What does Normal Distribution measure or decide?

A normal model is determined by μ and σ; probability is area, not curve height.

When should I use Normal Distribution?

Convert raw cutoffs to z-scores and treat probability as area under the curve.

Which conditions or design checks matter for Normal Distribution?

Use a normal model only when it is stated or defensible, and convert boundaries to standardized values before finding areas or percentiles.

Which formula or calculator output matters most for Normal Distribution?

Standardize each boundary with z=(x−μ)/σ, then use cumulative normal area or an inverse-normal calculation.

How should results from Normal Distribution 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 Normal Distribution?

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: Convert raw cutoffs to z-scores and treat probability as area under the curve.

Are the questions on this Normal Distribution page copied from past AP exams?

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

Is Normal Distribution required for the May 2027 AP Statistics exam?

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

Normal Distribution: Formula, Curve, Probabilities, and Examples

AP Statistics Practice

Normal Distribution: Formula, Curve, Probabilities, and Examples

Normal Distribution: Formula, Curve, Probabilities, 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

Normal Distribution: Formula, Curve, Probabilities, and Examples: formulas and targets

Standardize	z = (x−μ)σ
Central probability	P(a≤X≤b)=Φ(zb)−Φ(za)

Normal Distribution multiple-choice practice

Question 1. Normal Distribution

Model daily energy output at a solar installer in Westview during a winter readiness review as Normal(μ=79, σ=20). Find P(55≤X≤95) and interpret the area.

A. 0.115; use only the lower cumulative area.
B. 0.327; subtract the central area from 1.
C. 0.788; use only the upper cumulative area.
D. 0.673; about 67.3% of the modeled population lies between the cutoffs.

Answer: D

Standardize the bounds: zL=(55−79)20=-1.2 and zU=(95−79)20=0.8. P=0.788−0.115=0.673, so about 67.3% of modeled installations fall in the interval.

Why the other choices fail

Choice A: It omits a required part of the calculation. The correct comparison or result is: 0.673; about 67.3% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice B: It does not match the requested calculation. The correct comparison or result is: 0.673; about 67.3% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice C: It omits a required part of the calculation. The correct comparison or result is: 0.673; about 67.3% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 2. Normal Distribution

Model mobile-deposit adoption at a community bank in South Harbor during a follow-up evaluation period as Normal(μ=115, σ=13). Find P(99.4≤X≤129.3) and interpret the area.

A. 0.115; use only the lower cumulative area.
B. 0.749; about 74.9% of the modeled population lies between the cutoffs.
C. 0.251; subtract the central area from 1.
D. 0.864; use only the upper cumulative area.

Answer: B

Standardize the bounds: zL=(99.4−115)13=-1.2 and zU=(129.3−115)13=1.1. P=0.864−0.115=0.749, so about 74.9% of modeled customers fall in the interval.

Why the other choices fail

Choice A: It omits a required part of the calculation. The correct comparison or result is: 0.749; about 74.9% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice C: It does not match the requested calculation. The correct comparison or result is: 0.749; about 74.9% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice D: It omits a required part of the calculation. The correct comparison or result is: 0.749; about 74.9% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 3. Normal Distribution

Model response time at a municipal emergency dispatch center in New England network during a quarterly performance study as Normal(μ=89, σ=17). Find P(68.6≤X≤95.8) and interpret the area.

A. 0.655; use only the upper cumulative area.
B. 0.54; about 54.0% of the modeled population lies between the cutoffs.
C. 0.115; use only the lower cumulative area.
D. 0.46; subtract the central area from 1.

Answer: B

Standardize the bounds: zL=(68.6−89)17=-1.2 and zU=(95.8−89)17=0.4. P=0.655−0.115=0.54, so about 54.0% of modeled calls fall in the interval.

Why the other choices fail

Choice A: It omits a required part of the calculation. The correct comparison or result is: 0.54; about 54.0% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice C: It omits a required part of the calculation. The correct comparison or result is: 0.54; about 54.0% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice D: It does not match the requested calculation. The correct comparison or result is: 0.54; about 54.0% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 4. Normal Distribution

Model daily energy output at a solar installer in Pacific Northwest during a baseline measurement week as Normal(μ=82, σ=18). Find P(73≤X≤96.4) and interpret the area.

A. 0.48; about 48.0% of the modeled population lies between the cutoffs.
B. 0.788; use only the upper cumulative area.
C. 0.309; use only the lower cumulative area.
D. 0.52; subtract the central area from 1.

Answer: A

Standardize the bounds: zL=(73−82)18=-0.5 and zU=(96.4−82)18=0.8. P=0.788−0.309=0.48, so about 48.0% of modeled installations fall in the interval.

Why the other choices fail

Choice B: It omits a required part of the calculation. The correct comparison or result is: 0.48; about 48.0% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice C: It omits a required part of the calculation. The correct comparison or result is: 0.48; about 48.0% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice D: It does not match the requested calculation. The correct comparison or result is: 0.48; about 48.0% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 5. Normal Distribution

Model weekly material weight at a recycling program in South Harbor during a summer implementation review as Normal(μ=106, σ=6). Find P(98.8≤X≤108.4) and interpret the area.

A. 0.655; use only the upper cumulative area.
B. 0.115; use only the lower cumulative area.
C. 0.46; subtract the central area from 1.
D. 0.54; about 54.0% of the modeled population lies between the cutoffs.

Answer: D

Standardize the bounds: zL=(98.8−106)6=-1.2 and zU=(108.4−106)6=0.4. P=0.655−0.115=0.54, so about 54.0% of modeled households fall in the interval.

Why the other choices fail

Choice A: It omits a required part of the calculation. The correct comparison or result is: 0.54; about 54.0% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice B: It omits a required part of the calculation. The correct comparison or result is: 0.54; about 54.0% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice C: It does not match the requested calculation. The correct comparison or result is: 0.54; about 54.0% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 6. Normal Distribution

Model appointment completion at a regional hospital in Riverbend during a service-improvement study as Normal(μ=67, σ=18). Find P(45.4≤X≤74.2) and interpret the area.

A. 0.54; about 54.0% of the modeled population lies between the cutoffs.
B. 0.115; use only the lower cumulative area.
C. 0.46; subtract the central area from 1.
D. 0.655; use only the upper cumulative area.

Answer: A

Standardize the bounds: zL=(45.4−67)18=-1.2 and zU=(74.2−67)18=0.4. P=0.655−0.115=0.54, so about 54.0% of modeled patients fall in the interval.

Why the other choices fail

Choice B: It omits a required part of the calculation. The correct comparison or result is: 0.54; about 54.0% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice C: It does not match the requested calculation. The correct comparison or result is: 0.54; about 54.0% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice D: It omits a required part of the calculation. The correct comparison or result is: 0.54; about 54.0% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 7. Normal Distribution

Model appointment completion at a regional hospital in Lakeside district during a yearly program evaluation as Normal(μ=117, σ=17). Find P(108.5≤X≤135.7) and interpret the area.

A. 0.309; use only the lower cumulative area.
B. 0.444; subtract the central area from 1.
C. 0.864; use only the upper cumulative area.
D. 0.556; about 55.6% of the modeled population lies between the cutoffs.

Answer: D

Standardize the bounds: zL=(108.5−117)17=-0.5 and zU=(135.7−117)17=1.1. P=0.864−0.309=0.556, so about 55.6% of modeled patients fall in the interval.

Why the other choices fail

Choice A: It omits a required part of the calculation. The correct comparison or result is: 0.556; about 55.6% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice B: It does not match the requested calculation. The correct comparison or result is: 0.556; about 55.6% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice C: It omits a required part of the calculation. The correct comparison or result is: 0.556; about 55.6% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 8. Normal Distribution

Model program satisfaction at a city recreation department in North Valley during a baseline measurement week as Normal(μ=59, σ=7). Find P(50.6≤X≤64.6) and interpret the area.

A. 0.327; subtract the central area from 1.
B. 0.673; about 67.3% of the modeled population lies between the cutoffs.
C. 0.115; use only the lower cumulative area.
D. 0.788; use only the upper cumulative area.

Answer: B

Standardize the bounds: zL=(50.6−59)7=-1.2 and zU=(64.6−59)7=0.8. P=0.788−0.115=0.673, so about 67.3% of modeled participants fall in the interval.

Why the other choices fail

Choice A: It does not match the requested calculation. The correct comparison or result is: 0.673; about 67.3% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice C: It omits a required part of the calculation. The correct comparison or result is: 0.673; about 67.3% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice D: It omits a required part of the calculation. The correct comparison or result is: 0.673; about 67.3% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 9. Normal Distribution

Model vaccination appointment completion at a public health department in Riverbend during a service-improvement study as Normal(μ=64, σ=12). Find P(54.4≤X≤73.6) and interpret the area.

A. 0.576; about 57.6% of the modeled population lies between the cutoffs.
B. 0.424; subtract the central area from 1.
C. 0.788; use only the upper cumulative area.
D. 0.212; use only the lower cumulative area.

Answer: A

Standardize the bounds: zL=(54.4−64)12=-0.8 and zU=(73.6−64)12=0.8. P=0.788−0.212=0.576, so about 57.6% of modeled residents fall in the interval.

Why the other choices fail

Choice B: It does not match the requested calculation. The correct comparison or result is: 0.576; about 57.6% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice C: It omits a required part of the calculation. The correct comparison or result is: 0.576; about 57.6% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice D: It omits a required part of the calculation. The correct comparison or result is: 0.576; about 57.6% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 10. Normal Distribution

Model security wait time at a regional airport authority in North Valley during a yearly program evaluation as Normal(μ=64, σ=11). Find P(50.8≤X≤76.1) and interpret the area.

A. 0.864; use only the upper cumulative area.
B. 0.115; use only the lower cumulative area.
C. 0.251; subtract the central area from 1.
D. 0.749; about 74.9% of the modeled population lies between the cutoffs.

Answer: D

Standardize the bounds: zL=(50.8−64)11=-1.2 and zU=(76.1−64)11=1.1. P=0.864−0.115=0.749, so about 74.9% of modeled travelers fall in the interval.

Why the other choices fail

Choice A: It omits a required part of the calculation. The correct comparison or result is: 0.749; about 74.9% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice B: It omits a required part of the calculation. The correct comparison or result is: 0.749; about 74.9% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice C: It does not match the requested calculation. The correct comparison or result is: 0.749; about 74.9% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 11. Normal Distribution

Model sample concentration at a food safety laboratory in Lakeside district during a spring 2027 pilot as Normal(μ=100, σ=10). Find P(85≤X≤116.0) and interpret the area.

A. 0.122; subtract the central area from 1.
B. 0.945; use only the upper cumulative area.
C. 0.067; use only the lower cumulative area.
D. 0.878; about 87.8% of the modeled population lies between the cutoffs.

Answer: D

Standardize the bounds: zL=(85−100)10=-1.5 and zU=(116.0−100)10=1.6. P=0.945−0.067=0.878, so about 87.8% of modeled samples fall in the interval.

Why the other choices fail

Choice A: It does not match the requested calculation. The correct comparison or result is: 0.878; about 87.8% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice B: It omits a required part of the calculation. The correct comparison or result is: 0.878; about 87.8% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice C: It omits a required part of the calculation. The correct comparison or result is: 0.878; about 87.8% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 12. Normal Distribution

Model lesson completion at a digital learning platform in Midwest consortium during a pre-exam training cycle as Normal(μ=92, σ=13). Find P(85.5≤X≤97.2) and interpret the area.

A. 0.347; about 34.7% of the modeled population lies between the cutoffs.
B. 0.653; subtract the central area from 1.
C. 0.655; use only the upper cumulative area.
D. 0.309; use only the lower cumulative area.

Answer: A

Standardize the bounds: zL=(85.5−92)13=-0.5 and zU=(97.2−92)13=0.4. P=0.655−0.309=0.347, so about 34.7% of modeled learners fall in the interval.

Why the other choices fail

Choice B: It does not match the requested calculation. The correct comparison or result is: 0.347; about 34.7% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice C: It omits a required part of the calculation. The correct comparison or result is: 0.347; about 34.7% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice D: It omits a required part of the calculation. The correct comparison or result is: 0.347; about 34.7% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 13. Normal Distribution

Model crop yield at a farm cooperative in Metro East during a multiweek validation study as Normal(μ=98, σ=8). Find P(94≤X≤101.2) and interpret the area.

A. 0.347; about 34.7% of the modeled population lies between the cutoffs.
B. 0.309; use only the lower cumulative area.
C. 0.653; subtract the central area from 1.
D. 0.655; use only the upper cumulative area.

Answer: A

Standardize the bounds: zL=(94−98)8=-0.5 and zU=(101.2−98)8=0.4. P=0.655−0.309=0.347, so about 34.7% of modeled plots fall in the interval.

Why the other choices fail

Choice B: It omits a required part of the calculation. The correct comparison or result is: 0.347; about 34.7% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice C: It does not match the requested calculation. The correct comparison or result is: 0.347; about 34.7% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice D: It omits a required part of the calculation. The correct comparison or result is: 0.347; about 34.7% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 14. Normal Distribution

Model course completion at a community college in Prairie District during a follow-up evaluation period as Normal(μ=70, σ=16). Find P(62≤X≤82.8) and interpret the area.

A. 0.788; use only the upper cumulative area.
B. 0.309; use only the lower cumulative area.
C. 0.48; about 48.0% of the modeled population lies between the cutoffs.
D. 0.52; subtract the central area from 1.

Answer: C

Standardize the bounds: zL=(62−70)16=-0.5 and zU=(82.8−70)16=0.8. P=0.788−0.309=0.48, so about 48.0% of modeled enrolled learners fall in the interval.

Why the other choices fail

Choice A: It omits a required part of the calculation. The correct comparison or result is: 0.48; about 48.0% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice B: It omits a required part of the calculation. The correct comparison or result is: 0.48; about 48.0% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice D: It does not match the requested calculation. The correct comparison or result is: 0.48; about 48.0% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 15. Normal Distribution

Model program satisfaction at a city recreation department in Lakeside district during a regional benchmarking study as Normal(μ=76, σ=19). Find P(53.2≤X≤106.4) and interpret the area.

A. 0.945; use only the upper cumulative area.
B. 0.115; use only the lower cumulative area.
C. 0.17; subtract the central area from 1.
D. 0.83; about 83.0% of the modeled population lies between the cutoffs.

Answer: D

Standardize the bounds: zL=(53.2−76)19=-1.2 and zU=(106.4−76)19=1.6. P=0.945−0.115=0.83, so about 83.0% of modeled participants fall in the interval.

Why the other choices fail

Choice A: It omits a required part of the calculation. The correct comparison or result is: 0.83; about 83.0% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice B: It omits a required part of the calculation. The correct comparison or result is: 0.83; about 83.0% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice C: It does not match the requested calculation. The correct comparison or result is: 0.83; about 83.0% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 16. Normal Distribution

Model checkout time at a grocery cooperative in Midwest consortium during a baseline measurement week as Normal(μ=97, σ=17). Find P(83.4≤X≤103.8) and interpret the area.

A. 0.444; about 44.4% of the modeled population lies between the cutoffs.
B. 0.655; use only the upper cumulative area.
C. 0.556; subtract the central area from 1.
D. 0.212; use only the lower cumulative area.

Answer: A

Standardize the bounds: zL=(83.4−97)17=-0.8 and zU=(103.8−97)17=0.4. P=0.655−0.212=0.444, so about 44.4% of modeled customers fall in the interval.

Why the other choices fail

Choice B: It omits a required part of the calculation. The correct comparison or result is: 0.444; about 44.4% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice C: It does not match the requested calculation. The correct comparison or result is: 0.444; about 44.4% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice D: It omits a required part of the calculation. The correct comparison or result is: 0.444; about 44.4% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 17. Normal Distribution

Model lunch-program participation at a school district in Desert County during a randomized pilot period as Normal(μ=103, σ=12). Find P(97≤X≤107.8) and interpret the area.

A. 0.347; about 34.7% of the modeled population lies between the cutoffs.
B. 0.309; use only the lower cumulative area.
C. 0.655; use only the upper cumulative area.
D. 0.653; subtract the central area from 1.

Answer: A

Standardize the bounds: zL=(97−103)12=-0.5 and zU=(107.8−103)12=0.4. P=0.655−0.309=0.347, so about 34.7% of modeled students fall in the interval.

Why the other choices fail

Choice B: It omits a required part of the calculation. The correct comparison or result is: 0.347; about 34.7% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice C: It omits a required part of the calculation. The correct comparison or result is: 0.347; about 34.7% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice D: It does not match the requested calculation. The correct comparison or result is: 0.347; about 34.7% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 18. Normal Distribution

Model lunch-program participation at a school district in North Valley during a spring 2027 pilot as Normal(μ=104, σ=9). Find P(99.5≤X≤113.9) and interpret the area.

A. 0.444; subtract the central area from 1.
B. 0.556; about 55.6% of the modeled population lies between the cutoffs.
C. 0.309; use only the lower cumulative area.
D. 0.864; use only the upper cumulative area.

Answer: B

Standardize the bounds: zL=(99.5−104)9=-0.5 and zU=(113.9−104)9=1.1. P=0.864−0.309=0.556, so about 55.6% of modeled students fall in the interval.

Why the other choices fail

Choice A: It does not match the requested calculation. The correct comparison or result is: 0.556; about 55.6% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice C: It omits a required part of the calculation. The correct comparison or result is: 0.556; about 55.6% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice D: It omits a required part of the calculation. The correct comparison or result is: 0.556; about 55.6% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 19. Normal Distribution

Model daily energy output at a solar installer in New England network during a school-year data collection as Normal(μ=75, σ=7). Find P(69.4≤X≤86.2) and interpret the area.

A. 0.212; use only the lower cumulative area.
B. 0.267; subtract the central area from 1.
C. 0.733; about 73.3% of the modeled population lies between the cutoffs.
D. 0.945; use only the upper cumulative area.

Answer: C

Standardize the bounds: zL=(69.4−75)7=-0.8 and zU=(86.2−75)7=1.6. P=0.945−0.212=0.733, so about 73.3% of modeled installations fall in the interval.

Why the other choices fail

Choice A: It omits a required part of the calculation. The correct comparison or result is: 0.733; about 73.3% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice B: It does not match the requested calculation. The correct comparison or result is: 0.733; about 73.3% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice D: It omits a required part of the calculation. The correct comparison or result is: 0.733; about 73.3% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 20. Normal Distribution

Model crop yield at a farm cooperative in Sunbelt district during a winter readiness review as Normal(μ=115, σ=14). Find P(108.0≤X≤130.4) and interpret the area.

A. 0.864; use only the upper cumulative area.
B. 0.309; use only the lower cumulative area.
C. 0.556; about 55.6% of the modeled population lies between the cutoffs.
D. 0.444; subtract the central area from 1.

Answer: C

Standardize the bounds: zL=(108.0−115)14=-0.5 and zU=(130.4−115)14=1.1. P=0.864−0.309=0.556, so about 55.6% of modeled plots fall in the interval.

Why the other choices fail

Choice A: It omits a required part of the calculation. The correct comparison or result is: 0.556; about 55.6% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice B: It omits a required part of the calculation. The correct comparison or result is: 0.556; about 55.6% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice D: It does not match the requested calculation. The correct comparison or result is: 0.556; about 55.6% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 21. Normal Distribution

Model part diameter at a regional manufacturer in Prairie District during a yearly program evaluation as Normal(μ=106, σ=15). Find P(88≤X≤130.0) and interpret the area.

A. 0.115; use only the lower cumulative area.
B. 0.945; use only the upper cumulative area.
C. 0.83; about 83.0% of the modeled population lies between the cutoffs.
D. 0.17; subtract the central area from 1.

Answer: C

Standardize the bounds: zL=(88−106)15=-1.2 and zU=(130.0−106)15=1.6. P=0.945−0.115=0.83, so about 83.0% of modeled parts fall in the interval.

Why the other choices fail

Choice A: It omits a required part of the calculation. The correct comparison or result is: 0.83; about 83.0% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice B: It omits a required part of the calculation. The correct comparison or result is: 0.83; about 83.0% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice D: It does not match the requested calculation. The correct comparison or result is: 0.83; about 83.0% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 22. Normal Distribution

Model monthly household use at a municipal water office in Prairie District during a two-month observation window as Normal(μ=74, σ=12). Find P(68≤X≤78.8) and interpret the area.

A. 0.655; use only the upper cumulative area.
B. 0.347; about 34.7% of the modeled population lies between the cutoffs.
C. 0.309; use only the lower cumulative area.
D. 0.653; subtract the central area from 1.

Answer: B

Standardize the bounds: zL=(68−74)12=-0.5 and zU=(78.8−74)12=0.4. P=0.655−0.309=0.347, so about 34.7% of modeled accounts fall in the interval.

Why the other choices fail

Choice A: It omits a required part of the calculation. The correct comparison or result is: 0.347; about 34.7% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice C: It omits a required part of the calculation. The correct comparison or result is: 0.347; about 34.7% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice D: It does not match the requested calculation. The correct comparison or result is: 0.347; about 34.7% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 23. Normal Distribution

Model course completion at a community college in Pacific Northwest during a fall 2026 audit as Normal(μ=67, σ=17). Find P(46.6≤X≤80.6) and interpret the area.

A. 0.788; use only the upper cumulative area.
B. 0.115; use only the lower cumulative area.
C. 0.673; about 67.3% of the modeled population lies between the cutoffs.
D. 0.327; subtract the central area from 1.

Answer: C

Standardize the bounds: zL=(46.6−67)17=-1.2 and zU=(80.6−67)17=0.8. P=0.788−0.115=0.673, so about 67.3% of modeled enrolled learners fall in the interval.

Why the other choices fail

Choice A: It omits a required part of the calculation. The correct comparison or result is: 0.673; about 67.3% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice B: It omits a required part of the calculation. The correct comparison or result is: 0.673; about 67.3% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice D: It does not match the requested calculation. The correct comparison or result is: 0.673; about 67.3% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 24. Normal Distribution

Model course completion at a community college in South Harbor during a winter readiness review as Normal(μ=74, σ=10). Find P(66≤X≤78) and interpret the area.

A. 0.212; use only the lower cumulative area.
B. 0.444; about 44.4% of the modeled population lies between the cutoffs.
C. 0.655; use only the upper cumulative area.
D. 0.556; subtract the central area from 1.

Answer: B

Standardize the bounds: zL=(66−74)10=-0.8 and zU=(78−74)10=0.4. P=0.655−0.212=0.444, so about 44.4% of modeled enrolled learners fall in the interval.

Why the other choices fail

Choice A: It omits a required part of the calculation. The correct comparison or result is: 0.444; about 44.4% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice C: It omits a required part of the calculation. The correct comparison or result is: 0.444; about 44.4% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice D: It does not match the requested calculation. The correct comparison or result is: 0.444; about 44.4% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 25. Normal Distribution

Model sample concentration at a food safety laboratory in Westview during a weekday operations study as Normal(μ=105, σ=9). Find P(100.5≤X≤112.2) and interpret the area.

A. 0.309; use only the lower cumulative area.
B. 0.788; use only the upper cumulative area.
C. 0.52; subtract the central area from 1.
D. 0.48; about 48.0% of the modeled population lies between the cutoffs.

Answer: D

Standardize the bounds: zL=(100.5−105)9=-0.5 and zU=(112.2−105)9=0.8. P=0.788−0.309=0.48, so about 48.0% of modeled samples fall in the interval.

Why the other choices fail

Choice A: It omits a required part of the calculation. The correct comparison or result is: 0.48; about 48.0% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice B: It omits a required part of the calculation. The correct comparison or result is: 0.48; about 48.0% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice C: It does not match the requested calculation. The correct comparison or result is: 0.48; about 48.0% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 26. Normal Distribution

Model appointment wait time at a university advising center in Lakeside district during a school-year data collection as Normal(μ=68, σ=8). Find P(61.6≤X≤80.8) and interpret the area.

A. 0.733; about 73.3% of the modeled population lies between the cutoffs.
B. 0.945; use only the upper cumulative area.
C. 0.267; subtract the central area from 1.
D. 0.212; use only the lower cumulative area.

Answer: A

Standardize the bounds: zL=(61.6−68)8=-0.8 and zU=(80.8−68)8=1.6. P=0.945−0.212=0.733, so about 73.3% of modeled appointments fall in the interval.

Why the other choices fail

Choice B: It omits a required part of the calculation. The correct comparison or result is: 0.733; about 73.3% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice C: It does not match the requested calculation. The correct comparison or result is: 0.733; about 73.3% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice D: It omits a required part of the calculation. The correct comparison or result is: 0.733; about 73.3% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 27. Normal Distribution

Model response time at a municipal emergency dispatch center in Westview during a semester-long cohort study as Normal(μ=116, σ=9). Find P(111.5≤X≤123.2) and interpret the area.

A. 0.788; use only the upper cumulative area.
B. 0.309; use only the lower cumulative area.
C. 0.48; about 48.0% of the modeled population lies between the cutoffs.
D. 0.52; subtract the central area from 1.

Answer: C

Standardize the bounds: zL=(111.5−116)9=-0.5 and zU=(123.2−116)9=0.8. P=0.788−0.309=0.48, so about 48.0% of modeled calls fall in the interval.

Why the other choices fail

Choice A: It omits a required part of the calculation. The correct comparison or result is: 0.48; about 48.0% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice B: It omits a required part of the calculation. The correct comparison or result is: 0.48; about 48.0% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice D: It does not match the requested calculation. The correct comparison or result is: 0.48; about 48.0% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 28. Normal Distribution

Model checkout time at a grocery cooperative in Central County during a two-month observation window as Normal(μ=74, σ=18). Find P(59.6≤X≤88.4) and interpret the area.

A. 0.788; use only the upper cumulative area.
B. 0.576; about 57.6% of the modeled population lies between the cutoffs.
C. 0.424; subtract the central area from 1.
D. 0.212; use only the lower cumulative area.

Answer: B

Standardize the bounds: zL=(59.6−74)18=-0.8 and zU=(88.4−74)18=0.8. P=0.788−0.212=0.576, so about 57.6% of modeled customers fall in the interval.

Why the other choices fail

Choice A: It omits a required part of the calculation. The correct comparison or result is: 0.576; about 57.6% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice C: It does not match the requested calculation. The correct comparison or result is: 0.576; about 57.6% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice D: It omits a required part of the calculation. The correct comparison or result is: 0.576; about 57.6% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 29. Normal Distribution

Model weekly material weight at a recycling program in Pine Ridge during a yearly program evaluation as Normal(μ=69, σ=14). Find P(62≤X≤91.4) and interpret the area.

A. 0.945; use only the upper cumulative area.
B. 0.363; subtract the central area from 1.
C. 0.309; use only the lower cumulative area.
D. 0.637; about 63.7% of the modeled population lies between the cutoffs.

Answer: D

Standardize the bounds: zL=(62−69)14=-0.5 and zU=(91.4−69)14=1.6. P=0.945−0.309=0.637, so about 63.7% of modeled households fall in the interval.

Why the other choices fail

Choice A: It omits a required part of the calculation. The correct comparison or result is: 0.637; about 63.7% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice B: It does not match the requested calculation. The correct comparison or result is: 0.637; about 63.7% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice C: It omits a required part of the calculation. The correct comparison or result is: 0.637; about 63.7% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 30. Normal Distribution

Model daily energy output at a solar installer in Prairie District during a spring 2027 pilot as Normal(μ=116, σ=20). Find P(92≤X≤124.0) and interpret the area.

A. 0.54; about 54.0% of the modeled population lies between the cutoffs.
B. 0.655; use only the upper cumulative area.
C. 0.115; use only the lower cumulative area.
D. 0.46; subtract the central area from 1.

Answer: A

Standardize the bounds: zL=(92−116)20=-1.2 and zU=(124.0−116)20=0.4. P=0.655−0.115=0.54, so about 54.0% of modeled installations fall in the interval.

Why the other choices fail

Choice B: It omits a required part of the calculation. The correct comparison or result is: 0.54; about 54.0% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice C: It omits a required part of the calculation. The correct comparison or result is: 0.54; about 54.0% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice D: It does not match the requested calculation. The correct comparison or result is: 0.54; about 54.0% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 31. Normal Distribution

Model lunch-program participation at a school district in Sunbelt district during a summer implementation review as Normal(μ=105, σ=7). Find P(101.5≤X≤107.8) and interpret the area.

A. 0.655; use only the upper cumulative area.
B. 0.309; use only the lower cumulative area.
C. 0.653; subtract the central area from 1.
D. 0.347; about 34.7% of the modeled population lies between the cutoffs.

Answer: D

Standardize the bounds: zL=(101.5−105)7=-0.5 and zU=(107.8−105)7=0.4. P=0.655−0.309=0.347, so about 34.7% of modeled students fall in the interval.

Why the other choices fail

Choice A: It omits a required part of the calculation. The correct comparison or result is: 0.347; about 34.7% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice B: It omits a required part of the calculation. The correct comparison or result is: 0.347; about 34.7% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice C: It does not match the requested calculation. The correct comparison or result is: 0.347; about 34.7% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 32. Normal Distribution

Model crop yield at a farm cooperative in Desert County during a quarterly performance study as Normal(μ=75, σ=12). Find P(60.6≤X≤94.2) and interpret the area.

A. 0.83; about 83.0% of the modeled population lies between the cutoffs.
B. 0.945; use only the upper cumulative area.
C. 0.17; subtract the central area from 1.
D. 0.115; use only the lower cumulative area.

Answer: A

Standardize the bounds: zL=(60.6−75)12=-1.2 and zU=(94.2−75)12=1.6. P=0.945−0.115=0.83, so about 83.0% of modeled plots fall in the interval.

Why the other choices fail

Choice B: It omits a required part of the calculation. The correct comparison or result is: 0.83; about 83.0% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice C: It does not match the requested calculation. The correct comparison or result is: 0.83; about 83.0% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice D: It omits a required part of the calculation. The correct comparison or result is: 0.83; about 83.0% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 33. Normal Distribution

Model program satisfaction at a city recreation department in Cedar Grove during a multiweek validation study as Normal(μ=111, σ=19). Find P(95.8≤X≤141.4) and interpret the area.

A. 0.212; use only the lower cumulative area.
B. 0.733; about 73.3% of the modeled population lies between the cutoffs.
C. 0.267; subtract the central area from 1.
D. 0.945; use only the upper cumulative area.

Answer: B

Standardize the bounds: zL=(95.8−111)19=-0.8 and zU=(141.4−111)19=1.6. P=0.945−0.212=0.733, so about 73.3% of modeled participants fall in the interval.

Why the other choices fail

Choice A: It omits a required part of the calculation. The correct comparison or result is: 0.733; about 73.3% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice C: It does not match the requested calculation. The correct comparison or result is: 0.733; about 73.3% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice D: It omits a required part of the calculation. The correct comparison or result is: 0.733; about 73.3% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 34. Normal Distribution

Model weekly program attendance at a county library in Pacific Northwest during a summer implementation review as Normal(μ=70, σ=11). Find P(64.5≤X≤74.4) and interpret the area.

A. 0.653; subtract the central area from 1.
B. 0.347; about 34.7% of the modeled population lies between the cutoffs.
C. 0.655; use only the upper cumulative area.
D. 0.309; use only the lower cumulative area.

Answer: B

Standardize the bounds: zL=(64.5−70)11=-0.5 and zU=(74.4−70)11=0.4. P=0.655−0.309=0.347, so about 34.7% of modeled visitors fall in the interval.

Why the other choices fail

Choice A: It does not match the requested calculation. The correct comparison or result is: 0.347; about 34.7% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice C: It omits a required part of the calculation. The correct comparison or result is: 0.347; about 34.7% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice D: It omits a required part of the calculation. The correct comparison or result is: 0.347; about 34.7% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 35. Normal Distribution

Model appointment completion at a regional hospital in Desert County during a fall 2026 audit as Normal(μ=95, σ=13). Find P(79.4≤X≤105.4) and interpret the area.

A. 0.673; about 67.3% of the modeled population lies between the cutoffs.
B. 0.115; use only the lower cumulative area.
C. 0.327; subtract the central area from 1.
D. 0.788; use only the upper cumulative area.

Answer: A

Standardize the bounds: zL=(79.4−95)13=-1.2 and zU=(105.4−95)13=0.8. P=0.788−0.115=0.673, so about 67.3% of modeled patients fall in the interval.

Why the other choices fail

Choice B: It omits a required part of the calculation. The correct comparison or result is: 0.673; about 67.3% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice C: It does not match the requested calculation. The correct comparison or result is: 0.673; about 67.3% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice D: It omits a required part of the calculation. The correct comparison or result is: 0.673; about 67.3% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Question 36. Normal Distribution

Model part diameter at a regional manufacturer in Desert County during a fall 2026 audit as Normal(μ=82, σ=11). Find P(76.5≤X≤94.1) and interpret the area.

A. 0.444; subtract the central area from 1.
B. 0.556; about 55.6% of the modeled population lies between the cutoffs.
C. 0.864; use only the upper cumulative area.
D. 0.309; use only the lower cumulative area.

Answer: B

Standardize the bounds: zL=(76.5−82)11=-0.5 and zU=(94.1−82)11=1.1. P=0.864−0.309=0.556, so about 55.6% of modeled parts fall in the interval.

Why the other choices fail

Choice A: It does not match the requested calculation. The correct comparison or result is: 0.556; about 55.6% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice C: It omits a required part of the calculation. The correct comparison or result is: 0.556; about 55.6% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.
Choice D: It omits a required part of the calculation. The correct comparison or result is: 0.556; about 55.6% of the modeled population lies between the cutoffs. Key check: A normal model is determined by μ and σ; probability is area, not curve height.

Normal Distribution 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: Normal Distribution

Scenario. Model part diameter at a regional manufacturer in Capital Region during a monthly quality review as Normal(μ=115, σ=18). Find P(88≤X≤122.2) and interpret the area.

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

Standardize the bounds: zL=(88−115)18=-1.5 and zU=(122.2−115)18=0.4. P=0.655−0.067=0.589, so about 58.9% of modeled parts fall in the interval.

FRQ set 2: Normal Distribution

Scenario. Model vaccination appointment completion at a public health department in North Valley during a community outreach cycle as Normal(μ=56, σ=7). Find P(50.4≤X≤61.6) and interpret the area.

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

Standardize the bounds: zL=(50.4−56)7=-0.8 and zU=(61.6−56)7=0.8. P=0.788−0.212=0.576, so about 57.6% of modeled residents fall in the interval.

FRQ set 3: Normal Distribution

Scenario. Model lunch-program participation at a school district in Pine Ridge during a school-year data collection as Normal(μ=99, σ=19). Find P(76.2≤X≤106.6) and interpret the area.

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

Standardize the bounds: zL=(76.2−99)19=-1.2 and zU=(106.6−99)19=0.4. P=0.655−0.115=0.54, so about 54.0% of modeled students fall in the interval.

FRQ set 4: Normal Distribution

Scenario. Model algebra benchmark completion at a public high school in New England network during a winter readiness review as Normal(μ=60, σ=19). Find P(31.5≤X≤80.9) and interpret the area.

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

Standardize the bounds: zL=(31.5−60)19=-1.5 and zU=(80.9−60)19=1.1. P=0.864−0.067=0.798, so about 79.8% of modeled students fall in the interval.

FRQ set 5: Normal Distribution

Scenario. Model response time at a municipal emergency dispatch center in Lakeside district during a community outreach cycle as Normal(μ=85, σ=8). Find P(75.4≤X≤88.2) and interpret the area.

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

Standardize the bounds: zL=(75.4−85)8=-1.2 and zU=(88.2−85)8=0.4. P=0.655−0.115=0.54, so about 54.0% of modeled calls fall in the interval.

FRQ set 6: Normal Distribution

Scenario. Model weekly program attendance at a county library in Riverbend during a community outreach cycle as Normal(μ=118, σ=11). Find P(104.8≤X≤122.4) and interpret the area.

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

Standardize the bounds: zL=(104.8−118)11=-1.2 and zU=(122.4−118)11=0.4. P=0.655−0.115=0.54, so about 54.0% of modeled visitors fall in the interval.

FRQ set 7: Normal Distribution

Scenario. Model security wait time at a regional airport authority in Pine Ridge during a multiweek validation study as Normal(μ=114, σ=11). Find P(108.5≤X≤118.4) and interpret the area.

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

Standardize the bounds: zL=(108.5−114)11=-0.5 and zU=(118.4−114)11=0.4. P=0.655−0.309=0.347, so about 34.7% of modeled travelers fall in the interval.

FRQ set 8: Normal Distribution

Scenario. Model on-time arrival at a city transit agency in Desert County during a randomized pilot period as Normal(μ=114, σ=10). Find P(102.0≤X≤125.0) and interpret the area.

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

Standardize the bounds: zL=(102.0−114)10=-1.2 and zU=(125.0−114)10=1.1. P=0.864−0.115=0.749, so about 74.9% of modeled bus trips fall in the interval.

FRQ set 9: Normal Distribution

Scenario. Model program satisfaction at a city recreation department in South Harbor during a summer implementation review as Normal(μ=83, σ=18). Find P(74≤X≤111.8) and interpret the area.

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

Standardize the bounds: zL=(74−83)18=-0.5 and zU=(111.8−83)18=1.6. P=0.945−0.309=0.637, so about 63.7% of modeled participants fall in the interval.

FRQ set 10: Normal Distribution

Scenario. Model security wait time at a regional airport authority in Westview during a pre-exam training cycle as Normal(μ=78, σ=19). Find P(68.5≤X≤108.4) and interpret the area.

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

Standardize the bounds: zL=(68.5−78)19=-0.5 and zU=(108.4−78)19=1.6. P=0.945−0.309=0.637, so about 63.7% of modeled travelers fall in the interval.

FRQ set 11: Normal Distribution

Scenario. Model checkout time at a grocery cooperative in Desert County during a quarterly performance study as Normal(μ=70, σ=15). Find P(62.5≤X≤76) and interpret the area.

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

Standardize the bounds: zL=(62.5−70)15=-0.5 and zU=(76−70)15=0.4. P=0.655−0.309=0.347, so about 34.7% of modeled customers fall in the interval.

FRQ set 12: Normal Distribution

Scenario. Model appointment completion at a regional hospital in Pacific Northwest during a service-improvement study as Normal(μ=99, σ=13). Find P(83.4≤X≤104.2) and interpret the area.

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

Standardize the bounds: zL=(83.4−99)13=-1.2 and zU=(104.2−99)13=0.4. P=0.655−0.115=0.54, so about 54.0% of modeled patients fall in the interval.

FRQ set 13: Normal Distribution

Scenario. Model vaccination appointment completion at a public health department in Mountain Region during a pre-exam training cycle as Normal(μ=58, σ=14). Find P(51≤X≤73.4) and interpret the area.

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

Standardize the bounds: zL=(51−58)14=-0.5 and zU=(73.4−58)14=1.1. P=0.864−0.309=0.556, so about 55.6% of modeled residents fall in the interval.

FRQ set 14: Normal Distribution

Scenario. Model response time at a municipal emergency dispatch center in North Valley during a fall 2026 audit as Normal(μ=66, σ=16). Find P(46.8≤X≤91.6) and interpret the area.

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

Standardize the bounds: zL=(46.8−66)16=-1.2 and zU=(91.6−66)16=1.6. P=0.945−0.115=0.83, so about 83.0% of modeled calls fall in the interval.