Summary statistics for one quantitative variable

Learning goals

Calculates the mean of a data set.
Determines the median of a data set.
Calculates the range of a data set.
Calculates the interquartile range of a data set.
Calculates the standard deviation of a data set.
Chooses mean and standard deviation or median and IQR to summarize a data set.
Interprets summary statistics in the variable's context and units.
Finds percentiles in ordered data.

Difficulty: 1 – 5

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Summary statistics for one quantitative variable 1

For the data $8,11,12,14,15$, calculate the mean and median.

Hints

- Find the balance-point summary and the middle-position summary separately. - For the mean, divide the sum by the number of observations; for the median, use the ordered middle position.

Solution

1. The sum is $60$, so the mean is $\frac{60}{5}=12$. 2. The ordered middle value is $12$.

Answer

Mean: $12$. Median: $12$.

Summary statistics for one quantitative variable 2

Sample A has mean $72$ and standard deviation $4$. Sample B has mean $72$ and standard deviation $11$. Compare their centers and variability.

Hints

- Interpret each statistic according to what it measures. - Equal means imply equal centers under that measure; compare standard deviations for consistency.

Solution

1. The equal means show that the samples have the same measured center. 2. The larger standard deviation for Sample B indicates greater typical distance from the mean. 3. Sample B is more variable.

Answer

The samples have the same mean, but Sample B has greater variability.

Summary statistics for one quantitative variable 3

A sample has standard deviation $s=7.4$. Find the sample variance.

Hints

- Use the defining relationship between the two spread measures. - Variance is the square of standard deviation, so keep the units squared.

Solution

1. The sample variance is the square of the sample standard deviation. 2. $s^2=(7.4)^2=54.76$.

Answer

The sample variance is $54.76$.

Summary statistics for one quantitative variable 4

Sawyer's result is at the $72$nd percentile of a score distribution. Interpret this statement correctly. State two conclusions that do not follow from the percentile rank alone.

Hints

- A percentile describes relative position within a distribution. - Separate rank information from the measurement scale of the score. - Ask what numerical details would require the original score distribution.

Solution

1. A $72$nd-percentile result means approximately $72\%$ of the observations are at or below Sawyer's score. 2. It does not mean Sawyer answered $72\%$ of the questions correctly. 3. It does not identify the numerical score or the number of points separating Sawyer from another percentile.

Answer

Approximately $72\%$ of scores are at or below Sawyer's score. The percentile rank does not give the percent correct or the numerical score.

Summary statistics for one quantitative variable 5

A data set has mean $7$ and standard deviation $0$. What must be true about every observation? Explain.

Hints

- Standard deviation measures distance from the mean. - Determine when a sum of nonnegative squared distances can equal zero. - Use the stated mean to identify the common value.

Solution

1. Standard deviation $0$ means every squared deviation from the mean is $0$. 2. Therefore every observation has deviation $0$ from $7$. 3. Every observation must equal $7$.

Answer

Every observation equals $7$.

Summary statistics for one quantitative variable 6

Three siblings saved $\$5$, $\$7$, and $\$9$. If they shared their savings equally without changing the total, how much would each have? What statistic does that equal share represent?

Hints

- Equal sharing keeps the total unchanged. - Find the total and divide by the number of people. - Think about which center measure is an equal-share value.

Solution

1. The total is $5+7+9=21$ dollars. 2. Dividing evenly among three siblings gives $21\div3=7$ dollars each. 3. The equal-share amount is the arithmetic mean.

Answer

$\$7$ each; this is the mean.

Summary statistics for one quantitative variable 7

Five bike-repair times are $2,3,4,5,19$ minutes. Find the median. If only the longest repair instead took $29$ minutes, what would the new median be? Explain.

Hints

- Identify the middle position when five values are ordered. - The median depends on rank, not the sum of all times.

Solution

1. There are five times, so the median is the third ordered value. 2. The third value is $4$ minutes. 3. Raising the already largest time from $19$ to $29$ leaves the ordering of the other values unchanged, so the third value stays $4$ minutes.

Answer

Original median $4$ minutes; new median $4$ minutes. The last value remains largest, so the middle position does not change.

Summary statistics for one quantitative variable 8

Two samples have the same mean of $6$. Sample A is $4,6,8$, and Sample B is $2,6,10$. Without calculating a standard deviation, which sample has greater spread about the mean? Explain from the distances.

Hints

- Compare every value with the shared mean. - Decide which sample has greater typical distance from the center. - You do not need a full variance formula to compare these patterns.

Solution

1. A's values are at distances $2,0,2$ from $6$. 2. B's values are at distances $4,0,4$ from $6$. 3. The observations in B lie farther from the same mean, so B has greater standard deviation.

Answer

Sample B has greater spread and standard deviation because its values lie farther from the common mean.

Summary statistics for one quantitative variable 9

Use the median-of-halves method to find $Q_1$, the median, $Q_3$, the range, and the IQR for $4,7,9,10,13,17$.

Hints

- Order the values and separate the lower and upper halves. - After finding the quartiles, compute range from the extremes and IQR from the quartile difference.

Solution

1. The median is $\frac{9+10}{2}=9.5$. 2. The lower-half median is $Q_1=7$, and the upper-half median is $Q_3=13$. 3. The range is $17-4=13$. 4. The IQR is $13-7=6$.

Answer

$Q_1=7$, median $=9.5$, $Q_3=13$, range $=13$, IQR $=6$.

Summary statistics for one quantitative variable 10

Calculate the sample standard deviation of $3,5,7,8,12$. Round to two decimal places.

Hints

- Track deviations from the mean and remember that the requested spread is for a sample. - Square each deviation, divide their sum by $n-1$, and take the square root only at the end.

Solution

1. The sample mean is $\bar x=7.0$. 2. The sum of squared deviations is $46.0$. 3. The sample variance is $\frac{46.0}{5-1}=11.500$. 4. The sample standard deviation is $s\approx 3.39$.

Answer

$s\approx 3.39$.

Summary statistics for one quantitative variable 11

Four of five values are $12,15,17,19$. The mean of all five values is $16$. Find the missing value.

Hints

- Work backward from the total implied by the mean. - Use $\text{total}=n\bar{x}$, then compare that required total with the sum of the known values.

Solution

1. The required total is $5\cdot 16=80$. 2. The known values total $63$. 3. The missing value is $80-63=17$.

Answer

The missing value is $17$.

Summary statistics for one quantitative variable 12

Lengths measured in inches are converted to centimeters by multiplying every value by $2.54$. Describe the effect on the mean, median, range, IQR, and standard deviation.

Hints

- Consider how scaling all data values changes both positions and distances. - A positive scale factor multiplies both locations and distances by that factor.

Solution

1. Multiplying every value by a positive constant multiplies all location and spread measures by that constant. 2. Therefore, the mean, median, range, IQR, and standard deviation are each multiplied by $2.54$.

Answer

Each listed statistic is multiplied by $2.54$.

Summary statistics for one quantitative variable 13

A laboratory uses this screening rule: flag any measurement that is more than $2$ standard deviations from the mean for follow-up review. A distribution has mean $50$ and standard deviation $6$. Which of $37,41,62,64$ are flagged by this rule?

Hints

- Translate the stated screening distance into lower and upper boundaries around the mean. - Compare each candidate measurement with that interval. - Pay attention to the word “more than” when a value lies exactly on a boundary.

Solution

1. Two standard deviations is $2\cdot6=12$. 2. Measurements within two standard deviations of the mean lie in $[50-12,50+12]=[38,62]$. 3. The values $37$ and $64$ lie outside that interval, while $41$ and $62$ do not.

Answer

The screening rule flags $37$ and $64$.

Summary statistics for one quantitative variable 14

House-sale prices in a neighborhood are strongly right-skewed because of a few very expensive homes. Which pair is more appropriate for describing center and spread: mean and standard deviation, or median and IQR? Explain.

Hints

- Identify the skewness and the presence of extreme values. - Compare the resistant pair median and IQR with the nonresistant pair mean and standard deviation.

Solution

1. Extreme high prices can greatly affect the mean and standard deviation. 2. The median and IQR are resistant to those extreme values. 3. Therefore, median and IQR are more appropriate.

Answer

Median and IQR, because they are resistant to the high-price outliers.

Summary statistics for one quantitative variable 15

Nine observations have mean $14$. A tenth observation equal to $24$ is added. Find the new mean.

Hints

- Recover the original total from the old mean and sample size. - After adding the new observation, update both the total and the number of observations.

Solution

1. The original sum is $9\cdot14=126$. 2. The new sum is $126+24=150$. 3. The new mean is $\frac{150}{10}=15$.

Answer

The new mean is $15$.

Summary statistics for one quantitative variable 16

Twelve values have mean $18.5$. One value equal to $24$ is removed. Find the mean of the remaining values.

Hints

- Convert the original mean to a total, adjust the total, and update the sample size. - Remove the specified observation from the total and reduce the number of observations by one.

Solution

1. The original sum is $12\cdot18.5=222$. 2. The remaining sum is $222-24=198$. 3. The new mean is $\frac{198}{11}=18$.

Answer

The new mean is $18$.

Summary statistics for one quantitative variable 17

The ordered data are the even integers from $2$ through $40$. Find the median and explain its position.

Hints

- Use the number of ordered values to locate the middle position or positions. - Because the list has an even number of terms, average the two central ordered values.

Solution

1. There are $20$ values, so the median is the mean of the $10$th and $11$th values. 2. Those values are $20$ and $22$. 3. The median is $\frac{20+22}{2}=21$.

Answer

The median is $21$.

Summary statistics for one quantitative variable 18

Two data sets have the same minimum and the same quartiles, but one data set has a much larger maximum. Which spread measure must change, and which may remain unchanged: range and IQR?

Hints

- Identify which data positions each spread measure uses. - Range depends on the extremes, while IQR depends only on the two quartiles.

Solution

1. The range depends directly on the maximum and minimum, so a much larger maximum increases the range because the minimum is the same. 2. The IQR depends only on $Q_1$ and $Q_3$. 3. Since the quartiles are the same, the IQR remains unchanged.

Answer

The range must increase; the IQR remains unchanged.

Summary statistics for one quantitative variable 19

A distribution has mean $41.8$, median $35.2$, and a maximum far above the rest of the data. What does the relationship between the mean and median suggest, and which center is more resistant?

Hints

- Relate the direction of the mean–median difference to the likely tail. - An extreme high tail pulls the mean upward more strongly than it moves the median.

Solution

1. The mean is substantially greater than the median, consistent with right skew caused by high values. 2. The median is resistant to the extreme maximum, while the mean is not. 3. The median is the more resistant center.

Answer

The data are likely right-skewed, and the median is the more resistant measure of center.

Summary statistics for one quantitative variable 20

The ordered delivery times, in minutes, are $4,5,6,7,8,9,10,11,12,13,14,15,16,18,20,22,24,26,28,30$. Using the nearest-rank method $r=\lceil pn\rceil$, find the $65$th percentile and interpret it in context.

Hints

- Convert the percentile to a decimal and apply the stated nearest-rank rule to the sample size. - Use the resulting ordered position without interpolating between observations.

Solution

1. Here, $p=0.65$ and $n=20$, so the rank is $r=\lceil0.65\cdot20\rceil=13$. 2. The $13$th ordered value is $16$. 3. Therefore, the $65$th percentile is $16\,\text{minutes}$. In this sample, $65\%$ of the delivery times are at or below $16\,\text{minutes}$.

Answer

The $65$th percentile is $16\,\text{minutes}$; $65\%$ of the sampled delivery times are at or below this value.

Summary statistics for one quantitative variable 21

A boxplot in centimeters has five-number summary $(25.4,50.8,76.2,101.6,127.0)$. Convert the summary to inches using $1\,\text{in}=2.54\,\text{cm}$.

Hints

- Use the same unit conversion on every landmark. - Check that the converted summary remains increasing and is labeled in inches.

Solution

1. Divide each value by $2.54$. 2. The converted values are $10,20,30,40,50$ inches.

Answer

$(10,20,30,40,50)$ inches.

Summary statistics for one quantitative variable 22

During one week in October, a bike-repair shop records a median completion time of $2.6$ days for $41$ completed repairs. Write a complete interpretation of this statistic that does not overstate the population or time period represented.

Hints

- Name the group and time period from which the number was calculated. - Interpret the statistic using its position in the ordered data. - Avoid extending the conclusion beyond the observations described.

Solution

1. The value $2.6$ days is the sample median for the $41$ repairs completed during the stated October week. 2. A median interpretation states that about half of those recorded completion times were at or below $2.6$ days and about half were at or above it. 3. The statistic does not by itself describe repairs from other weeks or all repairs the shop will complete.

Answer

Among the $41$ repairs completed during that week in October, about half had completion times at or below $2.6$ days and about half had completion times at or above $2.6$ days. This statement does not automatically extend to other weeks.

Summary statistics for one quantitative variable 23

A dotplot's horizontal axis is labeled “Population (thousands).” Its smallest plotted value is $1.2$, and its largest is $1.8$. State the corresponding populations and the range in people.

Hints

- Read the unit attached to the axis label. - Convert both endpoints before subtracting. - State the final range in the requested unit.

Solution

1. The axis unit means each plotted unit represents $1000$ people. 2. The minimum population is $1.2\cdot1000=1200$, and the maximum is $1.8\cdot1000=1800$. 3. The range is $1800-1200=600$ people.

Answer

The populations are $1200$ and $1800$, and the range is $600$ people.

Summary statistics for one quantitative variable 24

A sample has $10$ observations, and the sum of squared deviations from its mean is $180$. Find the sample variance exactly and the sample standard deviation rounded to three decimal places.

Hints

- Check whether the data are described as a sample or an entire population. - Recall which denominator belongs in the sample variance formula. - Standard deviation is the positive square root of variance.

Solution

1. Use $n-1$ for the sample variance: $s^2=\frac{180}{10-1}=20$. 2. The sample standard deviation is $s=\sqrt{20}\approx4.472$.

Answer

Sample variance $s^2=20$; sample standard deviation $s=\sqrt{20}\approx4.472$.

Summary statistics for one quantitative variable 25

Use the percentile-position rule $L=\frac{p}{100}(n+1)$ for the ordered data $2,4,5,7,8,9,11,13,15,18,20$. Find the $75$th percentile.

Hints

- Use the stated percentile convention rather than a different software rule. - Calculate the position before reading the ordered list. - An integer position points directly to one observation.

Solution

1. Here $n=11$, so $L=0.75(11+1)=9$. 2. The ninth ordered value is $15$.

Answer

The $75$th percentile is $15$.

Summary statistics for one quantitative variable 26

A report states that a data set of $6$ observations has mean $8$ and total sum $43$. Determine whether both statements can be correct.

Hints

- Use the defining relationship among mean, sample size, and sum. - Compute the implied total from one statement. - Compare it with the independently reported total.

Solution

1. A mean of $8$ for $6$ observations requires a total of $6\cdot8=48$. 2. The stated total $43$ would instead give mean $\frac{43}{6}\approx7.167$. 3. Therefore the two reported summaries are inconsistent.

Answer

No. A mean of $8$ requires sum $48$, not $43$.

Summary statistics for one quantitative variable 27

A quantitative distribution is approximately symmetric with no strong outliers. A report must summarize its center and spread with one standard pair of statistics. Choose between mean with standard deviation and median with interquartile range. Justify the choice.

Hints

- Match the summary pair to the distribution's shape and unusual features. - Consider which center is naturally linked to standard deviation. - Resistant summaries are most valuable when extremes or skew are important.

Solution

1. For an approximately symmetric distribution without strong outliers, the mean represents the center well. 2. Standard deviation summarizes typical distance from the mean and uses all observations. 3. Therefore mean with standard deviation is the more natural pair; median with interquartile range would also be valid but is especially useful for skewed data or data with outliers.

Answer

Use the mean and standard deviation because the distribution is approximately symmetric and has no strong outliers.

Summary statistics for one quantitative variable 28

In a continuous distribution, value $a$ is at the $18$th percentile and value $b$ is at the $73$rd percentile. What proportion of observations lie between $a$ and $b$? Does this calculation require the distribution to be normal?

Hints

- Interpret each percentile as cumulative area to the left. - Find the area between two cumulative positions by subtraction. - Decide whether any feature specific to a normal curve was used.

Solution

1. The area to the left of $a$ is $0.18$, and the area to the left of $b$ is $0.73$. 2. The proportion between them is $0.73-0.18=0.55$. 3. The calculation uses only percentile definitions, so normality is not required.

Answer

The proportion is $0.55$, or $55\%$. The result does not require a normal distribution.

Summary statistics for one quantitative variable 29

A report about one quantitative distribution claims that a value of $42$ is at the $70$th percentile and a value of $50$ is at the $60$th percentile. Explain why the two claims cannot both be correct.

Hints

- Compare the order of the two raw values with the order of their claimed percentile ranks. - Think about whether the cumulative proportion at or below a cutoff can decrease when the cutoff moves upward. - This is an ordering question, not a distribution-model calculation.

Solution

1. A percentile rank describes the cumulative proportion at or below a value. 2. Within one distribution, moving to a larger raw value cannot decrease the cumulative proportion at or below the value. 3. Because $50>42$ but $60\%<70\%$, the reported ordering is impossible.

Answer

The claims are inconsistent. In the same distribution, the larger value $50$ cannot have a lower percentile rank than the smaller value $42$.

Summary statistics for one quantitative variable 30

A sample of rechargeable batteries has mean life $9.5\,\text{hours}$ and standard deviation $1.2\,\text{hours}$. a) Interpret the standard deviation in context. b) Explain why the standard deviation is measured in hours. c) A colleague says, “About $68\%$ of the batteries must last within $1.2\,\text{hours}$ of the mean.” Is that conclusion justified from the given information alone?

Hints

- Interpret standard deviation as a distance from the mean, not as a percentage or quartile width. - Think about the units of the original deviations after the standard-deviation calculation is completed. - Separate what standard deviation alone says from what requires a distribution-shape assumption.

Solution

1. A standard deviation of $1.2\,\text{hours}$ means individual battery lives typically differ from the sample mean of $9.5\,\text{hours}$ by about $1.2\,\text{hours}$. 2. Standard deviation is based on distances from the mean and is converted back from squared deviations, so it has the same units as the original battery-life measurements: hours. 3. The $68\%$ statement is associated with an approximately normal distribution. No distribution shape is given, so that conclusion is not justified here.

Answer

a) Battery lives typically differ from the mean by about $1.2\,\text{hours}$. b) Standard deviation has the same units as the original measurements, so its unit is hours. c) No. The $68\%$ conclusion needs an approximately normal model or comparable shape information.

Summary statistics for one quantitative variable 31

Eight checkout times, in minutes, are $3,4,4,5,6,7,8,12$. Using the median-of-halves convention: a) Find the median and interquartile range. b) Interpret both results in context.

Hints

- Find the overall middle before dividing the ordered data into halves. - The IQR is the distance from $Q_1$ to $Q_3$. - Interpret the numerical summaries using minutes and the middle positions of the data.

Solution

1. The median is $\frac{5+6}{2}=5.5$ minutes. 2. The lower half is $3,4,4,5$, so $Q_1=4$. The upper half is $6,7,8,12$, so $Q_3=7.5$. 3. The interquartile range is $7.5-4=3.5$ minutes. 4. A median of $5.5$ minutes means the ordered checkout times are split around $5.5$ minutes. The middle $50\%$ of the times run from $4$ to $7.5$ minutes, a span of $3.5$ minutes.

Answer

a) Median: $5.5\,\text{minutes}$; IQR: $3.5\,\text{minutes}$. b) The ordered times are centered at $5.5\,\text{minutes}$, and the middle $50\%$ lie from $4$ to $7.5\,\text{minutes}$, spanning $3.5\,\text{minutes}$.

Summary statistics for one quantitative variable 32

Find the sample standard deviation of $2,4,6$. Use $n-1$ in the sample variance.

Hints

- Find the mean before calculating deviations. - Square the deviations so negative and positive distances both contribute. - Remember the sample denominator is one less than the number of observations.

Solution

1. The mean is $\frac{2+4+6}{3}=4$. 2. The deviations are $-2,0,2$, whose squares add to $4+0+4=8$. 3. Divide by $3-1=2$ to get sample variance $4$. 4. The sample standard deviation is $s=\sqrt{4}=2$.

Answer

$s=2$.

Summary statistics for one quantitative variable 33

Four quiz scores are $6,7,9,x$, and their mean is $8$. Find the missing score. Explain how the mean tells you the total points.

Hints

- The mean is the total divided by the number of scores. - Work backward to find what total would give the stated mean. - Compare that total with the points already known.

Solution

1. Four scores with mean $8$ must total $4\cdot8=32$ points. 2. The known scores total $6+7+9=22$. 3. Thus $x=32-22=10$.

Answer

The four scores must total $4\cdot8=32$ points because the mean is $8$ over four scores. The other scores total $22$, so $x=32-22=10$.

Summary statistics for one quantitative variable 34

The five values are $2,4,5,7,8$. The largest value is corrected from $8$ to $20$. Find the original and corrected means and medians. Which measure changes more, and why?

Hints

- Recalculate the total after changing only the largest observation. - Find the middle position in both ordered lists. - Compare a total-based summary with a position-based summary.

Solution

1. The original mean is $\frac{26}{5}=5.2$, while the corrected mean is $\frac{38}{5}=7.6$. 2. Both ordered lists have middle value $5$, so both medians are $5$. 3. The mean increases by $2.4$, while the median stays the same because the corrected value remains in the highest position.

Answer

Mean: $5.2$ to $7.6$. Median: $5$ to $5$. The mean changes more because it depends on the magnitude of the large value.

Summary statistics for one quantitative variable 35

Four students have an average short shuttle-run time of $12$ seconds. After one time is removed, the other three have an average of $10$ seconds. What time was removed?

Hints

- What total corresponds to the original average? - What total remains after the time is removed? - How can you compare the two totals?

Solution

1. The total of all four times is $4\cdot12=48$ seconds. 2. The total of the three remaining times is $3\cdot10=30$ seconds. 3. The removed time is $48-30=18$ seconds.

Answer

$18$ seconds.

Summary statistics for one quantitative variable 36

Three students finish a warm-up in $2,4,6$ minutes. A coach predicts $4$ minutes for everyone. Another predicts $5$ minutes. For each prediction, add the squared errors. Which prediction has the smaller total?

Hints

- Find each difference between an observed time and a predicted time. - Square each difference before adding. - Compare the totals rather than the largest individual error.

Solution

1. For $4$ minutes, the squared errors are $(2-4)^2=4$, $(4-4)^2=0$, and $(6-4)^2=4$. The total is $8\,\text{min}^2$. 2. For $5$ minutes, they are $(2-5)^2=9$, $(4-5)^2=1$, and $(6-5)^2=1$. The total is $11\,\text{min}^2$. 3. Predicting $4$ minutes gives the smaller total; it is also the mean of the three times.

Answer

Squared-error totals: $8\,\text{min}^2$ for the $4$-minute prediction and $11\,\text{min}^2$ for the $5$-minute prediction. The $4$-minute prediction is better by this measure.

Summary statistics for one quantitative variable 37

Four potters spent $8,10,12,14$ minutes glazing small bowls. Find the mean time. Add the squared differences from the mean, then add the squared differences from $9$ minutes. How much larger is the second total? Compare the increase with four times the squared distance between $9$ and the mean.

Hints

- First locate the center of the four observed times. - Try writing out each difference before squaring and adding. - What role might the number of observations play in the increase?

Solution

1. The mean is $\frac{8+10+12+14}{4}=11$ minutes. 2. The squared differences from $11$ add to $9+1+1+9=20\,\text{min}^2$. 3. The squared differences from $9$ add to $1+1+9+25=36\,\text{min}^2$. 4. The increase is $36-20=16\,\text{min}^2$, which equals $4(11-9)^2=16\,\text{min}^2$.

Answer

Mean: $11\,\text{min}$. Squared-difference totals: $20\,\text{min}^2$ about the mean and $36\,\text{min}^2$ about $9\,\text{min}$. Increase: $16\,\text{min}^2$, equal to $4(11-9)^2$.

Summary statistics for one quantitative variable 38

In a robotics practice quiz, three students in the design group scored $6,6,6$ points, and three in the coding group scored $10,10,10$. Each group has a standard deviation of zero. Does the combined set of six scores also have standard deviation zero? Find its sample standard deviation to two decimal places.

Hints

- First find the center of all six scores together. - Are the two group centers the same as the combined center? - Measure the scores' distances from the combined center, not their separate group centers.

Solution

1. The six scores have mean $\frac{3\cdot6+3\cdot10}{6}=8$ points. 2. Every score is $2$ points from the combined mean, so the total of squared deviations is $6\cdot2^2=24$. 3. The combined sample standard deviation is $\sqrt{\frac{24}{6-1}}\approx2.19$ points, not zero.

Answer

No. The combined sample standard deviation is $\sqrt{\frac{24}{5}}\approx2.19$ points.

Summary statistics for one quantitative variable 39

Three students read $2,6,10$ books over the summer. Form the average for each pair of students. Find the mean of those three pair averages. Why does it match the mean for all three students?

Hints

- How many distinct pairs can you form from three students? - List the pairs before taking their averages. - Count how often each student's reading total appears in that list.

Solution

1. The pair averages are $\frac{2+6}{2}=4$, $\frac{2+10}{2}=6$, and $\frac{6+10}{2}=8$. 2. Their mean is $\frac{4+6+8}{3}=6$. The original mean is $\frac{2+6+10}{3}=6$. 3. Each student's count appears in two pair averages, with the same total weight. No student receives more weight than the others.

Answer

Pair averages $4,6,8$; their mean is $6$, the same as the original mean. Each count is included equally often in the pairs.

Summary statistics for one quantitative variable 40

The temperature changed by $-2,0,2$ degrees on three mornings. Find the square of the mean change and the mean of the three squared changes. Are they equal?

Hints

- In one calculation, average first and square afterward. - In the other, square each value before averaging. - Think about what happens to negative changes under squaring.

Solution

1. The mean change is $\frac{-2+0+2}{3}=0$ degrees, so its square is $0$. 2. The mean of the squared changes is $\frac{(-2)^2+0^2+2^2}{3}=\frac{8}{3}$ square degrees. 3. They are not equal: the positive and negative changes cancel before squaring, but not after.

Answer

Square of the mean: $0$. Mean of squares: $\frac{8}{3}$. They are not equal.

Summary statistics for one quantitative variable 41

Two test scores are missing from a list: $2,4,\square,\square$. The four scores have mean $5$ and sample variance $\frac{20}{3}$. A student suggests the missing scores are either $6,8$ or $5,9$. Which pair works? Explain why the mean alone cannot decide.

Hints

- Check the total required by the given mean. - Does that total rule out either suggested pair? - Compare the squared distances from the mean for the two possible lists.

Solution

1. Both pairs sum to $14$, so either would give a total of $20$ and mean $5$. 2. With $6,8$, the sum of squared deviations from $5$ is $9+1+1+9=20$, giving sample variance $\frac{20}{3}$. 3. With $5,9$, the squared deviations add to $9+1+0+16=26$, giving sample variance $\frac{26}{3}$. 4. Only $6,8$ matches both summaries. The mean fixes only the sum, not the spread.

Answer

The missing scores are $6,8$. Both choices match the mean, but only $6,8$ matches the sample variance.

Summary statistics for one quantitative variable 42

A filling line checks the masses of $100$ flour packages, rounded to the nearest gram. The results are shown below. <table> <tr> <td>Mass ($\text{g}$)</td> <td>497</td><td>498</td><td>499</td><td>500</td><td>501</td><td>502</td><td>503</td> </tr> <tr> <td>Number of packages</td> <td>3</td><td>12</td><td>25</td><td>30</td><td>18</td><td>9</td><td>3</td> </tr> </table> a) Describe the shape, center, and spread of the observed distribution qualitatively. b) Find the sample mean $\overline{x}$. c) Find the signed percent deviation of $\overline{x}$ from the target mass of $500\,\text{g}$, using the target mass as the reference value.

Hints

- Describe the overall pattern before calculating. - Use a weighted mean because values occur with different frequencies. - Divide the signed difference by the target value for the percent deviation.

Solution

1. The distribution is unimodal and approximately symmetric, with its center near $500\,\text{g}$ and relatively small spread. 2. The weighted total is $497(3)+498(12)+499(25)+500(30)+501(18)+502(9)+503(3)=49{,}987.$ Therefore, $\overline{x}=\frac{49{,}987}{100}=499.87\,\text{g}.$ 3. The signed percent deviation is $\frac{499.87-500}{500}\cdot100\%=-0.026\%.$ The sample mean is $0.026\%$ below the target.

Answer

a) Unimodal, approximately symmetric, centered near $500\,\text{g}$, with small spread. b) $\overline{x}=499.87\,\text{g}$. c) $-0.026\%$, meaning $0.026\%$ below the target.

Summary statistics for one quantitative variable 43

A factory measures the lengths of $200$ metal pins in millimeters. The results are shown below. <table><tr><td>Length ($\text{mm}$)</td><td>$24.8$</td><td>$24.9$</td><td>$25.0$</td><td>$25.1$</td><td>$25.2$</td></tr><tr><td>Count</td><td>12</td><td>45</td><td>82</td><td>48</td><td>13</td></tr></table> a) Describe the shape and center of the sample distribution. b) Find the sample mean $\bar{x}$ and sample standard deviation $s$, using $n-1$ for the variance denominator. Round $s$ to four decimal places. c) Find the signed percent difference between $\bar{x}$ and $25.0\,\text{mm}$, using $25.0\,\text{mm}$ as the reference.

Hints

- Compare the counts on either side of the central length. - The mean is a weighted total divided by the number of pins. - For a sample standard deviation, divide the sum of squared deviations by one less than the sample size. - A signed percent difference uses the stated reference length as its denominator.

Solution

1. The frequencies peak near $25.0\,\text{mm}$ and fall roughly symmetrically on both sides. 2. The weighted total is $24.8\cdot12+24.9\cdot45+25.0\cdot82+25.1\cdot48+25.2\cdot13=5000.5\,\text{mm}$. Thus $\bar{x}=\frac{5000.5}{200}=25.0025\,\text{mm}$. 3. The sum of squared deviations is $\sum f_i(x_i-\bar{x})^2=1.92875\,\text{mm}^2$. 4. The sample variance is $s^2=\frac{1.92875}{199}\approx0.0096922\,\text{mm}^2$, so $s\approx0.0984\,\text{mm}$. 5. The signed percent difference is $\frac{25.0025-25.0}{25.0}\cdot100\%=0.01\%$, positive.

Answer

a) Approximately symmetric and centered near $25.0\,\text{mm}$. b) $\bar{x}=25.0025\,\text{mm}$ and $s\approx0.0984\,\text{mm}$. c) $0.01\%$ above the reference length.

Summary statistics for one quantitative variable 44

Every temperature in a data set is converted from degrees Celsius to a new scale by adding $273.15$. Describe the effect on the mean, median, IQR, range, and standard deviation.

Hints

- Separate measures of location from measures based on differences between values. - Adding a constant shifts every ordered position by the same amount. - Measures built from differences between data values are unaffected by a common shift.

Solution

1. Adding the same constant to every value increases the mean and median by $273.15$. 2. Differences between values do not change. 3. Therefore, the IQR, range, and standard deviation remain unchanged.

Answer

Mean and median each increase by $273.15$; IQR, range, and standard deviation do not change.

Summary statistics for one quantitative variable 45

Using the median-of-halves method and the $1.5\cdot\text{IQR}$ rule, identify any potential outliers in $5,7,8,9,10,11,12,13,30$.

Hints

- Find the middle half first, then compare every extreme value with the resulting boundaries. - Use the stated quartile convention before calculating the IQR. - Form the lower and upper fences, then compare each extreme value using strict inequalities.

Solution

1. Excluding the overall median, $Q_1=7.5$ and $Q_3=12.5$. 2. The IQR is $5$. 3. The fences are $0$ and $20$. 4. Only $30>20$, so $30$ is a potential outlier.

Answer

The fences are $0$ and $20$. The value $30$ is a potential outlier.

Summary statistics for one quantitative variable 46

The grouped table summarizes commute times. <table><thead><tr><th>Time (minutes)</th><th>Frequency</th></tr></thead><tbody><tr><td>$[0,10)$</td><td>$4$</td></tr><tr><td>$[10,20)$</td><td>$9$</td></tr><tr><td>$[20,30)$</td><td>$5$</td></tr><tr><td>$[30,40)$</td><td>$2$</td></tr></tbody></table> Estimate the mean commute time using class midpoints. Round to one decimal place.

Hints

- Use each class midpoint as the representative value for that interval. - Multiply each midpoint by its frequency and add the products. - Divide by the total frequency; the result is an estimate because the exact values inside the classes are unknown.

Solution

1. The class midpoints are $5$, $15$, $25$, and $35$ minutes. 2. The weighted sum is $5\cdot4+15\cdot9+25\cdot5+35\cdot2=350$. 3. The total frequency is $4+9+5+2=20$. 4. The estimated mean is $\frac{350}{20}=17.5$ minutes.

Answer

The estimated mean commute time is $17.5\,\text{minutes}$.

Summary statistics for one quantitative variable 47

A variable $X$ has sample standard deviation $3.2$. A transformed variable is defined by $Y=4X-7$. Find the sample standard deviation of $Y$.

Hints

- Separate the effects of shifting and scaling. - A vertical shift changes location but not distances between observations. - Multiplication by a constant scales standard deviation by the constant’s absolute value.

Solution

1. Subtracting $7$ does not affect spread. 2. Multiplying by $4$ multiplies standard deviation by $4$. 3. The new standard deviation is $4\cdot3.2=12.8$.

Answer

The sample standard deviation of $Y$ is $12.8$.

Summary statistics for one quantitative variable 48

Can two data sets have the same median and IQR but different standard deviations? Explain without constructing a full example.

Hints

- Compare which observations influence each statistic. - Hold the central ordered positions fixed while imagining different values in the tails. - Ask which measure uses every distance from the mean.

Solution

1. The median and IQR depend only on central ordered positions. 2. Standard deviation uses every value and its distance from the mean. 3. Values outside the middle $50\%$ can differ while the median and quartiles remain fixed, producing different standard deviations.

Answer

Yes. The same middle $50\%$ can coexist with different tail values, which changes standard deviation but not the median or IQR.

Summary statistics for one quantitative variable 49

A temperature sensor used in a field study always reads $1.5\,\text{°C}$ above the true temperature. Explain the effect on estimates of mean temperature and temperature variability.

Hints

- Represent each recorded value as the true value plus a constant. - Consider how adding a constant changes center. - Consider whether distances between observations change.

Solution

1. Every recorded temperature equals the true value plus $1.5\,\text{°C}$. 2. The estimated mean is therefore biased upward by $1.5\,\text{°C}$. 3. Adding the same constant to every value does not change differences among values. 4. Measures of variability such as standard deviation are unchanged.

Answer

The mean temperature is biased upward by $1.5\,\text{°C}$. The variability is unchanged because the same constant is added to every reading.

Summary statistics for one quantitative variable 50

A noise monitor records a value only when sound exceeds $40\,\text{decibels}$; quieter periods are stored as missing. A researcher averages the recorded values to estimate the mean sound level over the entire day. Explain the bias.

Hints

- Identify which values can never appear in the dataset. - Compare the retained values with the full range of the response. - Decide how excluding one end of the distribution affects the mean.

Solution

1. All sound levels at or below $40\,\text{decibels}$ are excluded from the computed mean. 2. The missingness depends directly on the measured value, so the retained observations are systematically louder than the full set of periods. 3. The average of recorded values will tend to overestimate the true all-day mean sound level. 4. The study needs an instrument that records the full relevant range or a method that properly accounts for values below the detection threshold.

Answer

The monitor systematically omits quieter periods, so the recorded values are not representative of the entire day. Averaging only recorded values biases the estimated mean sound level upward.

Summary statistics for one quantitative variable 51

A salary survey records every salary above $\$200{,}000$ as $\$200{,}000$. A researcher uses the recorded numbers to estimate the mean salary. Will this make the mean too high, too low, or unchanged? Explain how this differs from rounding each salary to the nearest dollar.

Hints

- Compare each high salary's true value with what the file records. - Ask whether recording errors above the cutoff go upward or downward. - Think about why rounding to the nearest dollar behaves differently.

Solution

1. Salaries above $\$200{,}000$ are all recorded below their actual values. 2. This reduces the sum of recorded salaries and makes the estimated mean too low. 3. Ordinary rounding can move numbers either up or down; this rule moves only the largest salaries down.

Answer

The recorded mean will be too low if salaries exceed the cutoff. Unlike ordinary rounding, this recording rule always lowers those large values.

Summary statistics for one quantitative variable 52

A quantitative variable has the following frequency table. <table><thead><tr><th>Value</th><th>Frequency</th></tr></thead><tbody><tr><td>$1$</td><td>$2$</td></tr><tr><td>$2$</td><td>$5$</td></tr><tr><td>$3$</td><td>$4$</td></tr><tr><td>$4$</td><td>$1$</td></tr></tbody></table> Find the mean, median, and mode.

Hints

- Use frequencies as weights when finding the total. - Locate the two middle positions from cumulative frequencies. - The mode is tied to the greatest frequency, not the greatest value.

Solution

1. The sample size is $2+5+4+1=12$, and the total is $1\cdot2+2\cdot5+3\cdot4+4\cdot1=28$. 2. The mean is $\frac{28}{12}=\frac{7}{3}\approx2.33$. 3. The sixth and seventh ordered observations are both $2$, so the median is $2$. 4. The greatest frequency is $5$ at value $2$, so the mode is $2$.

Answer

Mean: $\frac{7}{3}\approx2.33$; median: $2$; mode: $2$.

Summary statistics for one quantitative variable 53

One group has $12$ observations with mean $18$. A second group has $8$ observations with mean $25$. Find the mean of all $20$ observations. Explain why averaging $18$ and $25$ directly would be incorrect.

Hints

- Convert each group mean into the corresponding group total. - Combine totals and sample sizes separately. - Equal weighting of group means is valid only for equal group sizes.

Solution

1. The first group total is $12\cdot18=216$, and the second group total is $8\cdot25=200$. 2. The combined total is $216+200=416$. 3. The combined mean is $\frac{416}{20}=20.8$. 4. The two group means cannot be averaged directly because the groups have different sizes.

Answer

The combined mean is $20.8$.

Summary statistics for one quantitative variable 54

Eight delivery times have deviations from $30$ minutes that sum to $-12$ minutes. a) Find the mean delivery time. b) One of the eight times was $38$ minutes. Find the mean of the other seven times, rounded to two decimal places.

Hints

- Add the signed deviations to the total you would have if each time were $30$ minutes. - Divide that reconstructed total by the number of deliveries. - After removing a delivery, both the total and the number of deliveries change.

Solution

1. The total time is $8\cdot30-12=228$ minutes. 2. The mean for all eight deliveries is $\frac{228}{8}=28.5$ minutes. 3. Removing the $38$-minute delivery leaves $228-38=190$ minutes. 4. The new mean is $\frac{190}{7}\approx27.14$ minutes.

Answer

a) $28.5$ minutes. b) Approximately $27.14$ minutes.

Summary statistics for one quantitative variable 55

A data set of $10$ scores has reported mean $50.4$. One score was entered as $83$ but should have been $38$. Find the corrected mean.

Hints

- Recover the original total from the reported mean. - Adjust the total by the difference between the incorrect and correct entries. - Keep the sample size unchanged.

Solution

1. The reported total is $10\cdot50.4=504$. 2. Correcting the entry reduces the total by $83-38=45$, giving $504-45=459$. 3. The corrected mean is $\frac{459}{10}=45.9$.

Answer

The corrected mean is $45.9$.

Summary statistics for one quantitative variable 56

A data set has mean $7$ and standard deviation $4$. Each observation is transformed using $y=-3x+10$. Find the mean and standard deviation of the transformed data.

Hints

- Apply the entire linear rule to the measure of center. - Standard deviation uses distances, so the sign of a scale factor does not make it negative. - A constant shift changes location but not spread.

Solution

1. The transformed mean is $-3\cdot7+10=-11$. 2. Multiplying by $-3$ multiplies all distances from the mean by $3$; adding $10$ does not change distances. 3. The transformed standard deviation is $3\cdot4=12$.

Answer

The transformed mean is $-11$, and the transformed standard deviation is $12$.

Summary statistics for one quantitative variable 57

A sample of $4$ observations has mean $10$, and the sum of squared deviations from the mean is $100$. A fifth observation equal to $10$ is added. Find the new mean and sample standard deviation.

Hints

- Consider how a value equal to the current mean changes the total. - Determine its contribution to squared deviation. - Update the sample degrees of freedom after increasing the sample size.

Solution

1. Adding a value equal to the original mean leaves the mean at $10$. 2. The added observation has deviation $0$, so the sum of squared deviations remains $100$. 3. With $5$ observations, the new sample variance is $\frac{100}{5-1}=25$. 4. The new sample standard deviation is $\sqrt{25}=5$.

Answer

The new mean is $10$, and the new sample standard deviation is $5$.

Summary statistics for one quantitative variable 58

For the data $2,4,6,8$, calculate both the population variance and standard deviation and the sample variance and standard deviation.

Hints

- The squared deviations are the same in both calculations. - The distinction is the divisor used for a population versus a sample. - Take the square root only after finding each variance.

Solution

1. The mean is $\frac{2+4+6+8}{4}=5$. 2. The squared deviations sum to $(2-5)^2+(4-5)^2+(6-5)^2+(8-5)^2=20$. 3. The population variance is $\frac{20}{4}=5$, and the population standard deviation is $\sqrt{5}\approx2.236$. 4. The sample variance is $\frac{20}{4-1}=\frac{20}{3}\approx6.667$, and the sample standard deviation is $\sqrt{\frac{20}{3}}\approx2.582$.

Answer

Population: variance $5$, standard deviation $\sqrt{5}\approx2.236$. Sample: variance $\frac{20}{3}\approx6.667$, standard deviation $\sqrt{\frac{20}{3}}\approx2.582$.

Summary statistics for one quantitative variable 59

Five deviations from a sample mean are $-4,-1,2,3,x$. Find $x$, then calculate the sample standard deviation.

Hints

- Use a defining property of deviations from the mean. - Find the missing deviation before calculating spread. - Apply the sample divisor to the squared deviations.

Solution

1. Deviations from the mean sum to $0$, so $-4-1+2+3+x=0$, giving $x=0$. 2. The sum of squared deviations is $16+1+4+9+0=30$. 3. The sample variance is $\frac{30}{5-1}=7.5$. 4. The sample standard deviation is $\sqrt{7.5}\approx2.739$.

Answer

$x=0$, and the sample standard deviation is $\sqrt{7.5}\approx2.739$.

Summary statistics for one quantitative variable 60

For the ordered data $1,2,3,4,5,6,7,8,9$, compute $Q_1$ and $Q_3$ using each convention: a) Exclude the overall median from both halves. b) Include the overall median in both halves. Explain why a reported five-number summary should identify its quartile convention when the distinction matters.

Hints

- Locate the overall median before forming the halves. - Follow each stated convention literally. - Compare the resulting lower- and upper-half medians.

Solution

1. The overall median is $5$. 2. Excluding the median gives lower half $1,2,3,4$ and upper half $6,7,8,9$, so $Q_1=2.5$ and $Q_3=7.5$. 3. Including the median gives lower half $1,2,3,4,5$ and upper half $5,6,7,8,9$, so $Q_1=3$ and $Q_3=7$. 4. Different accepted conventions can produce different quartiles for the same small data set, so the method should be stated.

Answer

a) $Q_1=2.5$, $Q_3=7.5$. b) $Q_1=3$, $Q_3=7$.

Summary statistics for one quantitative variable 61

Five deviations from the mean are $-3,-1,0,1,3$. The observation with deviation $-3$ equals $12$. Find the mean and all five observations.

Hints

- Translate the known observation and its deviation into an equation. - Once the center is known, each deviation locates one observation relative to it. - Verify that the recovered observations average to the stated center.

Solution

1. A deviation equals observation minus mean, so $12-\bar{x}=-3$. 2. Therefore $\bar{x}=15$. 3. Add each deviation to $15$ to obtain $12,14,15,16,18$.

Answer

The mean is $15$, and the observations are $12,14,15,16,18$.

Summary statistics for one quantitative variable 62

A sample has sum of squared deviations from its mean equal to $72$, and its sample standard deviation is $3$. Find the sample size.

Hints

- Square the standard deviation to obtain the variance. - Relate the variance to the given squared-deviation total. - Remember the sample degrees of freedom.

Solution

1. The sample variance is $s^2=3^2=9$. 2. Using $s^2=\frac{72}{n-1}$, solve $9=\frac{72}{n-1}$. 3. Then $n-1=8$, so $n=9$.

Answer

The sample size is $9$.

Summary statistics for one quantitative variable 63

Four paper airplanes flew different distances. The average distance of the other three was calculated after leaving out each flight in turn. Those four averages were $8,10,12,14$ feet. Find the average of all four flights and the four original distances.

Hints

- Think about how many of the four three-flight groups contain each original flight. - Turn a reported average into a total for that group. - Once you know the full total, how could you find the flight left out of one group?

Solution

1. In the four three-flight totals, each flight is counted three times. Dividing each total by three means the sum of the four reported averages equals the sum of the original distances. 2. The original total is $8+10+12+14=44$ feet, so the full average is $44\div4=11$ feet. 3. A reported average $m$ leaves out a flight of length $44-3m$. 4. The omitted distances are $20,14,8,2$ feet, so the original distances in increasing order are $2,8,14,20$ feet.

Answer

Overall mean $11$ feet; original lengths $2,8,14,20$ feet.

Summary statistics for one quantitative variable 64

Five walks have a mean time of $12$ minutes. The sum of their squared differences from $10$ minutes is $100$. Without knowing the five times, find their sample variance and sample standard deviation. Round the standard deviation to two decimal places.

Hints

- The squared differences were found from a reference other than the mean. - How far apart are that reference and the mean? - Think about what changes in the total squared differences when the reference moves to the mean.

Solution

1. The mean is $2$ minutes above the reference time of $10$. Changing the reference to the mean removes $5\cdot2^2=20$ from the squared-difference total. 2. The squared-difference total about the mean is $100-20=80$. 3. The sample variance is $s^2=\frac{80}{5-1}=20\,\text{min}^2$. 4. The sample standard deviation is $s=\sqrt{20}\approx4.47\,\text{min}$.

Answer

Sample variance $20\,\text{min}^2$; sample standard deviation about $4.47\,\text{min}$.

Summary statistics for one quantitative variable 65

A spreadsheet has four walking times. It shows $\sum x_i=20\,\text{min}$ and $\sum x_i^2=120\,\text{min}^2$, but not the individual times. Find the sample standard deviation. Round to two decimal places.

Hints

- What mean follows from the total and the number of times? - Can the total of squared differences from the mean be found from the two spreadsheet totals? - Remember which denominator is used for a sample standard deviation.

Solution

1. The mean is $\bar{x}=20\div4=5$ minutes. 2. Expanding the squared differences gives $\sum (x_i-\bar{x})^2=\sum x_i^2-4\bar{x}^2=120-4\cdot25=20\,\text{min}^2$. 3. The sample variance is $\frac{20}{4-1}=\frac{20}{3}\,\text{min}^2$. 4. The sample standard deviation is $\sqrt{\frac{20}{3}}\approx2.58\,\text{min}$.

Answer

Sample standard deviation $\sqrt{\frac{20}{3}}\approx2.58\,\text{min}$.

Summary statistics for one quantitative variable 66

A swim team's spreadsheet shows that five lap times total $100$ seconds and their squares total $2040\,\text{s}^2$. One time was entered as $18$ seconds but should be $20$ seconds. Find the corrected mean and sample standard deviation. Round the standard deviation to two decimal places.

Hints

- Which part of the original total changes when an entry is corrected? - How does a corrected value affect a sum of squares? - Use the new center when calculating the corrected spread.

Solution

1. Replace the wrong time in the total: $100-18+20=102$ seconds, so the corrected mean is $102\div5=20.4$ seconds. 2. Replace its square in the second total: $2040-18^2+20^2=2116\,\text{s}^2$. 3. The corrected sum of squared deviations is $2116-5(20.4)^2=35.2\,\text{s}^2$. 4. The sample variance is $35.2\div4=8.8\,\text{s}^2$, giving $s=\sqrt{8.8}\approx2.97$ seconds.

Answer

Corrected mean $20.4$ seconds; corrected sample standard deviation about $2.97$ seconds.

Summary statistics for one quantitative variable 67

Three members of the puzzle club took an average of $4$ minutes per puzzle, with a sample standard deviation of $1$ minute. Three other members averaged $8$ minutes, also with a sample standard deviation of $1$ minute. Find the mean and sample standard deviation for all six times together. Round the standard deviation to two decimal places.

Hints

- Find the mean for all six puzzle times using the sizes of the two groups. - How much spread comes from times differing within each group? - What changes when the group averages are measured from the shared center instead?

Solution

1. With equal group sizes, the combined mean is $\frac{3\cdot4+3\cdot8}{6}=6$ minutes. 2. Each group's squared-deviation total around its own mean is $(3-1)1^2=2\,\text{min}^2$, giving $4\,\text{min}^2$ within the groups. 3. Shifting each group center to the combined mean adds $3(4-6)^2+3(8-6)^2=24\,\text{min}^2$. 4. The combined sample standard deviation is $\sqrt{\frac{4+24}{6-1}}=\sqrt{5.6}\approx2.37$ minutes.

Answer

Combined mean $6$ minutes; combined sample standard deviation about $2.37$ minutes.

Summary statistics for one quantitative variable 68

A cyclist records speeds of $3,3,7,7$ miles per hour during four short rides. Treat these four speeds as the whole population. The root mean square (RMS) is the square root of the mean of their squared values. Find the population variance, population standard deviation, and RMS. Round the last two values to two decimal places.

Hints

- Find the mean speed before measuring deviations from it. - Use the population denominator for variance in this question. - Keep the order of squaring, averaging, and taking a square root straight.

Solution

1. The mean speed is $\frac{3+3+7+7}{4}=5$ miles per hour. 2. The mean squared distance from the mean is $\frac{(3-5)^2+(3-5)^2+(7-5)^2+(7-5)^2}{4}=4\,\text{(mi/h)}^2$. This is the population variance. 3. The population standard deviation is $\sqrt{4}=2.00$ miles per hour. 4. The mean of the squared speeds is $\frac{3^2+3^2+7^2+7^2}{4}=29\,\text{(mi/h)}^2$, so $\text{RMS}=\sqrt{29}\approx5.39$ miles per hour.

Answer

Population variance $4\,\text{(mi/h)}^2$; population standard deviation $2.00\,\text{mi/h}$; RMS $\sqrt{29}\approx5.39\,\text{mi/h}$.

Summary statistics for one quantitative variable 69

A sample contains the values $2,5,8$. A second sample is formed by recording each original value twice, giving $2,2,5,5,8,8$. Compare the mean, population standard deviation, and sample standard deviation of the two samples. Explain why the two standard-deviation results behave differently.

Hints

- Track both the sum of squared deviations and the number of observations. - Compare the denominators used for population and sample variability. - Repeating every observation does not change the relative-frequency pattern, but it does change the sample-size correction.

Solution

1. Both samples have mean $5$. 2. For $2,5,8$, the sum of squared deviations is $18$. The population standard deviation is $\sqrt{\frac{18}{3}}=\sqrt{6}$, and the sample standard deviation is $\sqrt{\frac{18}{2}}=3$. 3. Duplicating every value doubles the sum of squared deviations to $36$ and doubles the sample size to $6$. 4. The population standard deviation remains $\sqrt{\frac{36}{6}}=\sqrt{6}$. The sample standard deviation becomes $\sqrt{\frac{36}{5}}=\sqrt{7.2}\approx2.683$. 5. Population variance divides by $n$, so equal duplication leaves it unchanged. Sample variance divides by $n-1$, so duplication changes the correction factor.

Answer

Both means are $5$, and both population standard deviations are $\sqrt{6}\approx2.449$. The sample standard deviation changes from $3$ to $\sqrt{7.2}\approx2.683$ because the denominator changes from $n-1=2$ to $n-1=5$.

Summary statistics for one quantitative variable 70

A sample has $n=15$, mean $24$, and $\sum (x_i-20)^2=900$. Without reconstructing the individual observations, find the sample variance and sample standard deviation.

Hints

- Compare deviations from the stated reference value with deviations from the mean. - Account for the distance between the reference value and the mean across all observations. - Use the sample-size correction only after finding the squared deviations about the mean.

Solution

1. The mean is $24-20=4$ units above the reference value $20$. 2. Use $\sum (x_i-20)^2=\sum (x_i-\bar{x})^2+n(\bar{x}-20)^2$. 3. The sum of squared deviations from the sample mean is $900-15\cdot4^2=660$. 4. The sample variance is $s^2=\frac{660}{14}=\frac{330}{7}\approx47.143$. 5. The sample standard deviation is $s=\sqrt{\frac{330}{7}}\approx6.866$.

Answer

Sample variance: $\frac{330}{7}\approx47.143$ Sample standard deviation: $\sqrt{\frac{330}{7}}\approx6.866$

Summary statistics for one quantitative variable 71

The data are $4,7,10,13$. Form all six averages obtained by choosing two different observations at a time. Find the mean of those six pairwise averages, and explain why it equals the mean of the original four observations.

Hints

- List pair choices systematically so none are repeated or omitted. - Count how often each original observation appears among all pairs. - Compare the effective weight of each original value in the two means.

Solution

1. The six pairwise averages are $5.5,7,8.5,8.5,10,11.5$. 2. Their mean is $\frac{5.5+7+8.5+8.5+10+11.5}{6}=\frac{51}{6}=8.5$. 3. The original mean is $\frac{4+7+10+13}{4}=8.5$. 4. Each original observation appears in exactly three pairs. In the sum of all pairwise averages, each value therefore contributes $\frac{3}{2}$ times. Dividing by the six pairs gives the same weight, $\frac{1}{4}$, for each original observation.

Answer

The mean of the six pairwise averages is $8.5$, equal to the original mean. Each original observation receives equal total weight across all pairs.

Summary statistics for one quantitative variable 72

A frequency table has values $0,1,2,3$ with frequencies $2,4,x,1$. The mean is $1.5$. Find $x$ and identify the mode.

Hints

- Write both the total frequency and the weighted sum in terms of the unknown. - Use the definition of the mean to form an equation. - Recheck the largest frequency after solving.

Solution

1. The total frequency is $2+4+x+1=7+x$. 2. The weighted total is $0\cdot2+1\cdot4+2x+3\cdot1=7+2x$. 3. Set $\frac{7+2x}{7+x}=1.5$. Then $7+2x=10.5+1.5x$, so $x=7$. 4. Value $2$ has frequency $7$, the greatest frequency, so the mode is $2$.

Answer

$x=7$, and the mode is $2$.

Summary statistics for one quantitative variable 73

Twelve integer-valued observations are grouped as follows: $4$ are from $0$ through $9$, $6$ are from $10$ through $19$, and $2$ are from $20$ through $29$. Find the smallest and largest possible means consistent with this grouped information.

Hints

- To minimize the total, place every observation as low as its group permits. - To maximize the total, place every observation as high as its group permits. - Keep the frequency of each group fixed in both extreme cases.

Solution

1. The smallest total occurs when every observation is at the lower endpoint of its group: $4\cdot0+6\cdot10+2\cdot20=100$. 2. The smallest possible mean is $\frac{100}{12}=\frac{25}{3}\approx8.33$. 3. The largest total occurs when every observation is at the upper endpoint of its group: $4\cdot9+6\cdot19+2\cdot29=208$. 4. The largest possible mean is $\frac{208}{12}=\frac{52}{3}\approx17.33$.

Answer

Smallest possible mean: $\frac{25}{3}\approx8.33$ Largest possible mean: $\frac{52}{3}\approx17.33$

Summary statistics for one quantitative variable 74

A data set has $n=8$, mean $12$, and $\sum (x_i-12)^2=80$. Find $\sum (x_i-10)^2$ and $\sum (x_i-15)^2$. Which of the three reference values $10$, $12$, and $15$ gives the smallest total squared distance?

Hints

- Compare each proposed reference value with the mean. - Shifting the reference away from the mean adds the same squared offset for every observation. - Use the given squared-deviation total as the baseline.

Solution

1. For any reference value $c$, $\sum (x_i-c)^2=\sum (x_i-\bar{x})^2+n(\bar{x}-c)^2$. 2. For $c=10$, the total is $80+8(12-10)^2=80+32=112$. 3. For $c=15$, the total is $80+8(12-15)^2=80+72=152$. 4. The total at $c=12$ is the given $80$, which is smaller than $112$ and $152$. 5. The mean minimizes the sum of squared distances.

Answer

$\sum (x_i-10)^2=112$, and $\sum (x_i-15)^2=152$. The reference value $12$, the mean, gives the smallest total squared distance.

Summary statistics for one quantitative variable 75

Eight observations have mean $5$, and the mean of their squared values is $31$. Find the population variance, population standard deviation, and root mean square of the observations. Verify the relationship among these three quantities and the mean.

Hints

- Distinguish the mean of the squares from the square of the mean. - Population variance is the difference between those two quantities. - The root mean square is the square root of the stated mean square.

Solution

1. The mean square is $31$, and the square of the mean is $5^2=25$. 2. The population variance is $31-25=6$. 3. The population standard deviation is $\sqrt{6}\approx2.449$. 4. The root mean square is $\sqrt{31}\approx5.568$. 5. The relationship is $\text{RMS}^2=\mu^2+\sigma^2$, since $31=25+6$.

Answer

Population variance: $6$ Population standard deviation: $\sqrt{6}\approx2.449$ Root mean square: $\sqrt{31}\approx5.568$

Summary statistics for one quantitative variable 76

A data set has six observations. The means obtained after leaving out each observation, one at a time, are $12,13,14,15,16,17$. Find the mean of the complete data set and recover all six original observations.

Hints

- Express each leave-one-out mean using the full total and one omitted value. - Add all six equations and count how often each original observation appears. - Recover each omitted value from its corresponding reduced mean.

Solution

1. Let the full total be $S$. Each leave-one-out mean equals $\frac{S-x_i}{5}$. 2. Adding the six leave-one-out means counts the full total exactly once, so $12+13+14+15+16+17=S=87$. 3. The full mean is $\frac{87}{6}=14.5$. 4. For each leave-one-out mean $m_i$, the omitted observation is $x_i=87-5m_i$. 5. The omitted values are $27,22,17,12,7,2$, so the ordered data are $2,7,12,17,22,27$.

Answer

The complete mean is $14.5$. The six observations are $2,7,12,17,22,27$.

Summary statistics for one quantitative variable 77

A sample of $6$ observations has mean $20$ and sample standard deviation $4$. One observation was recorded as $18$ but should have been $24$. Find the corrected mean and corrected sample standard deviation without reconstructing the other five observations.

Hints

- Recover both the original sum and the original sum of squares from the given summaries. - A corrected observation changes its contribution to each total differently. - Recompute variability around the corrected mean rather than the original mean.

Solution

1. The original sum is $6\cdot20=120$. 2. From $s^2=\frac{\sum x_i^2-n\bar{x}^2}{n-1}$, the original sum of squares is $5\cdot4^2+6\cdot20^2=2480$. 3. Correcting the observation gives a new sum of $120-18+24=126$, so the corrected mean is $\frac{126}{6}=21$. 4. The corrected sum of squares is $2480-18^2+24^2=2732$. 5. The corrected sum of squared deviations is $2732-6\cdot21^2=86$. 6. The corrected sample standard deviation is $\sqrt{\frac{86}{5}}=\sqrt{17.2}\approx4.147$.

Answer

Corrected mean: $21$ Corrected sample standard deviation: $\sqrt{17.2}\approx4.147$

Summary statistics for one quantitative variable 78

Four observations are $2,4,x,y$. Their mean is $5$, and their sample variance is $\frac{20}{3}$. Given that $x\leq y$, find $x$ and $y$.

Hints

- Use the mean to obtain the sum of the two unknown observations. - Convert sample variance into a sum of squared deviations. - Relate the sum and sum of squares of two numbers to their product.

Solution

1. The mean condition gives $2+4+x+y=4\cdot5$, so $x+y=14$. 2. A sample variance of $\frac{20}{3}$ gives a sum of squared deviations of $3\cdot\frac{20}{3}=20$. 3. Using $\sum x_i^2=\sum (x_i-\bar{x})^2+n\bar{x}^2$, the sum of squares is $20+4\cdot5^2=120$. 4. Therefore, $2^2+4^2+x^2+y^2=120$, so $x^2+y^2=100$. 5. Since $(x+y)^2=x^2+y^2+2xy$, $14^2=100+2xy$, giving $xy=48$. 6. The numbers with sum $14$ and product $48$ are $6$ and $8$. Thus $x=6$ and $y=8$.

Answer

$x=6$ and $y=8$.

Summary statistics for one quantitative variable 79

Group 1 has $n_1=5$, mean $10$, and sample standard deviation $2$. Group 2 has $n_2=7$, mean $14$, and sample standard deviation $3$. Find the combined sample mean and sample standard deviation for all $12$ observations.

Hints

- Combine group totals to find the overall center. - Account for variation within each group and separation between group means. - Use the total sample degrees of freedom only after combining squared deviations.

Solution

1. The combined mean is $\bar{x}=\frac{5\cdot10+7\cdot14}{12}=\frac{37}{3}\approx12.333$. 2. The within-group squared-deviation total is $(5-1)2^2+(7-1)3^2=70$. 3. The between-group contribution is $5\left(10-\frac{37}{3}\right)^2+7\left(14-\frac{37}{3}\right)^2=\frac{140}{3}$. 4. The combined squared-deviation total is $70+\frac{140}{3}=\frac{350}{3}$. 5. The combined sample variance is $\frac{350/3}{12-1}=\frac{350}{33}\approx10.606$, so the sample standard deviation is $\sqrt{\frac{350}{33}}\approx3.257$.

Answer

The combined mean is $\frac{37}{3}\approx12.333$, and the combined sample standard deviation is $\sqrt{\frac{350}{33}}\approx3.257$.

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Summary statistics for one quantitative variable

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