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Standard scores and normal areas

Learning goals

  • Computes z-scores from observations, mean, and standard deviation.
  • Interprets a z-score as signed standard-deviation distance from the mean.
  • Uses normal areas or percentiles when a normal model is appropriate.
  • Converts between z-scores and original units.

Click problems to add them to your worksheet.

Standard scores and normal areas 1
A population has mean \(70\) and standard deviation \(8\). Find the z-score of \(86\) and interpret it.

Hints

- Compare the value’s deviation from the mean with the standard deviation. - Use \(z=\frac{x-\mu}{\sigma}\), then interpret the sign and magnitude in standard-deviation units.

Solution

1. \(z=\frac{86-70}{8}=2\). 2. The value \(86\) is \(2\) standard deviations above the population mean.

Answer

\(z=2\); the value is \(2\) standard deviations above the mean.
Standard scores and normal areas 2
The distribution of a measurement is approximately normal with mean \(50\) and standard deviation \(6\). Use the empirical rule to estimate the percentage between \(44\) and \(56\).

Hints

- Express each endpoint as a number of standard deviations from the mean. - Match the interval to the empirical-rule band centered at the mean.

Solution

1. The interval \([44,56]\) is within one standard deviation of the mean. 2. The empirical rule assigns approximately \(68\%\) to this interval.

Answer

Approximately \(68\%\).
Standard scores and normal areas 3
A standardized score has \(z=-0.8\). Is the original value above or below its distribution's mean? How far from the mean is it, in standard-deviation units?

Hints

- First focus on the sign of the z-score. - The magnitude tells you distance measured in standard deviations.

Solution

1. The negative sign shows the original value is below the mean. 2. The magnitude \(0.8\) means it is \(0.8\) standard deviations from the mean.

Answer

Below the mean by \(0.8\) standard deviations.
Standard scores and normal areas 4
The shaded region is under the standard normal curve. Which event does it represent: \(Z<1\), \(Z>1\), or \(-1<Z<1\)?
Figure for problem 563416

Hints

- Locate the boundary between shaded and unshaded areas. - Decide whether the shaded area lies to the left or right of that boundary.

Solution

1. The shaded region begins at \(1\) and extends to the right of that boundary. 2. It represents standard normal values greater than \(1\).

Answer

\(Z>1\).
Standard scores and normal areas 5
A population has mean \(120\) and standard deviation \(15\). Find the value with z-score \(-1.4\).

Hints

- Interpret the sign and magnitude of the z-score as a signed number of standard deviations from the mean. - Convert that standardized distance back to the original scale before combining it with the mean.

Solution

1. A z-score of \(-1.4\) means the value is \(1.4\) standard deviations below the mean. 2. The signed distance from the mean is \((-1.4)\cdot15=-21\). 3. Therefore, \(x=120-21=99\).

Answer

The value is \(99\).
Standard scores and normal areas 6
Student A scored \(78\) on a test with mean \(70\) and standard deviation \(4\). Student B scored \(640\) on a test with mean \(600\) and standard deviation \(50\). Who performed better relative to the test group?

Hints

- Put both performances on the same standardized scale. - The larger z-score represents the stronger performance relative to its own test distribution.

Solution

1. Student A: \(z_A=\frac{78-70}{4}=2\). 2. Student B: \(z_B=\frac{640-600}{50}=0.8\). 3. Since \(2>0.8\), Student A had the higher relative position.

Answer

Student A, with \(z=2\) compared with Student B’s \(z=0.8\).
Standard scores and normal areas 7
A normal distribution has mean \(100\) and standard deviation \(12\). Use the empirical rule to estimate the percentage below \(76\) or above \(124\).

Hints

- Express each endpoint as a number of standard deviations from the mean. - Choose the corresponding central empirical-rule percentage, then use its complement.

Solution

1. The endpoints are two standard deviations from the mean. 2. Approximately \(95\%\) lies within two standard deviations. 3. The percentage outside is \(100\%-95\%=5\%\).

Answer

Approximately \(5\%\).
Standard scores and normal areas 8
For an approximately normal distribution, estimate the percentage of observations between one and two standard deviations above the mean.

Hints

- Use nested central regions and symmetry. - Subtract the one-standard-deviation central area from the two-standard-deviation central area, then use symmetry.

Solution

1. Approximately \(95\%\) lies within two standard deviations and \(68\%\) within one. 2. The area between one and two standard deviations on both sides is \(95\%-68\%=27\%\). 3. By symmetry, the upper-side portion is \(\frac{27\%}{2}=13.5\%\).

Answer

Approximately \(13.5\%\).
Standard scores and normal areas 9
Scores are normally distributed with mean \(72\) and standard deviation \(10\). Find \(P(X<85)\). Round to four decimal places. Use a calculator or software with a normal CDF.

Hints

- Is the required probability to the left or to the right of the cutoff? - Express the cutoff as a number of standard deviations above or below the mean. - Look up the cumulative probability for that standardized position.

Solution

1. Standardize: \(z=\frac{85-72}{10}=1.30\). 2. The standard normal area to the left of \(1.30\) is approximately \(0.9032\).

Answer

\(P(X<85)\approx0.9032\).
Standard scores and normal areas 10
Package masses are normally distributed with mean \(500\,\text{g}\) and standard deviation \(40\,\text{g}\). Find the probability that a package exceeds \(560\,\text{g}\). Round to four decimal places. Use a normal CDF on a calculator or in software.

Hints

- Think about which side of the cutoff corresponds to “exceeds.” - Convert the cutoff to a z-score before using the normal distribution. - A normal CDF gives the area to the left; how can you find the area to the right?

Solution

1. Standardize: \(z=\frac{560-500}{40}=1.50\). 2. The upper-tail probability is \(1-\Phi(1.50)\approx0.0668\).

Answer

The probability is approximately \(0.0668\).
Standard scores and normal areas 11
A variable is normally distributed with mean \(30\) and standard deviation \(5\). Find \(P(24<X<36)\). Round to four decimal places. Use a normal CDF on a calculator or in software.

Hints

- Identify the lower and upper cutoffs for the requested interval. - Express both cutoffs in standard-deviation units. - Which two cumulative areas can you subtract to leave only the middle interval?

Solution

1. The standardized bounds are \(z=-1.20\) and \(z=1.20\). 2. Subtract left-tail areas: \(\Phi(1.20)-\Phi(-1.20)\approx0.7699\).

Answer

\(P(24<X<36)\approx0.7699\).
Standard scores and normal areas 12
A normal distribution has mean \(100\) and standard deviation \(15\). Find the \(90\)th percentile. Round to one decimal place. Use inverse normal on a calculator or in software.

Hints

- A percentile identifies the percentage of values below a cutoff. - Find the z-score that has the required cumulative probability. - Convert the standardized cutoff to the original distribution's scale.

Solution

1. Inverse normal with left-tail area \(0.90\) gives \(z\approx1.2816\). 2. Convert back to the original scale: \(x=100+1.2816\cdot15\approx119.2\).

Answer

The \(90\)th percentile is approximately \(119.2\).
Standard scores and normal areas 13
A variable is normally distributed with mean \(10\) and standard deviation \(2\). Use symmetry to find \(P(7<X<13)\) without looking up both endpoints separately. Round to four decimal places. Use a normal CDF on a calculator or in software.

Hints

- Compare each endpoint's distance from the mean. - How can symmetry let one cumulative probability replace two separate lookups? - Make sure your result represents the middle, not the tails.

Solution

1. The endpoints correspond to \(z=-1.5\) and \(z=1.5\). 2. By symmetry, the central area is \(2\Phi(1.5)-1\approx0.8664\).

Answer

\(P(7<X<13)\approx0.8664\).
Standard scores and normal areas 14
The value \(85\) is observed in two populations. Both have mean \(70\); Population A has standard deviation \(5\), and Population B has standard deviation \(10\). In which population is \(85\) more unusual?

Hints

- Compare distances from the means in standard-deviation units. - Compare absolute z-scores because unusualness depends on distance from the mean, not direction.

Solution

1. In A, \(z=\frac{15}{5}=3\). 2. In B, \(z=\frac{15}{10}=1.5\). 3. The larger absolute z-score occurs in A, so \(85\) is more unusual there.

Answer

Population A, because \(z=3\) there compared with \(z=1.5\) in Population B.
Standard scores and normal areas 15
A measurement is normally distributed with mean \(250\) and standard deviation \(30\). Find the cutoff exceeded by only the highest \(1\%\). Round to one decimal place. Use inverse normal on a calculator or in software.

Hints

- If only a small percentage is above the cutoff, what percentage is below it? - Use inverse normal for the appropriate cumulative probability. - Convert from the standard normal scale to the measurement scale.

Solution

1. The cutoff has a left-tail probability of \(0.99\). 2. Inverse normal gives \(z\approx2.3263\). 3. The cutoff is \(250+2.3263\cdot30\approx319.8\).

Answer

Approximately \(319.8\).
Standard scores and normal areas 16
Measurements are approximately normal with mean \(40\) and standard deviation \(3\). In \(1200\) observations, use the empirical rule to estimate how many lie outside \([31,49]\).

Hints

- Identify how many standard deviations the interval extends, then use the complementary percentage. - Use the empirical-rule percentage outside three standard deviations, then apply it to the sample size and round to a whole observation.

Solution

1. The interval endpoints are \(3\) standard deviations from the mean. 2. Approximately \(99.7\%\) lie inside, so \(0.3\%\) lie outside. 3. The expected outside count is \(0.003\cdot1200=3.6\), about \(4\).

Answer

About \(4\) observations.
Standard scores and normal areas 17
For a standard normal variable \(Z\), find \(P(-2<Z<1)\). Round to four decimal places. Use a normal CDF on a calculator or in software.

Hints

- Locate the two endpoints on the standard normal scale. - What does each cumulative probability count from the left? - Subtract the smaller left-tail area from the larger one.

Solution

1. The left-tail areas are \(\Phi(1)\approx0.84134\) and \(\Phi(-2)\approx0.02275\). 2. The area between them is \(0.84134-0.02275\approx0.8186\).

Answer

\(P(-2<Z<1)\approx0.8186\).
Standard scores and normal areas 18
A value \(30\) has z-score \(-1.5\) in a population with mean \(42\). Find the population standard deviation.

Hints

- Use the standardized-score equation and solve for the scale parameter. - Substitute the known mean, value, and z-score into the formula, then solve for the positive standard deviation.

Solution

1. Use \(-1.5=\frac{30-42}{\sigma}\). 2. Then \(-1.5=\frac{-12}{\sigma}\), so \(\sigma=8\).

Answer

\(\sigma=8\).
Standard scores and normal areas 19
Cheikh standardizes a score of \(82\) from a distribution with mean \(70\) and standard deviation \(6\) by writing \(z=82-70=12\). Identify the error, give the correct standard score, and explain why a standard score has no measurement unit.

Hints

- Distinguish a raw difference from a standardized difference. - Ask what quantity sets the scale for comparing deviations. - Check what happens to the units when one measured difference is divided by another.

Solution

1. Cheikh found the raw deviation but did not express it in standard-deviation units. 2. The correct calculation is \(z=\frac{82-70}{6}=2\). 3. Both the numerator and denominator are measured in score points, so their units cancel.

Answer

Cheikh forgot to divide by the standard deviation. The correct standard score is \(z=2\), and it is unitless because the score units cancel in the quotient.
Standard scores and normal areas 20
A distribution is strongly bimodal, with mean \(50\) and standard deviation \(10\). Vuk uses the empirical rule to claim that about \(68\%\) of observations lie between \(40\) and \(60\). Evaluate the claim.

Hints

- Identify the shape condition required by the stated percentage rule. - Separate what the mean and standard deviation locate from what they reveal about area. - Decide whether the distribution's described shape meets the rule's assumptions.

Solution

1. The interval \([40,60]\) is one standard deviation on either side of the mean. 2. The \(68\%\) empirical-rule estimate requires an approximately bell-shaped distribution. 3. A strongly bimodal distribution does not satisfy that shape condition, so the percentage cannot be inferred from the mean and standard deviation alone.

Answer

Vuk's claim is not justified. Although \([40,60]\) is within one standard deviation of the mean, the empirical rule does not apply to a strongly bimodal distribution.
Standard scores and normal areas 21
In a normal distribution with mean \(50\), the value \(34\) is at the \(12\)th percentile. Find the raw value at the \(88\)th percentile without using an inverse normal calculator.

Hints

- Where is the middle of a normal curve? - How are the \(12\)th and \(88\)th percentiles placed relative to the middle? - How could their locations help you find the missing value without a calculator?

Solution

1. The \(12\)th and \(88\)th percentiles are symmetric since their percentages add to \(100\%\). 2. The value \(34\) is \(16\) units below the mean. 3. The matching value is \(16\) units above the mean, or \(66\).

Answer

The \(88\)th percentile is \(66\).
Standard scores and normal areas 22
A lab flags a measurement if its lower-tail probability under the standard normal model is below \(1\%\). Should the marked z-score be flagged? Give its lower-tail probability to four decimal places and as a percentage to two decimal places. Use a normal CDF on a calculator or in software.
Figure for problem 548862

Hints

- Identify the z-score marked on the horizontal axis. - A lower-tail probability means area to the left of the score. - Compare that area with the lab's threshold, using matching units.

Solution

1. The marked value has \(z=-2.4\), so its lower-tail probability is \(\Phi(-2.4)\approx0.0082\). 2. As a percentage, this is about \(0.82\%\). 3. Since \(0.82\%<1\%\), the lab should flag the measurement.

Answer

Yes. Its lower-tail probability is about \(0.0082\), or \(0.82\%\), below the \(1\%\) cutoff.
Standard scores and normal areas 23
A school reports standardized test results using \(T=50+10z\), where \(z\) follows the standard normal distribution. Camilo receives \(T=68\). Find his z-score and percentile rank to two decimal places. Use a normal CDF on a calculator or in software.

Hints

- First undo the transformation that turned a z-score into a reported score. - Then use the standard normal model to find the percentage below that z-score. - Check whether a positive z-score should have a percentile rank above or below \(50\%\).

Solution

1. Solve \(68=50+10z\), giving \(z=1.8\). 2. The standard normal cumulative probability at \(1.8\) is approximately \(0.9641\). 3. This is a percentile rank of approximately \(96.41\%\).

Answer

\(z=1.8\); percentile rank approximately \(96.41\%\).
Standard scores and normal areas 24
The shaded tail contains \(10\%\) of a normal distribution. What percentile is marked by cutoff \(c\)? What percentage of observations are below \(c\)?
Figure for problem 563418

Hints

- A percentile tells you the share at or below a cutoff. - Check which side of the cutoff is shaded. - The two sides of the curve together make the whole distribution.

Solution

1. The shaded \(10\%\) is to the right of the cutoff. 2. The area to the left of \(c\) is \(100\%-10\%=90\%\). 3. The cutoff is therefore the \(90\)th percentile, with \(90\%\) of observations below it.

Answer

\(c\) is the \(90\)th percentile; \(90\%\) lie below it.
Standard scores and normal areas 25
The two graphs shade upper tails of a standard normal distribution. Use a normal calculator to find each shaded probability to four decimal places. Which is larger, and why?
Figure for problem 563419

Hints

- Which z-score is at each boundary of the shaded region? - Does a normal calculator return the area above or below a cutoff by default? - Compare which numbers are counted in the two right-hand tails.

Solution

1. Graph a) shades \(Z>1\), giving \(P(Z>1)\approx0.1587\). 2. Graph b) shades \(Z>2\), giving \(P(Z>2)\approx0.0228\). 3. Graph a) has more area because its cutoff is farther left and includes more possible values.

Answer

a) \(0.1587\) b) \(0.0228\) Graph a) has the larger probability because its cutoff is farther left, so more values are included in the right tail.
Standard scores and normal areas 26
A normal distribution has mean \(64\) and standard deviation \(7\). Find the interval containing the middle \(80\%\). Round both endpoints to one decimal place. Use inverse normal on a calculator or in software.

Hints

- Split the probability left outside the middle interval equally between the tails. - Find the lower and upper cumulative probabilities before using inverse normal. - Convert both z-scores back to the original scale and check symmetry around the mean.

Solution

1. The middle \(80\%\) leaves \(10\%\) in each tail. 2. Inverse normal gives standardized boundaries \(z\approx-1.2816\) and \(z\approx1.2816\). 3. Convert to the original scale: \(64-1.2816\cdot7\approx55.0\) and \(64+1.2816\cdot7\approx73.0\).

Answer

Approximately \([55.0, 73.0]\).
Standard scores and normal areas 27
Test scores are normally distributed with mean \(50\) and standard deviation \(8\). Find the cutoff for the lowest \(5\%\), rounded to one decimal place. In a group of \(800\) students, about how many scores are below that cutoff? Use inverse normal on a calculator or in software.

Hints

- Interpret “lowest” as a percentage below a cutoff. - Use inverse normal to locate that cutoff relative to the mean. - Apply the same probability to the group size to estimate the count.

Solution

1. Inverse normal for left-tail probability \(0.05\) gives \(z\approx-1.6449\). 2. The cutoff is \(50+(-1.6449)\cdot8\approx36.8\). 3. About \(0.05\cdot800=40\) scores are below the cutoff.

Answer

Cutoff: approximately \(36.8\). Expected count: about \(40\).
Standard scores and normal areas 28
A value of \(66\) comes from a distribution with standard deviation \(8\). The distribution's mean changes while the standard deviation stays \(8\). What new mean would make \(66\) have z-score \(1.25\)?

Hints

- Treat the mean as the unknown while the raw value and standard deviation remain fixed. - Interpret the positive z-score as a distance above the new mean in standard-deviation units. - Check whether your mean lies below \(66\), as a positive z-score requires.

Solution

1. A z-score of \(1.25\) means \(66\) is \(1.25\) standard deviations above the new mean. 2. That distance is \(1.25\cdot8=10\). 3. Therefore, the new mean is \(66-10=56\).

Answer

The new mean must be \(56\).
Standard scores and normal areas 29
For a standard normal variable \(Z\), compare the probabilities of the two equal-width intervals \([-1,1]\) and \([0,2]\). Explain why the interval centered at the mean contains more probability. Use a standard normal table or calculator.
Figure for problem 548843

Hints

- Both intervals have the same numerical width. - Compare their locations relative to the normal curve's peak. - Calculate each area from standard normal cumulative probabilities.

Solution

1. \(P(-1\leq Z\leq1)=\Phi(1)-\Phi(-1)\approx0.6827\). 2. \(P(0\leq Z\leq2)=\Phi(2)-\Phi(0)\approx0.4772\). 3. The first interval contains about \(0.2054\) more probability. 4. A normal density is highest at the mean and decreases as distance from the mean increases. Among these equal-width intervals, the centered interval covers more of the high-density region.

Answer

\(P(-1\leq Z\leq1)\approx0.6827\), while \(P(0\leq Z\leq2)\approx0.4772\). The centered interval contains about \(0.2054\) more probability.
Standard scores and normal areas 30
A variable is modeled by a normal distribution with mean \(50\) and standard deviation \(8\). The marked lines show a report's claimed first and third quartiles. Find the model's quartiles to two decimal places and explain whether the report is consistent with the model. Use inverse normal on a calculator or in software.
Figure for problem 548844

Hints

- Quartiles split off the lower and upper quarters of a distribution. - Locate the two quartile z-scores with inverse normal. - Convert each z-score back to the measurement scale and compare with the marked lines.

Solution

1. The first and third quartiles of a normal distribution have z-scores about \(-0.6745\) and \(0.6745\). 2. The modeled quartiles are \(50-0.6745\cdot8\approx44.60\) and \(50+0.6745\cdot8\approx55.40\). 3. The marked quartiles, \(43\) and \(59\), do not match the model and are not equally far from the mean.

Answer

The modeled quartiles are approximately \(Q_1=44.60\) and \(Q_3=55.40\). The report's markings are inconsistent with the normal model.
Standard scores and normal areas 31
Rod lengths are normally distributed with mean \(70\,\text{mm}\) and standard deviation \(3\,\text{mm}\). Each length is recorded to the nearest whole millimeter. Find the probability that a rod is recorded as \(72\,\text{mm}\). Round to four decimal places. Use a normal CDF on a calculator or in software.

Hints

- Think about which true lengths round to the stated recorded value. - Convert the lower and upper true-length limits to standard scores. - A probability between two values is the difference of their cumulative areas.

Solution

1. A recorded length of \(72\,\text{mm}\) corresponds to actual lengths from \(71.5\,\text{mm}\) up to but not including \(72.5\,\text{mm}\). 2. The standardized endpoints are \(0.5\) and approximately \(0.8333\). 3. The probability is \(\Phi(0.8333)-\Phi(0.5)\approx0.1062\).

Answer

Approximately \(0.1062\).
Standard scores and normal areas 32
A value of \(117\) comes from a normal distribution with mean \(100\) and standard deviation \(15\). Use the normal CDF on a calculator or in software to find its percentile rank with the unrounded z-score and with the z-score rounded to one decimal place. How much does rounding the z-score change the percentile rank? Give both ranks to three decimal places and the difference to two decimal places, using unrounded calculator results for the subtraction.

Hints

- Calculate the z-score before deciding how many digits to keep. - Find a left-tail normal probability with each version of the z-score. - Which values should you subtract if you want the rounding error itself rather than the difference between two displayed approximations?

Solution

1. The unrounded z-score is \(\frac{117-100}{15}=\frac{17}{15}\approx1.133333\). Its normal cumulative probability gives percentile rank \(87.146285\ldots\%\), or \(87.146\%\) to three decimal places. 2. Rounding the z-score first gives \(z=1.1\), with percentile rank \(86.433394\ldots\%\), or \(86.433\%\) to three decimal places. 3. Subtract the unrounded ranks: \(87.146285\ldots-86.433394\ldots\approx0.712891\) percentage points, which rounds to \(0.71\) percentage points. Rounding the z-score early lowers the rank.

Answer

Unrounded z-score: \(87.146\%\). Rounded z-score: \(86.433\%\). Change: approximately \(0.71\) percentage points lower after rounding the z-score.
Standard scores and normal areas 33
A store models the number of customer complaints per day with a normal distribution having mean \(3\) and standard deviation \(2\). Counts cannot be negative. Find the model's probability of a negative count, to four decimal places, and explain why it is a concern. Use a normal CDF on a calculator or in software.
Figure for problem 548851

Hints

- Identify which values the count can actually take. - Look at the area to the left of zero on the model's curve. - Does a positive modeled probability of an impossible value suggest a good model?

Solution

1. At zero, \(z=\frac{0-3}{2}=-1.5\). 2. The normal model assigns \(P(X<0)=\Phi(-1.5)\approx0.0668\) to negative counts. 3. This gives positive probability to impossible outcomes, so the model is questionable near zero.

Answer

Approximately \(0.0668\), or \(6.68\%\), for impossible negative counts. The normal model is a poor fit near the lower limit.
Standard scores and normal areas 34
A report claims that the complete population of four observations has standard scores \(-1,-0.5,0.5,1\), calculated using the population mean and population standard deviation. Determine whether this set of standard scores is possible. Justify the conclusion using both the mean and the population standard deviation of the listed scores.

Hints

- A complete standardized population must satisfy two summary conditions. - Check the average before checking the average squared distance from zero. - Passing the zero-mean condition alone is not sufficient.

Solution

1. The mean of the listed standard scores is \(\frac{-1-0.5+0.5+1}{4}=0\), which satisfies one required property. 2. Their population variance is \(\frac{(-1)^2+(-0.5)^2+(0.5)^2+1^2}{4}=\frac{2.5}{4}=0.625\). 3. Their population standard deviation is \(\sqrt{0.625}\approx0.791\), not \(1\). 4. A complete population standardized with its own population standard deviation must have mean \(0\) and population standard deviation \(1\). Therefore the report is impossible as stated.

Answer

The set is not possible as the complete population of z-scores. Its mean is \(0\), but its population standard deviation is \(\sqrt{0.625}\approx0.791\), not \(1\).
Standard scores and normal areas 35
The total mass of packages shipped each day is modeled by a normal distribution with mean \(200\,\text{kg}\) and standard deviation \(30\,\text{kg}\). A day's total is in the target range if it is between \(170\,\text{kg}\) and \(245\,\text{kg}\). Find the percentages within, below, and above the target range. Round each to two decimal places. Use a normal CDF on a calculator or in software.
Figure for problem 548856

Hints

- Divide the normal curve into three regions at the two output limits. - Standardize each limit to use a normal CDF. - The region between the limits is the difference of their cumulative areas. - Check whether your three percentages add to about \(100\%\).

Solution

1. Standardize the limits: \(z_L=\frac{170-200}{30}=-1\) and \(z_U=\frac{245-200}{30}=1.5\). 2. The acceptable percentage is \(100(\Phi(1.5)-\Phi(-1))\%\approx77.45\%\). 3. The percentage below the limit is approximately \(15.87\%\). 4. The percentage above the limit is approximately \(6.68\%\).

Answer

Acceptable: approximately \(77.45\%\). Below: approximately \(15.87\%\). Above: approximately \(6.68\%\).
Standard scores and normal areas 36
A juice bottling line (A) fills small bottles with mean \(50\,\text{mL}\) and standard deviation \(5\,\text{mL}\). An iced-tea bottling line (B) has mean \(55\,\text{mL}\) and standard deviation \(10\,\text{mL}\). Both fill volumes are normally distributed. Using the cutoff marked on the graphs, find the percentage above it for each drink and the difference in percentage points. Round the percentages and the difference to two decimal places, using unrounded probabilities for the difference. Use a normal CDF on a calculator or in software.
Figure for problem 548858

Hints

- Compare each model with the same marked volume cutoff. - The two bottling lines have different means and standard deviations, so compare them on a standard scale. - Subtract the unrounded tail probabilities before expressing the gap in percentage points.

Solution

1. The marked cutoff is \(60\,\text{mL}\). For A, \(z_A=\frac{60-50}{5}=2\), giving about \(2.28\%\) above it. 2. For B, \(z_B=\frac{60-55}{10}=0.5\), giving about \(30.85\%\) above it. 3. From unrounded probabilities, B's proportion is about \(28.58\) percentage points higher.

Answer

A: approximately \(2.28\%\). B: approximately \(30.85\%\). Difference: approximately \(28.58\) percentage points, with B higher.
Standard scores and normal areas 37
In a normal distribution, \(P(X<42)=P(X>68)\). Find the mean. Explain why the standard deviation is not needed.

Hints

- Use the symmetry of a normal distribution around its mean. - Ask where the center must lie between cutoffs with matching opposite-tail areas. - Separate the role of the center from the role of the spread.

Solution

1. Equal probabilities in opposite tails place the two cutoffs symmetrically about the mean. 2. The mean is the midpoint of the cutoffs: \(\mu=\frac{42+68}{2}=55\). 3. The standard deviation controls how large the equal tail probabilities are, but symmetry alone determines their center.

Answer

The mean is \(55\). The standard deviation is unnecessary because equal opposite-tail probabilities identify symmetric cutoffs.
Standard scores and normal areas 38
Temperatures are normally distributed with mean \(20\,{}^\circ\text{C}\) and standard deviation \(4\,{}^\circ\text{C}\). A safety interval is \(14\,{}^\circ\text{C}\) to \(27\,{}^\circ\text{C}\). Convert the distribution parameters and safety interval to degrees Fahrenheit, then verify that the standardized endpoints are unchanged.

Hints

- A shifted-and-scaled conversion affects the center and spread differently. - Convert both interval endpoints using the same temperature rule. - Standardize in each unit system to compare relative positions.

Solution

1. Convert the mean: \(1.8\cdot20+32=68\,{}^\circ\text{F}\). 2. Convert the standard deviation by the scale factor only: \(1.8\cdot4=7.2\,{}^\circ\text{F}\). 3. Convert the endpoints: \(14\,{}^\circ\text{C}=57.2\,{}^\circ\text{F}\) and \(27\,{}^\circ\text{C}=80.6\,{}^\circ\text{F}\). 4. In Celsius, the endpoint standard scores are \(\frac{14-20}{4}=-1.5\) and \(\frac{27-20}{4}=1.75\). 5. In Fahrenheit, they are \(\frac{57.2-68}{7.2}=-1.5\) and \(\frac{80.6-68}{7.2}=1.75\).

Answer

The Fahrenheit model has mean \(68\,{}^\circ\text{F}\) and standard deviation \(7.2\,{}^\circ\text{F}\). The safety interval is \([57.2,80.6]\,{}^\circ\text{F}\), and its endpoint standard scores remain \(-1.5\) and \(1.75\).
Standard scores and normal areas 39
A measurement \(X\) is normally distributed with mean \(65\) and standard deviation \(10\). Find the probability that a random measurement is closer to the lower marked value than to the higher one. Round to four decimal places. Use a normal CDF on a calculator or in software.
Figure for problem 548861

Hints

- Where on the number line are the distances to the two marked values equal? - Decide which side of that midpoint is closer to the lower value. - Convert the midpoint into a z-score and find the corresponding tail area.

Solution

1. The marked values are \(60\) and \(80\); their midpoint is \(70\). 2. A measurement below \(70\) is closer to \(60\); a measurement above \(70\) is closer to \(80\). 3. Standardize the midpoint: \(z=\frac{70-65}{10}=0.5\). 4. Thus \(P(X<70)=\Phi(0.5)\approx0.6915\).

Answer

Approximately \(0.6915\).
Standard scores and normal areas 40
Component masses are normally distributed with mean \(300\,\text{g}\) and standard deviation \(25\,\text{g}\). A system flags components below a whole-number mass cutoff and must flag at most \(2\%\). Which is the largest allowed cutoff: \(248\,\text{g}\) or \(249\,\text{g}\)? Give the percentage flagged at each cutoff to two decimal places. Use a normal CDF on a calculator or in software.

Hints

- Is the flagged region to the left or right of each cutoff? - A cutoff further to the right includes more observations on which side of the curve? - Check the proposed cutoff percentages against the allowed upper bound.

Solution

1. At \(248\,\text{g}\), \(z=\frac{248-300}{25}=-2.08\), so approximately \(1.88\%\) are below the cutoff. 2. At \(249\,\text{g}\), \(z=\frac{249-300}{25}=-2.04\), so approximately \(2.07\%\) are below it. 3. The higher cutoff exceeds the \(2\%\) limit, so choose \(248\,\text{g}\).

Answer

Use \(248\,\text{g}\); it flags about \(1.88\%\). The \(249\,\text{g}\) cutoff would flag about \(2.07\%\).
Standard scores and normal areas 41
The empirical rule estimates that \(95\%\) of a bell-shaped distribution lies within \(2\) standard deviations of the mean. For an exactly normal distribution, find that percentage using a normal CDF. How many percentage points does it differ from \(95\%\)? Round both results to two decimal places.
Figure for problem 548866

Hints

- What part of the normal curve does the shaded central band cover? - If your calculator gives the area to the left of a cutoff, how can you find an area between two cutoffs? - Should the exact normal-model result be close to the empirical-rule estimate?

Solution

1. The corresponding normal probability is \(P(-2<Z<2)\). 2. Using standard normal areas, this probability is approximately \(0.9545=95.45\%\). 3. The difference from the empirical-rule estimate is approximately \(95.45-95.00=0.45\) percentage points.

Answer

Exactly normal: approximately \(95.45\%\); difference: about \(0.45\) percentage points higher than the empirical-rule estimate.
Standard scores and normal areas 42
In one distribution, \(x=54\) has z-score \(-1\), and \(x=78\) has z-score \(2\). Find the population mean and standard deviation.

Hints

- Use the difference between two standardized positions to recover the scale. - Write one z-score equation for each given x-value. - Subtract the equations to eliminate the mean, then back-substitute to recover the center.

Solution

1. The x-value difference \(24\) represents \(3\) standard deviations. 2. \(\sigma=\frac{24}{3}=8\). 3. Using \(54=\mu-\sigma\), \(\mu=54+8=62\).

Answer

\(\mu=62\) and \(\sigma=8\).
Standard scores and normal areas 43
Distribution A is normal with mean \(60\) and standard deviation \(10\). Distribution B is normal with mean \(70\) and standard deviation \(5\). Find the raw value that has the same percentile rank in both distributions, and give that percentile rank. Use a standard normal table or calculator.

Hints

- Equal percentile positions correspond to the same location on the standardized normal scale. - Express the unknown raw value relative to each distribution's center and spread. - After finding the value, use either distribution to identify its percentile.

Solution

1. Equal percentile ranks in normal distributions require equal standard scores. 2. Set \(\frac{x-60}{10}=\frac{x-70}{5}\). 3. Solving gives \(x=80\). 4. The common standard score is \(z=\frac{80-60}{10}=2\), whose percentile rank is approximately \(97.72\%\).

Answer

The raw value is \(80\), with a percentile rank of approximately \(97.72\%\) in both distributions.
Standard scores and normal areas 44
A bakery models the weights of its cookie bags with a normal distribution having standard deviation \(4\,\text{g}\). It wants \(90\%\) of the bags to weigh more than \(70\,\text{g}\). What mean weight is needed? Round to two decimal places. Use inverse normal on a calculator or in software.

Hints

- First translate the percentage above the cutoff into a percentile below it. - Decide whether the cutoff lies above or below the unknown mean. - Convert the percentile's z-score into a distance in grams.

Solution

1. If \(90\%\) weigh more than \(70\,\text{g}\), then \(70\,\text{g}\) is the \(10\)th percentile. 2. The corresponding z-score is approximately \(-1.2816\). 3. Solve \(70=\mu-1.2816\cdot4\) to get \(\mu\approx75.13\,\text{g}\).

Answer

The mean must be approximately \(75.13\,\text{g}\).
Standard scores and normal areas 45
The weights of bread rolls are normally distributed with mean \(100\,\text{g}\). At least \(90\%\) must weigh between \(85\,\text{g}\) and \(115\,\text{g}\). What is the greatest allowed standard deviation? Round to two decimal places. Use inverse normal on a calculator or in software.

Hints

- The target interval has the mean at its midpoint. - Divide the percentage outside the interval equally between the tails. - Relate the resulting standard-normal cutoff to the fixed distance from the mean.

Solution

1. The weight interval extends \(15\,\text{g}\) on either side of the mean. 2. The middle \(90\%\) of a standard normal distribution lies between z-scores of approximately \(-1.6449\) and \(1.6449\). 3. At the largest allowed spread, \(\frac{15}{\sigma}=1.6449\). Hence \(\sigma\approx9.12\,\text{g}\).

Answer

The greatest allowed standard deviation is approximately \(9.12\,\text{g}\).
Standard scores and normal areas 46
In a normal distribution, the \(10\)th percentile is \(40\) and the \(90\)th percentile is \(60\). Find the mean exactly and the standard deviation to two decimal places. Use inverse normal on a calculator or in software.

Hints

- Think about where symmetric percentiles fall around the center of a normal curve. - After finding the center, use one percentile to determine the scale. - Convert the percentile to a standardized distance before solving for the standard deviation.

Solution

1. The two percentiles are symmetric about the mean, so \(\mu=\frac{40+60}{2}=50\). 2. The z-score for the \(90\)th percentile is approximately \(1.2816\). 3. Solve \(60=50+1.2816\sigma\) to obtain \(\sigma\approx7.80\).

Answer

Mean \(\mu=50\); standard deviation \(\sigma\approx7.80\).
Standard scores and normal areas 47
In a normal distribution, the values \(72\) and \(108\) have the same density. The combined probability below \(72\) or above \(108\) is \(0.1336\). Find the mean exactly and the standard deviation to two decimal places. Use inverse normal on a calculator or in software.

Hints

- A normal curve has equal density at points the same distance from the center. - Share the total outside probability equally between the two tails. - Compare the raw distance from the mean with the corresponding z-score distance.

Solution

1. Equal densities at different values on a normal curve mean the values are equally far from its mean, so \(\mu=\frac{72+108}{2}=90\). 2. The two equal tail probabilities are \(0.0668\) each. 3. Inverse normal gives a distance of approximately \(1.5\) standard deviations from the mean. 4. The raw distance is \(18\), so \(\sigma\approx\frac{18}{1.5}\approx12.00\).

Answer

Mean \(\mu=90\); standard deviation \(\sigma\approx12.00\).
Standard scores and normal areas 48
A normal variable \(X\) has mean \(50\) and standard deviation \(10\). Define \(Y=100-X\). Find the z-scores and percentile ranks, to two decimal places, of the observations marked on the two curves. Explain why their percentile ranks differ. Use a normal CDF on a calculator or in software.
Figure for problem 548867

Hints

- Read each marked value from its curve and standardize using the correct model. - How do the mean and standard deviation change when you subtract a variable from a constant? - Consider what happens to the order of two observations after this transformation.

Solution

1. The marked value on the first curve is \(X=65\), with \(z_X=\frac{65-50}{10}=1.5\) and percentile rank approximately \(93.32\%\). 2. The transformation gives \(Y=35\), mean \(50\), and standard deviation \(10\). 3. Thus \(z_Y=\frac{35-50}{10}=-1.5\), with percentile rank about \(6.68\%\). 4. Because \(Y=100-X\) reverses the order of values, corresponding percentile ranks add to \(100\%\).

Answer

\(X\): \(z=1.5\), rank approximately \(93.32\%\). \(Y\): \(z=-1.5\), rank approximately \(6.68\%\). The transformation reverses rank order.

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