours. 4. Saturday must account for \(39 - 30 = 9\) hours.

Answer

a) The mean is \(6\) hours per day. b) Saturday had \(9\) hours of sunshine.

5318256

Nine students recorded how many books they read last month. The horizontal bar graph shows the results. a) Find the median number of books read. b) Find the mean number of books read.

Figure for problem 531825

Hints

- Read one value for each student from the graph. - Order the values before finding the median. - Which position is in the middle of nine values? - Add all nine values and divide by \(9\) to find the mean.

Solution

1. Read the values from the graph: \(3, 6, 2, 5, 1, 7, 4, 6,\) and \(2\). 2. In order, the values are \(1, 2, 2, 3, 4, 5, 6, 6, 7\). With nine values, the fifth value is the median, so the median is \(4\) books. 3. The total is \(3 + 6 + 2 + 5 + 1 + 7 + 4 + 6 + 2 = 36\). The mean is \(36 \div 9 = 4\) books.

Answer

a) \(4\) books b) \(4\) books

5318646

The bar graph shows the weekly allowances of six friends: Anna, Ben, Clara, David, Emily, and Felix. a) Find the mean weekly allowance. b) Felix says, “If Clara received \(\$3\) less and David received \(\$3\) more, the mean would change.” Is Felix correct? Explain.

Figure for problem 531864

Hints

- Read each value from the bar graph. - Add the six values and divide by \(6\). - What happens to the total when one value decreases by \(\$3\) and another increases by \(\$3\)?

Solution

1. Read the amounts from the graph: Anna \(\$8\), Ben \(\$12\), Clara \(\$15\), David \(\$6\), Emily \(\$10\), and Felix \(\$9\). 2. The total is \(8 + 12 + 15 + 6 + 10 + 9 = 60\), so the mean is \(\$60 \div 6 = \$10\). 3. If Clara receives \(\$3\) less and David receives \(\$3\) more, the changes add to \(-\$3 + \$3 = \$0\). The total and the number of people stay the same, so the mean remains \(\$10\). Felix is not correct.

Answer

a) The mean weekly allowance is \(\$10\). b) No. The two changes cancel, so the total remains \(\$60\) and the mean remains \(\$10\).

5318726

Nine students recorded how many hours they exercised last week. The bar graph shows the results. a) Find the median number of exercise hours. b) Find the mean number of exercise hours per student.

Figure for problem 531872

Hints

- Read all nine values from the graph and order them. - Which value is in the middle of nine ordered values? - Add all values and divide by \(9\) to find the mean.

Solution

1. Read the values from the graph: \(4, 6, 2, 5, 3, 10, 3, 7,\) and \(5\). 2. In order, the values are \(2, 3, 3, 4, 5, 5, 6, 7, 10\). The middle value is \(5\), so the median is \(5\) hours. 3. The total is \(45\) hours, so the mean is \(45 \div 9 = 5\) hours.

Answer

a) \(5\) hours b) \(5\) hours

5318786

A class survey asked students how many hours per day they use a smartphone. The bar graph shows the results. a) How many students participated? b) Find the median daily usage time.

Figure for problem 531878

Hints

- Add the frequencies for all six bars. - For an even number of values, which two positions determine the median? - Use cumulative counts to locate the \(12\)th and \(13\)th values.

Solution

1. Add the frequencies: \(3 + 5 + 4 + 6 + 4 + 2 = 24\) students. 2. With \(24\) values, the median is the mean of the \(12\)th and \(13\)th values in order. 3. The cumulative frequencies show that the \(12\)th value is \(3\) hours and the \(13\)th value is \(4\) hours. The median is \(\frac{3 + 4}{2} = 3.5\) hours.

Answer

a) \(24\) students b) \(3.5\) hours

5350406

Jordan scored points for a basketball team in five consecutive games. The bar graph shows the results. a) How many points did Jordan score in Game 4? b) Find the difference between the highest and lowest scores. c) Find the mean number of points scored per game.

Figure for problem 535040

Hints

- Read the Game 4 bar using the y-axis. - Subtract the smallest score from the greatest score. - Add all five scores and divide by \(5\).

Solution

1. The graph shows scores of \(12\), \(18\), \(15\), \(20\), and \(10\) points. Jordan scored \(20\) points in Game 4. 2. The highest score is \(20\), and the lowest is \(10\), so the difference is \(20-10=10\) points. 3. The mean is \((12+18+15+20+10)\div5=75\div5=15\) points per game.

Answer

a) \(20\) points b) \(10\) points c) \(15\) points per game

5350576

A short math quiz was worth \(5\) points. The bar graph shows how many students earned each score. a) How many students took the quiz? b) Find the relative frequency of students who earned at least \(4\) points. Give the result as a percent. c) Find the mean quiz score. d) A student says, “Our mean score was greater than \(3\) points.” Determine whether the statement is correct.

Figure for problem 535057

Hints

- Add all bar heights to find the number of students. - “At least \(4\)” includes scores of \(4\) and \(5\). - Multiply each score by its frequency before finding the mean. - Compare the calculated mean with \(3\).

Solution

1. The number of students is \(1+3+5+7+6+3=25\). 2. At least \(4\) points includes the scores \(4\) and \(5\). There are \(6+3=9\) such students, so the relative frequency is \(\frac{9}{25}=36\%\). 3. The total score is \(0\times1+1\times3+2\times5+3\times7+4\times6+5\times3=73\). 4. The mean is \(73\div25=2.92\) points. 5. Since \(2.92<3\), the student’s statement is incorrect.

Answer

a) \(25\) students b) \(36\%\) c) \(2.92\) points d) The statement is incorrect because \(2.92<3\).

5350586

The bar graph shows monthly rainfall in a city during the first six months of a year. 1. Which month had the most rainfall? 2. What was the total rainfall during the six months? 3. Find the mean monthly rainfall for this period.

Figure for problem 535058

Hints

- Compare the heights of all six bars. - Add the six monthly amounts for the total. - Divide the total by \(6\) for the mean.

Solution

1. May has the highest bar, with \(3.2\) inches of rainfall. 2. The total is \(2.4+1.8+2.2+2.8+3.2+2.0=14.4\) inches. 3. The mean is \(14.4\div6=2.4\) inches per month.

Answer

1. May, with \(3.2\) inches 2. \(14.4\) inches 3. \(2.4\) inches per month

5350616

The bar graph shows the number of errors students made on a math test. a) How many students took the test? b) Find the mean number of errors. Round to the nearest tenth. c) How many students made fewer than \(3\) errors?

Figure for problem 535061

Hints

- Add the bar heights to find the number of students. - Multiply each error count by its frequency. - Divide the total errors by the number of students. - “Fewer than \(3\)” includes \(1\) and \(2\).

Solution

1. The graph shows \(4\) students with \(1\) error, \(9\) with \(2\), \(7\) with \(3\), \(3\) with \(4\), and \(2\) with \(5\). 2. The number of students is \(4+9+7+3+2=25\). 3. The total number of errors is \(4\times1+9\times2+7\times3+3\times4+2\times5=65\). 4. The mean is \(65\div25=2.6\) errors. 5. Fewer than \(3\) errors means \(1\) or \(2\) errors, so \(4+9=13\) students.

Answer

a) \(25\) students b) \(2.6\) errors c) \(13\) students

5350706

The bar graph shows the number of visitors to a public swimming pool during one week in July. a) Find the mean number of visitors per day. b) Find the median number of visitors per day. c) Which measure is greater? Explain how the weekend attendance affects the comparison.

Figure for problem 535070

Hints

- Add all seven values and divide by \(7\) to find the mean. - Put the seven attendance values in order to find the middle value. - Consider how unusually large values affect the mean.

Solution

1. The graph shows Monday, \(1500\); Tuesday, \(2000\); Wednesday, \(500\); Thursday, \(1500\); Friday, \(2500\); Saturday, \(4500\); and Sunday, \(5000\) visitors. 2. The weekly total is \(1500+2000+500+1500+2500+4500+5000=17{,}500\). 3. The mean is \(17{,}500\div7=2500\) visitors per day. 4. In order, the values are \(500, 1500, 1500, 2000, 2500, 4500, 5000\). The middle value is \(2000\), so the median is \(2000\) visitors. 5. The mean is greater than the median because the two large weekend values raise the mean.

Answer

a) Mean: \(2500\) visitors per day b) Median: \(2000\) visitors per day c) The mean is greater because the high Saturday and Sunday attendance pulls it upward.

5350786

A weather station in a northern Alaska town recorded the noon temperature during a cold March week. The bar graph shows the results. 1. Find the mean noon temperature for the week. 2. Find the median noon temperature for the week. 3. Which measure is greater, and by how much?

Figure for problem 535078

Hints

- Add all seven temperatures and divide by \(7\) to find the mean. - Put the seven temperatures in order to identify the middle value. - Subtract the mean from the median to compare them.

Solution

1. The graph shows Monday, \(16^\circ\text{F}\); Tuesday, \(14^\circ\text{F}\); Wednesday, \(8^\circ\text{F}\); Thursday, \(7^\circ\text{F}\); Friday, \(12^\circ\text{F}\); Saturday, \(14^\circ\text{F}\); and Sunday, \(13^\circ\text{F}\). 2. The sum is \(16+14+8+7+12+14+13=84\), so the mean is \(84\div7=12^\circ\text{F}\). 3. In order, the temperatures are \(7, 8, 12, 13, 14, 14, 16\). The middle value is \(13^\circ\text{F}\), so the median is \(13^\circ\text{F}\). 4. The median is greater than the mean by \(13^\circ\text{F}-12^\circ\text{F}=1^\circ\text{F}\).

Answer

1. Mean: \(12^\circ\text{F}\) 2. Median: \(13^\circ\text{F}\) 3. The median is greater by \(1^\circ\text{F}\).

5351086

The bar graph shows the frequency of scores on a math test. a) How many students took the test? b) Find the median score. c) Find the mean score.

Figure for problem 535108

Hints

- A frequency tells how many times each score occurs. - Use cumulative frequencies to locate the middle score. - Multiply each score by its frequency before finding the mean.

Solution

1. Add the frequencies: \(1 + 2 + 4 + 4 + 3 + 1 = 15\) students. 2. With \(15\) scores, the median is the \(8\)th score in order. The cumulative frequency through \(6\) points is \(7\), and through \(7\) points it is \(11\), so the \(8\)th score is \(7\). 3. The total number of points is \((4 \times 1) + (5 \times 2) + (6 \times 4) + (7 \times 4) + (8 \times 3) + (9 \times 1) = 99\). The mean is \(99 \div 15 = 6.6\) points.

Answer

a) \(15\) students b) \(7\) points c) \(6.6\) points

5351166

A survey recorded the number of bicycles owned by each of \(40\) households. The bar graph shows the results. Find the mean number of bicycles per household.

Figure for problem 535116

Hints

- Record the frequency for each possible number of bicycles. - How can you use each value and its frequency to find the total number of bicycles? - Divide the total number of bicycles by the number of households.

Solution

1. The graph shows \(4\) households with \(0\) bicycles, \(8\) with \(1\), \(12\) with \(2\), \(10\) with \(3\), \(4\) with \(4\), and \(2\) with \(5\). 2. The total number of bicycles is \((4 \times 0) + (8 \times 1) + (12 \times 2) + (10 \times 3) + (4 \times 4) + (2 \times 5) = 88\). 3. The mean is \(88 \div 40 = 2.2\) bicycles per household.

Answer

The mean is \(2.2\) bicycles per household.

5351196

A class survey asked students how many siblings they have. The bar graph shows the results. a) How many students participated in the survey? b) How many students have more than one sibling? c) Find the mean number of siblings per student.

Figure for problem 535119

Hints

- Add all bar heights to find the total number of students. - Which bars represent more than one sibling? - To find the mean, first determine the total number of siblings represented. - Divide that total by the number of students.

Solution

1. Add the bar heights to find the number of students: \(2 + 6 + 9 + 6 + 2 = 25\). 2. More than one sibling means \(2\), \(3\), or \(4\) siblings. The number of students is \(9 + 6 + 2 = 17\). 3. The total number of siblings represented is \((2 \times 0) + (6 \times 1) + (9 \times 2) + (6 \times 3) + (2 \times 4) = 50\). 4. The mean is \(50 \div 25 = 2\) siblings per student.

Answer

a) \(25\) students b) \(17\) students c) \(2\) siblings per student

5351296

Forestry students counted five tree species in a marked area of a forest. The bar graph shows the results. 1) Which species was most common and which was least common? Give each count. 2) How many more oak trees than beech trees were counted? 3) How many times as many birch trees as pine trees were counted? 4) Find the mean number of trees across the five species.

Figure for problem 535129

Hints

- Read each bar from the y-axis. - “How many more” calls for subtraction. - “How many times as many” calls for division. - Add all five counts and divide by \(5\) for the mean.

Solution

1. The graph shows oak, \(24\); beech, \(10\); pine, \(8\); spruce, \(18\); and birch, \(40\). 2. Birch is most common with \(40\) trees, and pine is least common with \(8\). 3. The difference between oak and beech is \(24-10=14\). 4. The ratio of birch to pine is \(40\div8=5\), so there are five times as many birch trees. 5. The mean is \(\frac{24+10+8+18+40}{5}=\frac{100}{5}=20\) trees per species.

Answer

1) Most common: birch, \(40\); least common: pine, \(8\) 2) \(14\) more oak trees 3) \(5\) times as many 4) \(20\) trees per species

5351336

An amusement park recorded the number of admission tickets sold each day during one week. The bar graph shows the data. a) On which day were the most tickets sold? b) How many tickets were sold during the weekend, Saturday and Sunday, in all? c) What was the mean number of tickets sold per day that week?

Figure for problem 535133

Hints

- Compare all bar heights to find the greatest value. - Add the Saturday and Sunday values. - Add all seven values and divide by \(7\).

Solution

1. The tallest bar is Sunday, with \(1320\) tickets. 2. Weekend sales were \(1150+1320=2470\) tickets. 3. Weekly sales were \(420+380+450+310+590+1150+1320=4620\) tickets. 4. The mean was \(4620\div7=660\) tickets per day.

Answer

a) Sunday b) \(2470\) tickets c) \(660\) tickets per day

5351346

The bar graph shows enrollment in five school clubs. 1) Which club has the most members? 2) Which clubs have exactly the same number of members? 3) Find the mean enrollment across the five clubs. 4) The Chess Club enrollment is what percent of the Cooking Club enrollment?

Figure for problem 535134

Hints

- Read each bar using the y-axis scale. - Compare the bar heights for parts 1 and 2. - Add all five enrollments and divide by \(5\). - For part 4, compare Chess with Cooking as a fraction.

Solution

1. The graph shows Dance, \(10\); Chess, \(8\); Cooking, \(20\); Robotics, \(14\); and Choir, \(8\) members. Cooking has the most members. 2. Chess and Choir each have \(8\) members. 3. The mean is \((10+8+20+14+8)\div5=60\div5=12\) members. 4. The Chess Club enrollment is \(\frac{8}{20}=40\%\) of the Cooking Club enrollment.

Answer

1) Cooking Club, with \(20\) members 2) Chess Club and Choir, with \(8\) members each 3) \(12\) members 4) \(40\%\)

5351386

Two small groups, Group A and Group B, took a math quiz. The scores of the five students in each group are shown in graphs a) and b). Which group has the higher mean score? Support your answer with calculations.

Figure for problem 535138

Hints

- Read all five scores from each graph. - Add the scores in each group and divide each sum by \(5\). - Compare the two means.

Solution

1. Group A has scores \(2, 4, 5, 5,\) and \(9\). Their sum is \(25\), so the mean is \(25 \div 5 = 5\). 2. Group B has scores \(3, 4, 6, 7,\) and \(10\). Their sum is \(30\), so the mean is \(30 \div 5 = 6\). 3. Since \(6 > 5\), Group B has the higher mean score.

Answer

Group B has the higher mean score. Group A’s mean is \(5\) points, and Group B’s mean is \(6\) points.

5387146

A rowing team recorded stroke rates during several training intervals. <table> <tr><th>Strokes per minute</th><th>\(18\)</th><th>\(20\)</th><th>\(22\)</th><th>\(24\)</th><th>\(26\)</th></tr> <tr><td>Number of measurements</td><td>\(3\)</td><td>\(5\)</td><td>\(6\)</td><td>\(4\)</td><td>\(2\)</td></tr> </table> Find the mean stroke rate.

Hints

- Some values occur more than once. - Multiply each stroke rate by its frequency. - Confirm the total number of measurements before dividing.

Solution

1. The number of measurements is \(3 + 5 + 6 + 4 + 2 = 20\). 2. The total of all recorded stroke rates is \((18 \times 3) + (20 \times 5) + (22 \times 6) + (24 \times 4) + (26 \times 2) = 434\). 3. The mean is \(434 \div 20 = 21.7\) strokes per minute.

Answer

The mean stroke rate is \(21.7\) strokes per minute.

5387156

Six museum audio recordings should have a mean length of \(7.8\) minutes. Five of the lengths are \(7.4\), \(8.1\), \(6.9\), \(7.8\), and \(8.3\) minutes. How long must the sixth recording be?

Hints

- What total length corresponds to the given mean for six recordings? - Add the five known lengths. - Subtract the known total from the required total.

Solution

1. The required total length is \(7.8 \times 6 = 46.8\) minutes. 2. The five known recordings total \(7.4 + 8.1 + 6.9 + 7.8 + 8.3 = 38.5\) minutes. 3. The missing length is \(46.8 - 38.5 = 8.3\) minutes.

Answer

The sixth recording must be \(8.3\) minutes long.

5387166

A drone club averaged \(30.5\) points over four obstacle-course runs. The club scored \(38\) points on the fifth run. Find the new mean score.

Hints

- The original mean represents four scores. - Find the original total first. - After adding the fifth score, divide by the new number of scores.

Solution

1. The total for the first four runs is \(30.5 \times 4 = 122\) points. 2. The new total is \(122 + 38 = 160\) points. 3. The new mean is \(160 \div 5 = 32\) points.

Answer

The new mean score is \(32\) points.

5387176

The mean firing time for six ceramic tiles was calculated as \(46\) minutes. Later, one time was found to have been entered as \(52\) minutes instead of \(46\) minutes. Find the corrected mean without adding all six original values again.

Hints

- Use the original mean to find the original total. - Only one value in the total needs to be replaced. - The number of measurements does not change.

Solution

1. The incorrect total was \(46 \times 6 = 276\) minutes. 2. Correct the total by replacing \(52\) with \(46\): \(276 - 52 + 46 = 270\) minutes. 3. The corrected mean is \(270 \div 6 = 45\) minutes.

Answer

The corrected mean is \(45\) minutes.

5387186

At a community cleanup, a group of \(12\) people collected a mean of \(6.5\,\text{kg}\) per person. A second group of \(18\) people collected a mean of \(8\,\text{kg}\) per person. What was the mean amount collected per person when both groups are combined?

Hints

- The two groups are different sizes. - Find the total amount collected by each group. - Divide the combined total by the combined number of people.

Solution

1. The first group collected \(12 \times 6.5 = 78\,\text{kg}\). 2. The second group collected \(18 \times 8 = 144\,\text{kg}\). 3. Together, the groups collected \(78 + 144 = 222\,\text{kg}\), and there were \(12 + 18 = 30\) people. 4. The combined mean is \(222 \div 30 = 7.4\,\text{kg}\) per person.

Answer

The combined mean is \(7.4\,\text{kg}\) per person.

5387206

The graph shows eight water-level measurements at a small dam. Find the mean water level.

Figure for problem 538720

Hints

- Use the grid lines to read each marked point. - The order of the measurements does not affect the mean. - Keep track of the decimal places.

Solution

1. Read the eight values from the graph: \(3.2\,\text{m}, 3.6\,\text{m}, 3.4\,\text{m}, 3.9\,\text{m}, 3.7\,\text{m}, 3.5\,\text{m}, 3.8\,\text{m},\) and \(3.3\,\text{m}\). 2. Their sum is \(28.4\,\text{m}\). 3. The mean is \(28.4 \div 8 = 3.55\,\text{m}\).

Answer

The mean water level is \(3.55\,\text{m}\).

5387216

The bar graph shows how 20 participants rated a new escape room from 1 to 4 points. Find the mean rating.

Figure for problem 538721

Hints

- Read the frequency for each rating. - Multiply each rating by its frequency. - Divide the total points by the total number of ratings.

Solution

1. The frequencies are \(3\) ratings of \(1\), \(7\) ratings of \(2\), \(6\) ratings of \(3\), and \(4\) ratings of \(4\). 2. The total number of points is \(3 \times 1 + 7 \times 2 + 6 \times 3 + 4 \times 4 = 51\). 3. The mean is \(51 \div 20 = 2.55\).

Answer

The mean rating is \(2.55\) points.

5387226

At a charity run, organizers recorded how many laps each student completed. <table> <tr><th>Laps</th><th>\(2\)</th><th>\(3\)</th><th>\(4\)</th><th>\(5\)</th><th>\(6\)</th></tr> <tr><td>Number of students</td><td>\(4\)</td><td>\(7\)</td><td>\(5\)</td><td>\(3\)</td><td>\(1\)</td></tr> </table> Find the mean number of laps.

Hints

- The first row gives the values, and the second row gives their frequencies. - Find both the total number of students and the total number of laps. - A mean does not have to be a whole number.

Solution

1. The number of students is \(4 + 7 + 5 + 3 + 1 = 20\). 2. The total number of laps is \((2 \times 4) + (3 \times 7) + (4 \times 5) + (5 \times 3) + (6 \times 1) = 70\). 3. The mean is \(70 \div 20 = 3.5\) laps.

Answer

The mean is \(3.5\) laps.

5387246

A library recorded how many times students visited in one week. <table> <tr><th>Number of visits</th><th>\(1\)</th><th>\(2\)</th><th>\(3\)</th><th>\(4\)</th></tr> <tr><td>Frequency</td><td>\(8\)</td><td>\(6\)</td><td>\(4\)</td><td>\(2\)</td></tr> </table> Mara calculates \((1 + 2 + 3 + 4) \div 4 = 2.5\). Explain her error and find the correct mean.

Hints

- What does the second row of the table represent? - Does Mara’s calculation include every student? - Imagine writing each number of visits as many times as its frequency.

Solution

1. Mara counted each number of visits only once and ignored the frequencies. 2. The table represents \(8 + 6 + 4 + 2 = 20\) students. 3. The total number of visits is \((1 \times 8) + (2 \times 6) + (3 \times 4) + (4 \times 2) = 40\). 4. The correct mean is \(40 \div 20 = 2\) visits.

Answer

Mara ignored the frequencies. The correct mean is \(2\) visits.

5387286

After eight rounds, an archer has a mean of \(15\) hits per round. After a ninth round, the mean increases to \(16\) hits per round. How many hits did she make in the ninth round?

Hints

- Compare the totals before and after the ninth round. - The new mean represents one more round. - The difference between the two totals is the ninth-round result.

Solution

1. Her total after eight rounds was \(15 \times 8 = 120\) hits. 2. Her total after nine rounds was \(16 \times 9 = 144\) hits. 3. The ninth-round score was \(144 - 120 = 24\) hits.

Answer

She made \(24\) hits in the ninth round.

5387316

In a robotics challenge, organizers recorded how many failed attempts each robot made before reaching the goal. <table> <tr><th>Failed attempts</th><th>\(0\)</th><th>\(1\)</th><th>\(2\)</th><th>\(3\)</th><th>\(4\)</th></tr> <tr><td>Number of robots</td><td>\(3\)</td><td>\(6\)</td><td>\(8\)</td><td>\(5\)</td><td>\(2\)</td></tr> </table> Find the median number of failed attempts.

Hints

- Add the frequencies first. - Identify the two middle positions. - Use cumulative frequencies; you do not need to write every value.

Solution

1. The table represents \(3 + 6 + 8 + 5 + 2 = 24\) robots. 2. The median is the mean of the \(12\)th and \(13\)th values in order. 3. The cumulative frequency through \(1\) failed attempt is \(9\), and through \(2\) failed attempts it is \(17\). Therefore, both middle values are \(2\).

Answer

The median is \(2\) failed attempts.

5387326

A theater group summarized the lengths of its rehearsals. <table> <tr><th>Rehearsal length in hours</th><th>\(1.5\)</th><th>\(2\)</th><th>\(2.5\)</th><th>\(3\)</th><th>\(3.5\)</th></tr> <tr><td>Frequency</td><td>\(2\)</td><td>\(5\)</td><td>\(7\)</td><td>\(4\)</td><td>\(3\)</td></tr> </table> Find the median rehearsal length.

Hints

- Which position is in the middle of \(21\) values? - Add the frequencies from left to right. - Keep the unit in your answer.

Solution

1. The table represents \(2 + 5 + 7 + 4 + 3 = 21\) rehearsals. 2. The median is the \(11\)th value in order. 3. The cumulative frequency through \(2\) hours is \(7\), and through \(2.5\) hours it is \(14\). Therefore, the \(11\)th value is \(2.5\) hours.

Answer

The median rehearsal length is \(2.5\) hours.

5387406

The pie chart summarizes \(24\) trips to music lessons. The numbers inside the slices show how many trips are in each category. Find the median number of transfers.

Figure for problem 538740

Hints

- Convert the pie slices into frequencies. - Order the categories by number of transfers. - Use cumulative frequencies to locate the two middle trips.

Solution

1. There are \(24\) values, so the median is determined by positions \(12\) and \(13\). 2. The first \(6\) values are \(0\) transfers, and the next \(10\) values are \(1\) transfer. 3. Both middle positions are in the \(1\)-transfer category, so the median is \(1\) transfer.

Answer

The median is \(1\) transfer.

5387416

The bar graph shows the star ratings given by 20 visitors to an exhibit. Find the median rating.

Figure for problem 538741

Hints

- Read the frequency for each star rating. - Place the ratings in order using their frequencies. - For \(20\) values, identify positions \(10\) and \(11\).

Solution

1. There are \(4\) one-star ratings, \(7\) two-star ratings, \(5\) three-star ratings, and \(4\) four-star ratings. 2. With \(20\) data values, the median is the mean of the values in positions \(10\) and \(11\) when the ratings are ordered. 3. Positions \(1\) through \(4\) are one-star ratings, and positions \(5\) through \(11\) are two-star ratings. Thus, both middle values are \(2\). 4. The median is \((2 + 2) \div 2 = 2\).

Answer

The median rating is \(2\) stars.

5387426

The graph shows nine water-depth measurements at a stream crossing. Find the median depth.

Figure for problem 538742

Hints

- Read all nine marked points. - The measurement order does not affect the median. - Order the depths and find the middle value.

Solution

1. Read the nine depths from the graph: \(42\,\text{cm}, 38\,\text{cm}, 45\,\text{cm}, 41\,\text{cm}, 39\,\text{cm}, 44\,\text{cm}, 40\,\text{cm}, 43\,\text{cm},\) and \(37\,\text{cm}\). 2. Order the values: \(37, 38, 39, 40, 41, 42, 43, 44, 45\) centimeters. 3. The fifth value is \(41\,\text{cm}\), so the median is \(41\,\text{cm}\).

Answer

The median depth is \(41\,\text{cm}\).

5387436

A workshop recorded how many specialty tools were checked out on \(40\) days. <table> <tr><th>Tools checked out</th><th>\(2\)</th><th>\(3\)</th><th>\(4\)</th><th>\(5\)</th><th>\(6\)</th></tr> <tr><td>Number of days</td><td>\(5\)</td><td>\(9\)</td><td>\(12\)</td><td>\(8\)</td><td>\(6\)</td></tr> </table> Find the median.

Hints

- Use cumulative frequencies. - With an even number of observations, identify two middle positions. - Determine which value contains both positions.

Solution

1. With \(40\) observations, the median is determined by positions \(20\) and \(21\). 2. The cumulative frequency through \(3\) tools is \(14\). 3. The next \(12\) observations have a value of \(4\), so both middle positions are \(4\).

Answer

The median is \(4\) tools checked out.

5387486

Give an ordered data set of exactly seven different integers with a median of \(20\) and a least value of \(4\).

Hints

- Place the median in the middle position first. - Put the same number of values on each side. - Check that all seven integers are different.

Solution

1. With seven ordered values, the fourth value must be \(20\). 2. Choose three different integers less than \(20\) and three different integers greater than \(20\), while keeping \(4\) as the least value. 3. One possible data set is \(4, 11, 17, 20, 23, 28, 35\).

Answer

One possible data set is \(4, 11, 17, 20, 23, 28, 35\).

5387506

The bar graph shows how many checkpoints teams completed in a city scavenger hunt. Find the median number of completed checkpoints.

Figure for problem 538750

Hints

- The bar heights are frequencies. - Find the total number of teams first. - Use cumulative frequencies from smaller to larger values.

Solution

1. Add the frequencies: \(2 + 5 + 9 + 6 + 2 = 24\) teams. 2. The median is determined by positions \(12\) and \(13\). 3. The cumulative frequency through \(2\) checkpoints is \(7\), and through \(3\) checkpoints it is \(16\). Therefore, both middle positions are \(3\).

Answer

The median is \(3\) completed checkpoints.

5387546

The value \(6\) was accidentally left out of an ordered data set. Without it, the data set is \(4, 8, 9, 11, 13, 15\). a) Find the median of the incomplete data set. b) Insert the missing value and find the correct median.

Hints

- Treat the two data sets separately. - Notice that the number of values changes from even to odd. - Insert the missing value in the correct position.

Solution

1. The incomplete data set has six values. Its middle values are \(9\) and \(11\), so its median is \((9 + 11) \div 2 = 10\). 2. The complete ordered data set is \(4, 6, 8, 9, 11, 13, 15\). 3. With seven values, the median is the fourth value, \(9\).

Answer

a) The incomplete median is \(10\). b) The correct median is \(9\).

5387756

Six repairs to a bicycle trailer took \(12, 14, 15, 13, 16, 56\) minutes. Find the mean and median. Which measure better describes a typical repair time?

Hints

- Compare where most of the values lie with each measure. - Look for a value far from the others. - A useful typical value should represent most observations.

Solution

1. The mean is \((12 + 14 + 15 + 13 + 16 + 56) \div 6 = 126 \div 6 = 21\) minutes. 2. In order, the times are \(12, 13, 14, 15, 16, 56\). The median is \((14 + 15) \div 2 = 14.5\) minutes. 3. The value \(56\) is much greater than the other times and pulls the mean upward. 4. The median better describes a typical repair time.

Answer

Mean: \(21\) minutes; median: \(14.5\) minutes. The median better describes a typical repair.

5387766

Seven street-music groups collected these amounts during one afternoon: \(\$0, \$28, \$30, \$31, \$32, \$33, \$35\). One group had to stop immediately because of rain and collected \(\$0\). Decide whether the mean or median better describes the groups' usual earnings.

Hints

- Consider why one value is much lower than the others. - Compare both measures with the interval containing most values. - The purpose of the summary determines which measure is more useful.

Solution

1. The mean is \((0 + 28 + 30 + 31 + 32 + 33 + 35) \div 7 = 189 \div 7 = \$27\). 2. The median of the ordered data is \(\$31\). 3. The unusual value \(\$0\) pulls the mean downward. 4. The median is closer to most of the amounts and better describes the usual earnings.

Answer

Mean: \(\$27\); median: \(\$31\). The median better describes the groups' usual earnings.

5387886

The bar graph shows wait times at a station. Find the mean and median, and decide which value better describes a usual wait time.

Figure for problem 538788

Hints

- The bar heights show how often each wait time occurs. - Compare the rare large values with the main cluster. - Base your decision on the meaning of “usual.”

Solution

1. There are \(8 + 10 + 6 + 2 = 26\) groups. The total wait time is \(2 \times 8 + 3 \times 10 + 4 \times 6 + 17 \times 2 = 104\) minutes. 2. The mean is \(104 \div 26 = 4\) minutes. 3. The middle positions, \(13\) and \(14\), both have a value of \(3\), so the median is \(3\) minutes. 4. The two waits of \(17\) minutes increase the mean. The median better describes a usual wait time.

Answer

Mean: \(4\) minutes; median: \(3\) minutes. The median better describes a usual wait time.

5387906

A survey uses the responses “dissatisfied,” “neutral,” and “satisfied.” A student replaces them with \(1\), \(2\), and \(3\) and reports a mean. What limitation must the student state?

Figure for problem 538790

Hints

- An order does not automatically create measurable equal distances. - Identify the extra assumption introduced by the numerical codes. - State the conclusion cautiously rather than treating the mean as exact measurement.

Solution

1. The responses have an order, but the distances between the categories are not measured and may not be equal. 2. Coding the responses as \(1, 2, 3\) assumes equal spacing between the levels. 3. The resulting mean depends on that assumption. 4. The category percentages and the median category are safer to interpret.

Answer

The student must state that the mean depends on an artificial coding system and the assumption that the response levels are equally spaced.

5387946

A small outdoor museum recorded \(92, 105, 98, 101, 96, 455, 103\) visitors on seven Saturdays. A free special festival was held on the sixth Saturday. Which measure is better for planning attendance on a normal Saturday?

Hints

- Consider the effect of the special event. - Compare both measures with the normal Saturdays. - The situation being planned determines which values are relevant.

Solution

1. The mean is \((92 + 105 + 98 + 101 + 96 + 455 + 103) \div 7 = 1050 \div 7 = 150\) visitors. 2. In order, the values are \(92, 96, 98, 101, 103, 105, 455\), so the median is \(101\) visitors. 3. The festival attendance of \(455\) is not typical of a normal Saturday and greatly increases the mean. 4. The median is close to the six normal Saturdays and is better for planning.

Answer

The median of \(101\) visitors is better for planning a normal Saturday than the mean of \(150\) visitors.

5387966

A boat rental averages \(14\) rentals per weekday and \(28\) rentals per weekend day. Tom calculates the daily average for a full week as \((14 + 28) \div 2 = 21\). Evaluate Tom's calculation and find the correct daily average for the week.

Hints

- Count how many days each average represents. - Find the total rentals for the week. - Divide the weekly total by \(7\).

Solution

1. The weekday mean applies to \(5\) days, while the weekend mean applies to \(2\) days. 2. The total number of weekly rentals is \(5 \times 14 + 2 \times 28 = 126\). 3. The daily average is \(126 \div 7 = 18\) rentals. 4. Tom's calculation incorrectly gives the weekday and weekend means equal weight.

Answer

Tom's calculation is incorrect. The correct daily average is \(18\) boat rentals.

5387976

In Group A, \(80\%\) of \(10\) people solve a problem. In Group B, \(40\%\) of \(30\) people solve it. Mia averages the two percentages and gets \(60\%\). Is \(60\%\) the percent of all people who solved the problem?

Hints

- The two percentages describe groups of different sizes. - Convert each percentage to a number of people first. - Find the overall percent using all participants.

Solution

1. In Group A, \(0.8 \times 10 = 8\) people solved the problem. 2. In Group B, \(0.4 \times 30 = 12\) people solved the problem. 3. Altogether, \(8 + 12 = 20\) of \(10 + 30 = 40\) people solved it. 4. The overall percent is \(20 \div 40 = 50\%\), not \(60\%\).

Answer

No. Altogether, \(20\) of \(40\) people solved the problem, so the overall percent is \(50\%\).

5387986

A kitchen is planning portions for seven groups with sizes \(12, 12, 13, 13, 14, 14, 41\). For the question “How many portions are needed altogether?” the mean and median are available. Which measure directly helps with the total planning?

Hints

- Distinguish a typical group size from the total amount needed. - Identify which measure, together with the number of groups, determines the sum. - A large group is still important when planning the total.

Solution

1. The total number of people is \(12 + 12 + 13 + 13 + 14 + 14 + 41 = 119\). 2. The mean group size is \(119 \div 7 = 17\) people. 3. The total can be reconstructed from the mean and number of groups: \(17 \times 7 = 119\). 4. The median of \(13\) describes a typical group but does not determine the total.

Answer

The mean is more useful for the total planning. With \(7\) groups, it gives \(17 \times 7 = 119\) portions.

5388006

Seven teenagers spent \(\$4, \$5, \$5, \$6, \$6, \$7,\) and \(\$44\) at a flea market. A newspaper reports, “A teenager spent an average of about \(\$11\).” Is this a good description of a typical purchase?

Hints

- Compare the newspaper's value with most individual purchases. - Look for one purchase that strongly affects the mean. - Distinguish a correct calculation from an appropriate interpretation.

Solution

1. The mean is \((4 + 5 + 5 + 6 + 6 + 7 + 44) \div 7 = 77 \div 7 = \$11\). 2. The median is \(\$6\). 3. Six purchases are between \(\$4\) and \(\$7\), while one purchase is \(\$44\). 4. The reported mean is mathematically correct but misleading as a description of a typical purchase.

Answer

The report is mathematically correct but misleading. The median of \(\$6\) better describes a typical purchase.

5115716

A weather station recorded these temperatures at \(8{:}00\) a.m. on six consecutive days: Monday: \(54^\circ\text{F}\), Tuesday: \(58^\circ\text{F}\), Wednesday: \(53^\circ\text{F}\), Thursday: \(60^\circ\text{F}\), Friday: \(56^\circ\text{F}\), Saturday: \(62^\circ\text{F}\). The mean temperature for all seven days was exactly \(58^\circ\text{F}\). a) Find Sunday’s temperature. b) Suppose Monday’s reading was actually \(4^\circ\text{F}\) too high and Tuesday’s reading was \(4^\circ\text{F}\) too low. Explain without recomputing the mean whether the weekly mean changes.

Hints

- Use the weekly mean to find the total for seven days. - Add the six known temperatures and subtract from the required total. - For part b), consider only the net change in the sum.

Solution

1. A mean of \(58^\circ\text{F}\) for \(7\) days requires a total of \(58\times7=406\). 2. The Monday-through-Saturday total is \(54+58+53+60+56+62=343\). 3. Sunday’s temperature is \(406-343=63^\circ\text{F}\). 4. Correcting Monday subtracts \(4\), while correcting Tuesday adds \(4\). The total change is \(-4+4=0\), so the total and the mean remain unchanged.

Answer

a) \(63^\circ\text{F}\) b) The mean does not change because the two corrections have a net effect of \(0\) on the total.

5115836

These noon temperatures were recorded during one week: \(54^\circ\text{F}\), \(56^\circ\text{F}\), \(52^\circ\text{F}\), \(58^\circ\text{F}\), \(54^\circ\text{F}\), \(56^\circ\text{F}\), and \(90^\circ\text{F}\). The \(90^\circ\text{F}\) reading was probably a measurement error because the thermometer was briefly in direct sunlight. a) Find the mean of all seven readings. b) Find the mean of only the six realistic readings. c) Which mean better describes the typical weather that week? Explain.

Hints

- Find one mean with all seven values and another without the suspected error. - Compare both means with the cluster of realistic temperatures. - Consider how one unusually large value affects a mean.

Solution

1. The sum of all readings is \(54+56+52+58+54+56+90=420\), so the mean is \(420\div7=60^\circ\text{F}\). 2. Without the suspected error, the sum is \(420-90=330\), so the mean is \(330\div6=55^\circ\text{F}\). 3. The six realistic readings range from \(52^\circ\text{F}\) to \(58^\circ\text{F}\). The mean of \(55^\circ\text{F}\) lies within this cluster, while the \(90^\circ\text{F}\) outlier pulls the overall mean upward. Therefore, \(55^\circ\text{F}\) better represents the typical weather.

Answer

a) \(60^\circ\text{F}\) b) \(55^\circ\text{F}\) c) \(55^\circ\text{F}\) better describes the typical weather because the suspected outlier raises the mean of all seven readings.

5148166

Create a data set of exactly five different positive integers that meets all three conditions: 1. The median is \(10\). 2. The mean is exactly \(20\). 3. No value is repeated. Give your data set and show that it meets all three conditions.

Hints

- What total must five values have if their mean is \(20\)? - Which position contains the median when five values are ordered? - Set the median first, then choose the other values so the total is correct. - Order the values and check that none are repeated.

Solution

1. For five ordered values \(x_1 < x_2 < x_3 < x_4 < x_5\), the median is the middle value, so \(x_3 = 10\). 2. A mean of \(20\) requires a total of \(5 \times 20 = 100\). 3. Choose two different positive integers less than \(10\), such as \(1\) and \(2\). Their total with the median is \(1 + 2 + 10 = 13\). 4. The last two values must total \(100 - 13 = 87\). Choosing \(11\) gives the final value \(87 - 11 = 76\). 5. The data set \(1, 2, 10, 11, 76\) has median \(10\), mean \((1 + 2 + 10 + 11 + 76) \div 5 = 20\), and five different positive-integer values.

Answer

One possible data set is \(1, 2, 10, 11, 76\). Its median is \(10\), its mean is \(20\), and all five values are different positive integers.

5318186

A soccer team scored different numbers of goals in its first seven games. The bar graph shows the goals for Games G1 through G7. a) Find the median number of goals for the seven games. b) How many goals could the team score in Game 8 so that the median for all eight games is exactly \(2\) goals? Give all possible whole-number answers.

Figure for problem 531818

Hints

- Read all seven values from the graph and order them. - How do you find the median when there is an odd number of values? - With eight values, which two positions determine the median? - Test possible whole-number scores for the eighth game.

Solution

1. The graph shows \(2, 1, 4, 1, 3, 2,\) and \(5\) goals. 2. In order, the values are \(1, 1, 2, 2, 3, 4, 5\). The middle value is \(2\), so the median is \(2\) goals. 3. With eight values, the median is the mean of the fourth and fifth values in order. If the eighth score is \(0\), \(1\), or \(2\), those middle values are both \(2\), so the median is \(2\). 4. If the eighth score is \(3\) or more, the middle values are \(2\) and \(3\), giving a median of \(2.5\). Therefore, the possible eighth-game scores are \(0, 1,\) and \(2\).

Answer

a) \(2\) goals b) \(0\), \(1\), or \(2\) goals

5387276

Seven ratings must have a mean of \(10\) points. Five ratings are \(6, 8, 9, 11,\) and \(14\). The two missing ratings are different whole numbers from \(5\) through \(15\). Give one possible pair of missing ratings.

Hints

- Find the total needed for seven values to have a mean of \(10\). - More than one pair can work. - Check every condition on the two missing ratings.

Solution

1. The required total is \(10 \times 7 = 70\) points. 2. The five known ratings total \(6 + 8 + 9 + 11 + 14 = 48\) points. 3. The two missing ratings must total \(70 - 48 = 22\). 4. One possible pair is \(10\) and \(12\). They are different whole numbers in the required range and have a sum of \(22\).

Answer

One possible pair is \(10\) and \(12\).

5387366

A nonnegative whole number \(x\) is added to the data set \(3, 6, 8, 12, 15\). The median of the six values must be \(10\). Describe all possible values of \(x\).

Hints

- Consider where the new value can appear in the ordered list. - Analyze ranges of values rather than testing only a few examples. - Check the boundary where the two middle positions change.

Solution

1. With six values, the median is the mean of the third and fourth values in order. 2. If \(x \ge 12\), the third and fourth values remain \(8\) and \(12\). 3. Their mean is \((8 + 12) \div 2 = 10\). 4. If \(x < 12\), the median is less than \(10\). Therefore, every nonnegative whole number \(x \ge 12\) works.

Answer

All nonnegative whole numbers \(x \ge 12\) are possible.

5387376

Exactly one value is removed from the ordered data set \(2, 4, 5, 6, 7, 8, 9, 10, 12\). After the removal, the median must be \(7.5\). Which values can be removed?

Hints

- After the removal, there is an even number of values. - First test removals to the left of the middle. - More than one value may work.

Solution

1. After one value is removed, eight values remain, so the median is the mean of the fourth and fifth values. 2. If \(2\), \(4\), \(5\), or \(6\) is removed, the middle values are \(7\) and \(8\). 3. Their mean is \((7 + 8) \div 2 = 7.5\). 4. Removing \(7\) or any greater value produces different middle values, so no other removal works.

Answer

\(2\), \(4\), \(5\), or \(6\) can be removed.

5387446

The ordered data set is \(3, 5, 8, a, b, 17, 20, 22\). Its median is \(12\), and \(b = a + 4\). Find \(a\) and \(b\).

Hints

- Which two values determine the median of eight data values? - Translate both conditions into equations. - Check that the answers preserve the ordered list.

Solution

1. With eight values, the median is the mean of the fourth and fifth values: \(\frac{a + b}{2} = 12\). Therefore, \(a + b = 24\). 2. Substitute \(b = a + 4\): \(a + (a + 4) = 24\). 3. Solve: \(2a = 20\), so \(a = 10\). Then \(b = 14\). 4. Both values fit the stated order.

Answer

\(a = 10\) and \(b = 14\)

5387536

A frequency table has four values equal to \(1\), \(x\) values equal to \(2\), five values equal to \(3\), and three values equal to \(4\). The value of \(x\) is a nonnegative whole number. What is the least possible value of \(x\) for which the median is \(2\)?

Hints

- Test values near the point where the middle position moves from the threes to the twos. - Track both the total number of values and the middle position. - Show why a smaller value does not work.

Solution

1. If \(x=4\), there are \(4 + 4 + 5 + 3 = 16\) values. The middle values, in positions \(8\) and \(9\), are \(2\) and \(3\), so the median is \((2 + 3) \div 2 = 2.5\). 2. If \(x=5\), there are \(17\) values. The median is the ninth value, and the block of twos occupies positions \(5\) through \(9\). 3. Therefore, the median is \(2\) when \(x=5\). Since every smaller value of \(x\) gives a median greater than \(2\), this is the least possible value.

Answer

The least possible value is \(x=5\).

All problems may be used, copied and printed free of charge for school and tutoring, including paid tutoring. Commercial adaptations as well as publication or redistribution on the internet are not permitted.