What does Percentiles and z-Scores measure or decide?

A z-score reports how many standard deviations a value lies from the mean.

When should I use Percentiles and z-Scores?

Standardize with z=(x−μ)/σ and interpret sign and magnitude.

Which conditions or design checks matter for Percentiles and z-Scores?

Confirm that the value, mean, and standard deviation use the same scale and units; a z-score is meaningful only relative to the stated distribution.

Which formula or calculator output matters most for Percentiles and z-Scores?

Use z=(x−μ)/σ and report the direction and distance from the mean.

How should results from Percentiles and z-Scores be interpreted?

Name the population, process, groups, variable, and units from the prompt. Then explain what the calculated value or design result means without claiming causation or generalization beyond what the study supports.

What mistake most often changes the answer in Percentiles and z-Scores?

The most damaging error is solving a nearby question with a different denominator, group order, tail, parameter, standard error, or scope. Use this correction rule: Standardize with z=(x−μ)/σ and interpret sign and magnitude.

Are the questions on this Percentiles and z-Scores page copied from past AP exams?

No. Released AP Statistics materials are used to identify authentic skills and response demands. Every Percentiles and z-Scores practice item here is an original parallel problem with changed data, context, target, and solution.

Is Percentiles and z-Scores required for the May 2027 AP Statistics exam?

Yes. Percentiles and z-Scores is included in the current five-unit course and is linked here to Unit 1 practice, formulas, MCQs, and FRQs.

Percentiles and Z-Scores: Formula, Meaning, and Examples

AP Statistics Practice

Percentiles and Z-Scores: Formula, Meaning, and Examples

Percentiles and Z-Scores: Formula, Meaning, and Examples practice bank: Solve the visible multiple-choice and free-response questions, then compare every step with the worked answers.

MCQs36

FRQ sets14

AnswersVisible

Topic links73 pages

Percentiles and Z-Scores: Formula, Meaning, and Examples: formulas and targets

Standard score	z = (x−μ)σ
Reverse standardization	x = μ + zσ

Percentiles and z-Scores multiple-choice practice

Question 1. Percentiles and z-Scores

At a county library in Atlantic Corridor during a winter readiness review, weekly program attendance has mean 104 and standard deviation 8. A value is 109.6. Find and interpret its z-score.

A. z=13.7; ignore the mean.
B. z=0.7; the value is 0.7 standard deviations above the mean.
C. z=5.6; do not divide by the standard deviation.
D. z=-0.7; reverse the numerator.

Answer: B

z=(109.6−104)8=0.7. The value is 0.7 standard deviations above the mean.

Why the other choices fail

Choice A: It omits a required part of the calculation. The correct comparison or result is: z=0.7; the value is 0.7 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice C: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=0.7; the value is 0.7 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice D: It reverses the stated order or sign. The correct comparison or result is: z=0.7; the value is 0.7 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 2. Percentiles and z-Scores

At a regional hospital in Pacific Northwest during a community outreach cycle, appointment completion has mean 84 and standard deviation 13. A value is 68.4. Find and interpret its z-score.

A. z=-15.6; do not divide by the standard deviation.
B. z=-1.2; the value is 1.2 standard deviations below the mean.
C. z=1.2; reverse the numerator.
D. z=5.262; ignore the mean.

Answer: B

z=(68.4−84)13=-1.2. The value is 1.2 standard deviations below the mean.

Why the other choices fail

Choice A: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=-1.2; the value is 1.2 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice C: It reverses the stated order or sign. The correct comparison or result is: z=-1.2; the value is 1.2 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice D: It omits a required part of the calculation. The correct comparison or result is: z=-1.2; the value is 1.2 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 3. Percentiles and z-Scores

At a grocery cooperative in Desert County during a winter readiness review, checkout time has mean 102 and standard deviation 18. A value is 125.4. Find and interpret its z-score.

A. z=-1.3; reverse the numerator.
B. z=23.4; do not divide by the standard deviation.
C. z=1.3; the value is 1.3 standard deviations above the mean.
D. z=6.967; ignore the mean.

Answer: C

z=(125.4−102)18=1.3. The value is 1.3 standard deviations above the mean.

Why the other choices fail

Choice A: It reverses the stated order or sign. The correct comparison or result is: z=1.3; the value is 1.3 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice B: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=1.3; the value is 1.3 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice D: It omits a required part of the calculation. The correct comparison or result is: z=1.3; the value is 1.3 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 4. Percentiles and z-Scores

At a regional manufacturer in Atlantic Corridor during a community outreach cycle, part diameter has mean 68 and standard deviation 17. A value is 100.3. Find and interpret its z-score.

A. z=1.9; the value is 1.9 standard deviations above the mean.
B. z=32.3; do not divide by the standard deviation.
C. z=-1.9; reverse the numerator.
D. z=5.9; ignore the mean.

Answer: A

z=(100.3−68)17=1.9. The value is 1.9 standard deviations above the mean.

Why the other choices fail

Choice B: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=1.9; the value is 1.9 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice C: It reverses the stated order or sign. The correct comparison or result is: z=1.9; the value is 1.9 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice D: It omits a required part of the calculation. The correct comparison or result is: z=1.9; the value is 1.9 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 5. Percentiles and z-Scores

At a city recreation department in Coastal Plains during a multiweek validation study, program satisfaction has mean 71 and standard deviation 14. A value is 97.6. Find and interpret its z-score.

A. z=-1.9; reverse the numerator.
B. z=6.971; ignore the mean.
C. z=1.9; the value is 1.9 standard deviations above the mean.
D. z=26.6; do not divide by the standard deviation.

Answer: C

z=(97.6−71)14=1.9. The value is 1.9 standard deviations above the mean.

Why the other choices fail

Choice A: It reverses the stated order or sign. The correct comparison or result is: z=1.9; the value is 1.9 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice B: It omits a required part of the calculation. The correct comparison or result is: z=1.9; the value is 1.9 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice D: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=1.9; the value is 1.9 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 6. Percentiles and z-Scores

At a regional airport authority in Metro East during a community outreach cycle, security wait time has mean 50 and standard deviation 12. A value is 28.4. Find and interpret its z-score.

A. z=-21.6; do not divide by the standard deviation.
B. z=2.367; ignore the mean.
C. z=-1.8; the value is 1.8 standard deviations below the mean.
D. z=1.8; reverse the numerator.

Answer: C

z=(28.4−50)12=-1.8. The value is 1.8 standard deviations below the mean.

Why the other choices fail

Choice A: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=-1.8; the value is 1.8 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice B: It omits a required part of the calculation. The correct comparison or result is: z=-1.8; the value is 1.8 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice D: It reverses the stated order or sign. The correct comparison or result is: z=-1.8; the value is 1.8 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 7. Percentiles and z-Scores

At a public health department in Midwest consortium during a spring 2027 pilot, vaccination appointment completion has mean 62 and standard deviation 16. A value is 92.4. Find and interpret its z-score.

A. z=1.9; the value is 1.9 standard deviations above the mean.
B. z=30.4; do not divide by the standard deviation.
C. z=-1.9; reverse the numerator.
D. z=5.775; ignore the mean.

Answer: A

z=(92.4−62)16=1.9. The value is 1.9 standard deviations above the mean.

Why the other choices fail

Choice B: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=1.9; the value is 1.9 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice C: It reverses the stated order or sign. The correct comparison or result is: z=1.9; the value is 1.9 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice D: It omits a required part of the calculation. The correct comparison or result is: z=1.9; the value is 1.9 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 8. Percentiles and z-Scores

At a wildlife clinic in Cedar Grove during a randomized pilot period, recovery time has mean 65 and standard deviation 15. A value is 47. Find and interpret its z-score.

A. z=-18; do not divide by the standard deviation.
B. z=-1.2; the value is 1.2 standard deviations below the mean.
C. z=3.133; ignore the mean.
D. z=1.2; reverse the numerator.

Answer: B

z=(47−65)15=-1.2. The value is 1.2 standard deviations below the mean.

Why the other choices fail

Choice A: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=-1.2; the value is 1.2 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice C: It omits a required part of the calculation. The correct comparison or result is: z=-1.2; the value is 1.2 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice D: It reverses the stated order or sign. The correct comparison or result is: z=-1.2; the value is 1.2 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 9. Percentiles and z-Scores

At a university advising center in Lakeside district during a spring 2027 pilot, appointment wait time has mean 77 and standard deviation 11. A value is 70.4. Find and interpret its z-score.

A. z=-0.6; the value is 0.6 standard deviations below the mean.
B. z=6.4; ignore the mean.
C. z=-6.6; do not divide by the standard deviation.
D. z=0.6; reverse the numerator.

Answer: A

z=(70.4−77)11=-0.6. The value is 0.6 standard deviations below the mean.

Why the other choices fail

Choice B: It omits a required part of the calculation. The correct comparison or result is: z=-0.6; the value is 0.6 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice C: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=-0.6; the value is 0.6 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice D: It reverses the stated order or sign. The correct comparison or result is: z=-0.6; the value is 0.6 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 10. Percentiles and z-Scores

At a county election office in Great Lakes during a summer implementation review, ballot-processing time has mean 76 and standard deviation 12. A value is 54.4. Find and interpret its z-score.

A. z=4.533; ignore the mean.
B. z=-21.6; do not divide by the standard deviation.
C. z=-1.8; the value is 1.8 standard deviations below the mean.
D. z=1.8; reverse the numerator.

Answer: C

z=(54.4−76)12=-1.8. The value is 1.8 standard deviations below the mean.

Why the other choices fail

Choice A: It omits a required part of the calculation. The correct comparison or result is: z=-1.8; the value is 1.8 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice B: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=-1.8; the value is 1.8 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice D: It reverses the stated order or sign. The correct comparison or result is: z=-1.8; the value is 1.8 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 11. Percentiles and z-Scores

At a county election office in Desert County during a school-year data collection, ballot-processing time has mean 109 and standard deviation 17. A value is 78.4. Find and interpret its z-score.

A. z=-30.6; do not divide by the standard deviation.
B. z=4.612; ignore the mean.
C. z=1.8; reverse the numerator.
D. z=-1.8; the value is 1.8 standard deviations below the mean.

Answer: D

z=(78.4−109)17=-1.8. The value is 1.8 standard deviations below the mean.

Why the other choices fail

Choice A: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=-1.8; the value is 1.8 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice B: It omits a required part of the calculation. The correct comparison or result is: z=-1.8; the value is 1.8 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice C: It reverses the stated order or sign. The correct comparison or result is: z=-1.8; the value is 1.8 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 12. Percentiles and z-Scores

At a grocery cooperative in Metro East during a spring 2027 pilot, checkout time has mean 68 and standard deviation 17. A value is 100.3. Find and interpret its z-score.

A. z=-1.9; reverse the numerator.
B. z=1.9; the value is 1.9 standard deviations above the mean.
C. z=5.9; ignore the mean.
D. z=32.3; do not divide by the standard deviation.

Answer: B

z=(100.3−68)17=1.9. The value is 1.9 standard deviations above the mean.

Why the other choices fail

Choice A: It reverses the stated order or sign. The correct comparison or result is: z=1.9; the value is 1.9 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice C: It omits a required part of the calculation. The correct comparison or result is: z=1.9; the value is 1.9 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice D: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=1.9; the value is 1.9 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 13. Percentiles and z-Scores

At a city transit agency in Capital Region during a spring 2027 pilot, on-time arrival has mean 99 and standard deviation 10. A value is 112.0. Find and interpret its z-score.

A. z=11.2; ignore the mean.
B. z=13; do not divide by the standard deviation.
C. z=-1.3; reverse the numerator.
D. z=1.3; the value is 1.3 standard deviations above the mean.

Answer: D

z=(112.0−99)10=1.3. The value is 1.3 standard deviations above the mean.

Why the other choices fail

Choice A: It omits a required part of the calculation. The correct comparison or result is: z=1.3; the value is 1.3 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice B: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=1.3; the value is 1.3 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice C: It reverses the stated order or sign. The correct comparison or result is: z=1.3; the value is 1.3 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 14. Percentiles and z-Scores

At a community bank in Westview during a community outreach cycle, mobile-deposit adoption has mean 68 and standard deviation 13. A value is 44.6. Find and interpret its z-score.

A. z=-1.8; the value is 1.8 standard deviations below the mean.
B. z=3.431; ignore the mean.
C. z=1.8; reverse the numerator.
D. z=-23.4; do not divide by the standard deviation.

Answer: A

z=(44.6−68)13=-1.8. The value is 1.8 standard deviations below the mean.

Why the other choices fail

Choice B: It omits a required part of the calculation. The correct comparison or result is: z=-1.8; the value is 1.8 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice C: It reverses the stated order or sign. The correct comparison or result is: z=-1.8; the value is 1.8 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice D: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=-1.8; the value is 1.8 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 15. Percentiles and z-Scores

At a grocery cooperative in Atlantic Corridor during a semester-long cohort study, checkout time has mean 100 and standard deviation 6. A value is 96.4. Find and interpret its z-score.

A. z=0.6; reverse the numerator.
B. z=16.067; ignore the mean.
C. z=-3.6; do not divide by the standard deviation.
D. z=-0.6; the value is 0.6 standard deviations below the mean.

Answer: D

z=(96.4−100)6=-0.6. The value is 0.6 standard deviations below the mean.

Why the other choices fail

Choice A: It reverses the stated order or sign. The correct comparison or result is: z=-0.6; the value is 0.6 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice B: It omits a required part of the calculation. The correct comparison or result is: z=-0.6; the value is 0.6 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice C: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=-0.6; the value is 0.6 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 16. Percentiles and z-Scores

At a grocery cooperative in Great Lakes during a baseline measurement week, checkout time has mean 66 and standard deviation 11. A value is 52.8. Find and interpret its z-score.

A. z=-13.2; do not divide by the standard deviation.
B. z=-1.2; the value is 1.2 standard deviations below the mean.
C. z=4.8; ignore the mean.
D. z=1.2; reverse the numerator.

Answer: B

z=(52.8−66)11=-1.2. The value is 1.2 standard deviations below the mean.

Why the other choices fail

Choice A: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=-1.2; the value is 1.2 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice C: It omits a required part of the calculation. The correct comparison or result is: z=-1.2; the value is 1.2 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice D: It reverses the stated order or sign. The correct comparison or result is: z=-1.2; the value is 1.2 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 17. Percentiles and z-Scores

At a solar installer in Atlantic Corridor during a yearly program evaluation, daily energy output has mean 68 and standard deviation 13. A value is 92.7. Find and interpret its z-score.

A. z=24.7; do not divide by the standard deviation.
B. z=-1.9; reverse the numerator.
C. z=1.9; the value is 1.9 standard deviations above the mean.
D. z=7.131; ignore the mean.

Answer: C

z=(92.7−68)13=1.9. The value is 1.9 standard deviations above the mean.

Why the other choices fail

Choice A: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=1.9; the value is 1.9 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice B: It reverses the stated order or sign. The correct comparison or result is: z=1.9; the value is 1.9 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice D: It omits a required part of the calculation. The correct comparison or result is: z=1.9; the value is 1.9 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 18. Percentiles and z-Scores

At a university advising center in Mountain Region during a quarterly performance study, appointment wait time has mean 66 and standard deviation 7. A value is 61.8. Find and interpret its z-score.

A. z=-0.6; the value is 0.6 standard deviations below the mean.
B. z=-4.2; do not divide by the standard deviation.
C. z=0.6; reverse the numerator.
D. z=8.829; ignore the mean.

Answer: A

z=(61.8−66)7=-0.6. The value is 0.6 standard deviations below the mean.

Why the other choices fail

Choice B: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=-0.6; the value is 0.6 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice C: It reverses the stated order or sign. The correct comparison or result is: z=-0.6; the value is 0.6 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice D: It omits a required part of the calculation. The correct comparison or result is: z=-0.6; the value is 0.6 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 19. Percentiles and z-Scores

At a regional airport authority in Coastal Plains during a fall 2026 audit, security wait time has mean 64 and standard deviation 10. A value is 71. Find and interpret its z-score.

A. z=-0.7; reverse the numerator.
B. z=0.7; the value is 0.7 standard deviations above the mean.
C. z=7.1; ignore the mean.
D. z=7; do not divide by the standard deviation.

Answer: B

z=(71−64)10=0.7. The value is 0.7 standard deviations above the mean.

Why the other choices fail

Choice A: It reverses the stated order or sign. The correct comparison or result is: z=0.7; the value is 0.7 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice C: It omits a required part of the calculation. The correct comparison or result is: z=0.7; the value is 0.7 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice D: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=0.7; the value is 0.7 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 20. Percentiles and z-Scores

At a grocery cooperative in Pine Ridge during a baseline measurement week, checkout time has mean 58 and standard deviation 17. A value is 69.9. Find and interpret its z-score.

A. z=0.7; the value is 0.7 standard deviations above the mean.
B. z=11.9; do not divide by the standard deviation.
C. z=4.112; ignore the mean.
D. z=-0.7; reverse the numerator.

Answer: A

z=(69.9−58)17=0.7. The value is 0.7 standard deviations above the mean.

Why the other choices fail

Choice B: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=0.7; the value is 0.7 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice C: It omits a required part of the calculation. The correct comparison or result is: z=0.7; the value is 0.7 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice D: It reverses the stated order or sign. The correct comparison or result is: z=0.7; the value is 0.7 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 21. Percentiles and z-Scores

At a grocery cooperative in Atlantic Corridor during a community outreach cycle, checkout time has mean 50 and standard deviation 7. A value is 37.4. Find and interpret its z-score.

A. z=1.8; reverse the numerator.
B. z=-12.6; do not divide by the standard deviation.
C. z=5.343; ignore the mean.
D. z=-1.8; the value is 1.8 standard deviations below the mean.

Answer: D

z=(37.4−50)7=-1.8. The value is 1.8 standard deviations below the mean.

Why the other choices fail

Choice A: It reverses the stated order or sign. The correct comparison or result is: z=-1.8; the value is 1.8 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice B: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=-1.8; the value is 1.8 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice C: It omits a required part of the calculation. The correct comparison or result is: z=-1.8; the value is 1.8 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 22. Percentiles and z-Scores

At a county election office in Prairie District during a quarterly performance study, ballot-processing time has mean 77 and standard deviation 9. A value is 71.6. Find and interpret its z-score.

A. z=7.956; ignore the mean.
B. z=-5.4; do not divide by the standard deviation.
C. z=-0.6; the value is 0.6 standard deviations below the mean.
D. z=0.6; reverse the numerator.

Answer: C

z=(71.6−77)9=-0.6. The value is 0.6 standard deviations below the mean.

Why the other choices fail

Choice A: It omits a required part of the calculation. The correct comparison or result is: z=-0.6; the value is 0.6 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice B: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=-0.6; the value is 0.6 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice D: It reverses the stated order or sign. The correct comparison or result is: z=-0.6; the value is 0.6 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 23. Percentiles and z-Scores

At a community college in Sunbelt district during a semester-long cohort study, course completion has mean 93 and standard deviation 15. A value is 103.5. Find and interpret its z-score.

A. z=0.7; the value is 0.7 standard deviations above the mean.
B. z=-0.7; reverse the numerator.
C. z=6.9; ignore the mean.
D. z=10.5; do not divide by the standard deviation.

Answer: A

z=(103.5−93)15=0.7. The value is 0.7 standard deviations above the mean.

Why the other choices fail

Choice B: It reverses the stated order or sign. The correct comparison or result is: z=0.7; the value is 0.7 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice C: It omits a required part of the calculation. The correct comparison or result is: z=0.7; the value is 0.7 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice D: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=0.7; the value is 0.7 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 24. Percentiles and z-Scores

At a university advising center in Metro East during a winter readiness review, appointment wait time has mean 89 and standard deviation 16. A value is 60.2. Find and interpret its z-score.

A. z=-1.8; the value is 1.8 standard deviations below the mean.
B. z=3.763; ignore the mean.
C. z=1.8; reverse the numerator.
D. z=-28.8; do not divide by the standard deviation.

Answer: A

z=(60.2−89)16=-1.8. The value is 1.8 standard deviations below the mean.

Why the other choices fail

Choice B: It omits a required part of the calculation. The correct comparison or result is: z=-1.8; the value is 1.8 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice C: It reverses the stated order or sign. The correct comparison or result is: z=-1.8; the value is 1.8 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice D: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=-1.8; the value is 1.8 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 25. Percentiles and z-Scores

At a municipal emergency dispatch center in Central County during a spring 2027 pilot, response time has mean 97 and standard deviation 10. A value is 91. Find and interpret its z-score.

A. z=9.1; ignore the mean.
B. z=0.6; reverse the numerator.
C. z=-6; do not divide by the standard deviation.
D. z=-0.6; the value is 0.6 standard deviations below the mean.

Answer: D

z=(91−97)10=-0.6. The value is 0.6 standard deviations below the mean.

Why the other choices fail

Choice A: It omits a required part of the calculation. The correct comparison or result is: z=-0.6; the value is 0.6 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice B: It reverses the stated order or sign. The correct comparison or result is: z=-0.6; the value is 0.6 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice C: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=-0.6; the value is 0.6 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 26. Percentiles and z-Scores

At a wildlife clinic in Desert County during a quarterly performance study, recovery time has mean 57 and standard deviation 13. A value is 49.2. Find and interpret its z-score.

A. z=0.6; reverse the numerator.
B. z=3.785; ignore the mean.
C. z=-0.6; the value is 0.6 standard deviations below the mean.
D. z=-7.8; do not divide by the standard deviation.

Answer: C

z=(49.2−57)13=-0.6. The value is 0.6 standard deviations below the mean.

Why the other choices fail

Choice A: It reverses the stated order or sign. The correct comparison or result is: z=-0.6; the value is 0.6 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice B: It omits a required part of the calculation. The correct comparison or result is: z=-0.6; the value is 0.6 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice D: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=-0.6; the value is 0.6 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 27. Percentiles and z-Scores

At a school district in New England network during a weekday operations study, lunch-program participation has mean 85 and standard deviation 17. A value is 117.3. Find and interpret its z-score.

A. z=6.9; ignore the mean.
B. z=-1.9; reverse the numerator.
C. z=32.3; do not divide by the standard deviation.
D. z=1.9; the value is 1.9 standard deviations above the mean.

Answer: D

z=(117.3−85)17=1.9. The value is 1.9 standard deviations above the mean.

Why the other choices fail

Choice A: It omits a required part of the calculation. The correct comparison or result is: z=1.9; the value is 1.9 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice B: It reverses the stated order or sign. The correct comparison or result is: z=1.9; the value is 1.9 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice C: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=1.9; the value is 1.9 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 28. Percentiles and z-Scores

At a school district in Metro East during a pre-exam training cycle, lunch-program participation has mean 72 and standard deviation 10. A value is 66. Find and interpret its z-score.

A. z=-0.6; the value is 0.6 standard deviations below the mean.
B. z=0.6; reverse the numerator.
C. z=6.6; ignore the mean.
D. z=-6; do not divide by the standard deviation.

Answer: A

z=(66−72)10=-0.6. The value is 0.6 standard deviations below the mean.

Why the other choices fail

Choice B: It reverses the stated order or sign. The correct comparison or result is: z=-0.6; the value is 0.6 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice C: It omits a required part of the calculation. The correct comparison or result is: z=-0.6; the value is 0.6 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice D: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=-0.6; the value is 0.6 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 29. Percentiles and z-Scores

At a housing authority in North Valley during a school-year data collection, application processing time has mean 100 and standard deviation 12. A value is 78.4. Find and interpret its z-score.

A. z=-21.6; do not divide by the standard deviation.
B. z=1.8; reverse the numerator.
C. z=6.533; ignore the mean.
D. z=-1.8; the value is 1.8 standard deviations below the mean.

Answer: D

z=(78.4−100)12=-1.8. The value is 1.8 standard deviations below the mean.

Why the other choices fail

Choice A: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=-1.8; the value is 1.8 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice B: It reverses the stated order or sign. The correct comparison or result is: z=-1.8; the value is 1.8 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice C: It omits a required part of the calculation. The correct comparison or result is: z=-1.8; the value is 1.8 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 30. Percentiles and z-Scores

At a municipal water office in Midwest consortium during a pre-exam training cycle, monthly household use has mean 74 and standard deviation 9. A value is 68.6. Find and interpret its z-score.

A. z=-0.6; the value is 0.6 standard deviations below the mean.
B. z=-5.4; do not divide by the standard deviation.
C. z=0.6; reverse the numerator.
D. z=7.622; ignore the mean.

Answer: A

z=(68.6−74)9=-0.6. The value is 0.6 standard deviations below the mean.

Why the other choices fail

Choice B: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=-0.6; the value is 0.6 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice C: It reverses the stated order or sign. The correct comparison or result is: z=-0.6; the value is 0.6 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice D: It omits a required part of the calculation. The correct comparison or result is: z=-0.6; the value is 0.6 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 31. Percentiles and z-Scores

At a public health department in Metro East during a service-improvement study, vaccination appointment completion has mean 69 and standard deviation 13. A value is 93.7. Find and interpret its z-score.

A. z=7.208; ignore the mean.
B. z=1.9; the value is 1.9 standard deviations above the mean.
C. z=24.7; do not divide by the standard deviation.
D. z=-1.9; reverse the numerator.

Answer: B

z=(93.7−69)13=1.9. The value is 1.9 standard deviations above the mean.

Why the other choices fail

Choice A: It omits a required part of the calculation. The correct comparison or result is: z=1.9; the value is 1.9 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice C: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=1.9; the value is 1.9 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice D: It reverses the stated order or sign. The correct comparison or result is: z=1.9; the value is 1.9 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 32. Percentiles and z-Scores

At a municipal water office in Coastal Plains during a quarterly performance study, monthly household use has mean 86 and standard deviation 13. A value is 78.2. Find and interpret its z-score.

A. z=-7.8; do not divide by the standard deviation.
B. z=0.6; reverse the numerator.
C. z=-0.6; the value is 0.6 standard deviations below the mean.
D. z=6.015; ignore the mean.

Answer: C

z=(78.2−86)13=-0.6. The value is 0.6 standard deviations below the mean.

Why the other choices fail

Choice A: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=-0.6; the value is 0.6 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice B: It reverses the stated order or sign. The correct comparison or result is: z=-0.6; the value is 0.6 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice D: It omits a required part of the calculation. The correct comparison or result is: z=-0.6; the value is 0.6 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 33. Percentiles and z-Scores

At a digital learning platform in South Harbor during a school-year data collection, lesson completion has mean 100 and standard deviation 7. A value is 91.6. Find and interpret its z-score.

A. z=1.2; reverse the numerator.
B. z=-1.2; the value is 1.2 standard deviations below the mean.
C. z=13.086; ignore the mean.
D. z=-8.4; do not divide by the standard deviation.

Answer: B

z=(91.6−100)7=-1.2. The value is 1.2 standard deviations below the mean.

Why the other choices fail

Choice A: It reverses the stated order or sign. The correct comparison or result is: z=-1.2; the value is 1.2 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice C: It omits a required part of the calculation. The correct comparison or result is: z=-1.2; the value is 1.2 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice D: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=-1.2; the value is 1.2 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 34. Percentiles and z-Scores

At a municipal water office in Sunbelt district during a baseline measurement week, monthly household use has mean 99 and standard deviation 7. A value is 112.3. Find and interpret its z-score.

A. z=-1.9; reverse the numerator.
B. z=1.9; the value is 1.9 standard deviations above the mean.
C. z=13.3; do not divide by the standard deviation.
D. z=16.043; ignore the mean.

Answer: B

z=(112.3−99)7=1.9. The value is 1.9 standard deviations above the mean.

Why the other choices fail

Choice A: It reverses the stated order or sign. The correct comparison or result is: z=1.9; the value is 1.9 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice C: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=1.9; the value is 1.9 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice D: It omits a required part of the calculation. The correct comparison or result is: z=1.9; the value is 1.9 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 35. Percentiles and z-Scores

At a recycling program in North Valley during a quarterly performance study, weekly material weight has mean 54 and standard deviation 8. A value is 69.2. Find and interpret its z-score.

A. z=1.9; the value is 1.9 standard deviations above the mean.
B. z=-1.9; reverse the numerator.
C. z=15.2; do not divide by the standard deviation.
D. z=8.65; ignore the mean.

Answer: A

z=(69.2−54)8=1.9. The value is 1.9 standard deviations above the mean.

Why the other choices fail

Choice B: It reverses the stated order or sign. The correct comparison or result is: z=1.9; the value is 1.9 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice C: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=1.9; the value is 1.9 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice D: It omits a required part of the calculation. The correct comparison or result is: z=1.9; the value is 1.9 standard deviations above the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Question 36. Percentiles and z-Scores

At a state park in Midwest consortium during a two-month observation window, trail-use duration has mean 94 and standard deviation 14. A value is 68.8. Find and interpret its z-score.

A. z=4.914; ignore the mean.
B. z=-1.8; the value is 1.8 standard deviations below the mean.
C. z=1.8; reverse the numerator.
D. z=-25.2; do not divide by the standard deviation.

Answer: B

z=(68.8−94)14=-1.8. The value is 1.8 standard deviations below the mean.

Why the other choices fail

Choice A: It omits a required part of the calculation. The correct comparison or result is: z=-1.8; the value is 1.8 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice C: It reverses the stated order or sign. The correct comparison or result is: z=-1.8; the value is 1.8 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.
Choice D: It uses the wrong denominator or omits a required scale adjustment. The correct comparison or result is: z=-1.8; the value is 1.8 standard deviations below the mean. Key check: A z-score reports how many standard deviations a value lies from the mean.

Percentiles and z-Scores free-response practice

10-point analytic study rubric used for the sets below
Evidence	Points
Correct target, notation, direction, or group order	2
Correct method/model and defensible conditions	2
Correct setup and execution	3
Contextual interpretation and scope/limitation	2
Clear communication with units and labels	1

FRQ set 1: Percentiles and z-Scores

Scenario. At a municipal emergency dispatch center in New England network during a school-year data collection, response time has mean 52 and standard deviation 18. A value is 19.6. Find and interpret its z-score.

Define the variable, units, groups, and requested distribution or model feature.
Show the required calculation or graphical/model reasoning with labeled quantities.
Interpret the numerical result in the context of the data rather than as an isolated number.
Identify an unusual feature, limitation, or condition that affects the conclusion.

Model response

z=(19.6−52)18=-1.8. The value is 1.8 standard deviations below the mean.

FRQ set 2: Percentiles and z-Scores

Scenario. At a municipal emergency dispatch center in Riverbend during a school-year data collection, response time has mean 86 and standard deviation 6. A value is 78.8. Find and interpret its z-score.

Define the variable, units, groups, and requested distribution or model feature.
Show the required calculation or graphical/model reasoning with labeled quantities.
Interpret the numerical result in the context of the data rather than as an isolated number.
Identify an unusual feature, limitation, or condition that affects the conclusion.

Model response

z=(78.8−86)6=-1.2. The value is 1.2 standard deviations below the mean.

FRQ set 3: Percentiles and z-Scores

Scenario. At a recycling program in Coastal Plains during a weekday operations study, weekly material weight has mean 68 and standard deviation 12. A value is 60.8. Find and interpret its z-score.

Define the variable, units, groups, and requested distribution or model feature.
Show the required calculation or graphical/model reasoning with labeled quantities.
Interpret the numerical result in the context of the data rather than as an isolated number.
Identify an unusual feature, limitation, or condition that affects the conclusion.

Model response

z=(60.8−68)12=-0.6. The value is 0.6 standard deviations below the mean.

FRQ set 4: Percentiles and z-Scores

Scenario. At a university advising center in Capital Region during a two-month observation window, appointment wait time has mean 90 and standard deviation 18. A value is 124.2. Find and interpret its z-score.

Define the variable, units, groups, and requested distribution or model feature.
Show the required calculation or graphical/model reasoning with labeled quantities.
Interpret the numerical result in the context of the data rather than as an isolated number.
Identify an unusual feature, limitation, or condition that affects the conclusion.

Model response

z=(124.2−90)18=1.9. The value is 1.9 standard deviations above the mean.

FRQ set 5: Percentiles and z-Scores

Scenario. At a county library in Pacific Northwest during a semester-long cohort study, weekly program attendance has mean 96 and standard deviation 6. A value is 88.8. Find and interpret its z-score.

Define the variable, units, groups, and requested distribution or model feature.
Show the required calculation or graphical/model reasoning with labeled quantities.
Interpret the numerical result in the context of the data rather than as an isolated number.
Identify an unusual feature, limitation, or condition that affects the conclusion.

Model response

z=(88.8−96)6=-1.2. The value is 1.2 standard deviations below the mean.

FRQ set 6: Percentiles and z-Scores

Scenario. At a public health department in Lakeside district during a pre-exam training cycle, vaccination appointment completion has mean 73 and standard deviation 14. A value is 82.8. Find and interpret its z-score.

Define the variable, units, groups, and requested distribution or model feature.
Show the required calculation or graphical/model reasoning with labeled quantities.
Interpret the numerical result in the context of the data rather than as an isolated number.
Identify an unusual feature, limitation, or condition that affects the conclusion.

Model response

z=(82.8−73)14=0.7. The value is 0.7 standard deviations above the mean.

FRQ set 7: Percentiles and z-Scores

Scenario. At a regional airport authority in Sunbelt district during a service-improvement study, security wait time has mean 50 and standard deviation 17. A value is 19.4. Find and interpret its z-score.

Define the variable, units, groups, and requested distribution or model feature.
Show the required calculation or graphical/model reasoning with labeled quantities.
Interpret the numerical result in the context of the data rather than as an isolated number.
Identify an unusual feature, limitation, or condition that affects the conclusion.

Model response

z=(19.4−50)17=-1.8. The value is 1.8 standard deviations below the mean.

FRQ set 8: Percentiles and z-Scores

Scenario. At a community bank in Westview during a pre-exam training cycle, mobile-deposit adoption has mean 98 and standard deviation 10. A value is 105.0. Find and interpret its z-score.

Define the variable, units, groups, and requested distribution or model feature.
Show the required calculation or graphical/model reasoning with labeled quantities.
Interpret the numerical result in the context of the data rather than as an isolated number.
Identify an unusual feature, limitation, or condition that affects the conclusion.

Model response

z=(105.0−98)10=0.7. The value is 0.7 standard deviations above the mean.

FRQ set 9: Percentiles and z-Scores

Scenario. At a housing authority in Sunbelt district during a randomized pilot period, application processing time has mean 84 and standard deviation 7. A value is 79.8. Find and interpret its z-score.

Define the variable, units, groups, and requested distribution or model feature.
Show the required calculation or graphical/model reasoning with labeled quantities.
Interpret the numerical result in the context of the data rather than as an isolated number.
Identify an unusual feature, limitation, or condition that affects the conclusion.

Model response

z=(79.8−84)7=-0.6. The value is 0.6 standard deviations below the mean.

FRQ set 10: Percentiles and z-Scores

Scenario. At a recycling program in Midwest consortium during a school-year data collection, weekly material weight has mean 107 and standard deviation 18. A value is 141.2. Find and interpret its z-score.

Define the variable, units, groups, and requested distribution or model feature.
Show the required calculation or graphical/model reasoning with labeled quantities.
Interpret the numerical result in the context of the data rather than as an isolated number.
Identify an unusual feature, limitation, or condition that affects the conclusion.

Model response

z=(141.2−107)18=1.9. The value is 1.9 standard deviations above the mean.

FRQ set 11: Percentiles and z-Scores

Scenario. At a county election office in Pine Ridge during a quarterly performance study, ballot-processing time has mean 81 and standard deviation 10. A value is 63. Find and interpret its z-score.

Define the variable, units, groups, and requested distribution or model feature.
Show the required calculation or graphical/model reasoning with labeled quantities.
Interpret the numerical result in the context of the data rather than as an isolated number.
Identify an unusual feature, limitation, or condition that affects the conclusion.

Model response

z=(63−81)10=-1.8. The value is 1.8 standard deviations below the mean.

FRQ set 12: Percentiles and z-Scores

Scenario. At a county library in Atlantic Corridor during a service-improvement study, weekly program attendance has mean 67 and standard deviation 10. A value is 80. Find and interpret its z-score.

Define the variable, units, groups, and requested distribution or model feature.
Show the required calculation or graphical/model reasoning with labeled quantities.
Interpret the numerical result in the context of the data rather than as an isolated number.
Identify an unusual feature, limitation, or condition that affects the conclusion.

Model response

z=(80−67)10=1.3. The value is 1.3 standard deviations above the mean.

FRQ set 13: Percentiles and z-Scores

Scenario. At a regional hospital in Metro East during a six-week field trial, appointment completion has mean 110 and standard deviation 17. A value is 121.9. Find and interpret its z-score.

Define the variable, units, groups, and requested distribution or model feature.
Show the required calculation or graphical/model reasoning with labeled quantities.
Interpret the numerical result in the context of the data rather than as an isolated number.
Identify an unusual feature, limitation, or condition that affects the conclusion.

Model response

z=(121.9−110)17=0.7. The value is 0.7 standard deviations above the mean.

FRQ set 14: Percentiles and z-Scores

Scenario. At a county election office in Sunbelt district during a quarterly performance study, ballot-processing time has mean 91 and standard deviation 14. A value is 65.8. Find and interpret its z-score.

Define the variable, units, groups, and requested distribution or model feature.
Show the required calculation or graphical/model reasoning with labeled quantities.
Interpret the numerical result in the context of the data rather than as an isolated number.
Identify an unusual feature, limitation, or condition that affects the conclusion.

Model response

z=(65.8−91)14=-1.8. The value is 1.8 standard deviations below the mean.