Sampling Distributions: Center, Spread, and Shape

AP Statistics Topic Guide

Sampling Distributions: Center, Spread, and Shape

Understand what a sampling distribution is, how its center and standard error behave, and why repeated-sample variation is different from variation among individual observations.

StatusCurrent AP Statistics core

Main keywordsampling distributions

Worked analysis24

Practice30 MCQs + 8 FRQs

Study progress0 completed

Sampling Distributions: direct answer

Understand what a sampling distribution is, how its center and standard error behave, and why repeated-sample variation is different from variation among individual observations.

This page uses sampling distributions as its single primary search focus. Every instructional block and every retained practice item is tied to that title intent rather than to a generic AP Statistics question-bank template.

Quick reference for Sampling Distributions: Center, Spread, and Shape

Sampling Distributions: Center, Spread, and Shape quick reference
Object that varies	A statistic computed from repeated samples.
Center	The parameter when the statistic is unbiased.
Spread	Standard error: sampling variability of the statistic.
Shape	Depends on the statistic, population/model, and sample size.
Design warning	Small standard error does not remove selection bias.

Concept mastery: Sampling Distributions: Center, Spread, and Shape

A sampling distribution is a distribution of statistics

Imagine repeatedly drawing samples by the same design and calculating the same statistic each time. The resulting distribution of those statistics is a sampling distribution. It is not the histogram of the original population and not the distribution of observations inside one sample.

Center connects statistics to parameters

An unbiased statistic has a sampling distribution centered at the parameter it estimates. Sample means are centered at μ and sample proportions are centered at p under the usual random-sampling framework. Bias is a systematic displacement of that center, not merely random scatter.

Spread is measured by standard error

Standard error quantifies how much the statistic varies from sample to sample. Larger samples generally reduce standard error because they average or aggregate more information. The formulas differ by statistic, but the interpretation is always about repeated-sample variability.

Shape depends on model and sample size

Some sampling distributions are exactly described under strong assumptions; others become approximately normal under suitable sample-size and population conditions. Never infer normality solely from a formula for standard error.

The sampling design controls what inference means

A mathematically small standard error does not compensate for selection bias. Randomization and representativeness determine whether the repeated-sampling model supports population conclusions; standard error only quantifies sampling variability under the assumed design.

Worked analysis for Sampling Distributions: Center, Spread, and Shape

Sampling-distribution case 1: Urban Transit Riders

A population of urban transit riders has mean 56 and standard deviation 9. Imagine taking many independent random samples of size 25 and recording each sample mean. The center of that sampling distribution is 56, while its standard error is 9/√25≈1.800. The population SD describes individual variation; the standard error describes sample-to-sample variation in x̄.

For a cutoff 57.13, standardization gives z=(57.13−56)/1.800≈0.63. The probability question is about the statistic across repeated samples, not about a randomly selected individual. This distinction is the main interpretive checkpoint in case 1.

Sampling-distribution case 2: Regional Hospital Visits

A population of regional hospital visits has mean 60 and standard deviation 10. Imagine taking many independent random samples of size 36 and recording each sample mean. The center of that sampling distribution is 60, while its standard error is 10/√36≈1.667. The population SD describes individual variation; the standard error describes sample-to-sample variation in x̄.

For a cutoff 61.35, standardization gives z=(61.35−60)/1.667≈0.81. The probability question is about the statistic across repeated samples, not about a randomly selected individual. This distinction is the main interpretive checkpoint in case 2.

Sampling-distribution case 3: Community College Placement Scores

A population of community college placement scores has mean 64 and standard deviation 11. Imagine taking many independent random samples of size 49 and recording each sample mean. The center of that sampling distribution is 64, while its standard error is 11/√49≈1.571. The population SD describes individual variation; the standard error describes sample-to-sample variation in x̄.

For a cutoff 65.56, standardization gives z=(65.56−64)/1.571≈0.99. The probability question is about the statistic across repeated samples, not about a randomly selected individual. This distinction is the main interpretive checkpoint in case 3.

Sampling-distribution case 4: Warehouse Order Times

A population of warehouse order times has mean 68 and standard deviation 12. Imagine taking many independent random samples of size 64 and recording each sample mean. The center of that sampling distribution is 68, while its standard error is 12/√64≈1.500. The population SD describes individual variation; the standard error describes sample-to-sample variation in x̄.

For a cutoff 68.68, standardization gives z=(68.68−68)/1.500≈0.45. The probability question is about the statistic across repeated samples, not about a randomly selected individual. This distinction is the main interpretive checkpoint in case 4.

Sampling-distribution case 5: Public-Library Checkouts

A population of public-library checkouts has mean 72 and standard deviation 8. Imagine taking many independent random samples of size 81 and recording each sample mean. The center of that sampling distribution is 72, while its standard error is 8/√81≈0.889. The population SD describes individual variation; the standard error describes sample-to-sample variation in x̄.

For a cutoff 72.56, standardization gives z=(72.56−72)/0.889≈0.63. The probability question is about the statistic across repeated samples, not about a randomly selected individual. This distinction is the main interpretive checkpoint in case 5.

Sampling-distribution case 6: Solar-Panel Output Readings

A population of solar-panel output readings has mean 76 and standard deviation 9. Imagine taking many independent random samples of size 16 and recording each sample mean. The center of that sampling distribution is 76, while its standard error is 9/√16≈2.250. The population SD describes individual variation; the standard error describes sample-to-sample variation in x̄.

For a cutoff 77.82, standardization gives z=(77.82−76)/2.250≈0.81. The probability question is about the statistic across repeated samples, not about a randomly selected individual. This distinction is the main interpretive checkpoint in case 6.

Sampling-distribution case 7: School Attendance Rates

A population of school attendance rates has mean 52 and standard deviation 10. Imagine taking many independent random samples of size 25 and recording each sample mean. The center of that sampling distribution is 52, while its standard error is 10/√25≈2.000. The population SD describes individual variation; the standard error describes sample-to-sample variation in x̄.

For a cutoff 53.98, standardization gives z=(53.98−52)/2.000≈0.99. The probability question is about the statistic across repeated samples, not about a randomly selected individual. This distinction is the main interpretive checkpoint in case 7.

Sampling-distribution case 8: Restaurant Service Times

A population of restaurant service times has mean 56 and standard deviation 11. Imagine taking many independent random samples of size 36 and recording each sample mean. The center of that sampling distribution is 56, while its standard error is 11/√36≈1.833. The population SD describes individual variation; the standard error describes sample-to-sample variation in x̄.

For a cutoff 56.82, standardization gives z=(56.82−56)/1.833≈0.45. The probability question is about the statistic across repeated samples, not about a randomly selected individual. This distinction is the main interpretive checkpoint in case 8.

Sampling-distribution case 9: Wildlife Tag Measurements

A population of wildlife tag measurements has mean 60 and standard deviation 12. Imagine taking many independent random samples of size 49 and recording each sample mean. The center of that sampling distribution is 60, while its standard error is 12/√49≈1.714. The population SD describes individual variation; the standard error describes sample-to-sample variation in x̄.

For a cutoff 61.08, standardization gives z=(61.08−60)/1.714≈0.63. The probability question is about the statistic across repeated samples, not about a randomly selected individual. This distinction is the main interpretive checkpoint in case 9.

Sampling-distribution case 10: Municipal Water-Use Records

A population of municipal water-use records has mean 64 and standard deviation 8. Imagine taking many independent random samples of size 64 and recording each sample mean. The center of that sampling distribution is 64, while its standard error is 8/√64≈1.000. The population SD describes individual variation; the standard error describes sample-to-sample variation in x̄.

For a cutoff 64.81, standardization gives z=(64.81−64)/1.000≈0.81. The probability question is about the statistic across repeated samples, not about a randomly selected individual. This distinction is the main interpretive checkpoint in case 10.

Sampling-distribution case 11: Online Course Completion Times

A population of online course completion times has mean 68 and standard deviation 9. Imagine taking many independent random samples of size 81 and recording each sample mean. The center of that sampling distribution is 68, while its standard error is 9/√81≈1.000. The population SD describes individual variation; the standard error describes sample-to-sample variation in x̄.

For a cutoff 68.99, standardization gives z=(68.99−68)/1.000≈0.99. The probability question is about the statistic across repeated samples, not about a randomly selected individual. This distinction is the main interpretive checkpoint in case 11.

Sampling-distribution case 12: Farm Yield Measurements

A population of farm yield measurements has mean 72 and standard deviation 10. Imagine taking many independent random samples of size 16 and recording each sample mean. The center of that sampling distribution is 72, while its standard error is 10/√16≈2.500. The population SD describes individual variation; the standard error describes sample-to-sample variation in x̄.

For a cutoff 73.12, standardization gives z=(73.12−72)/2.500≈0.45. The probability question is about the statistic across repeated samples, not about a randomly selected individual. This distinction is the main interpretive checkpoint in case 12.

Sampling-distribution case 13: Clinic Appointment Waits

A population of clinic appointment waits has mean 76 and standard deviation 11. Imagine taking many independent random samples of size 25 and recording each sample mean. The center of that sampling distribution is 76, while its standard error is 11/√25≈2.200. The population SD describes individual variation; the standard error describes sample-to-sample variation in x̄.

For a cutoff 77.39, standardization gives z=(77.39−76)/2.200≈0.63. The probability question is about the statistic across repeated samples, not about a randomly selected individual. This distinction is the main interpretive checkpoint in case 13.

Sampling-distribution case 14: Manufacturing Fill Weights

A population of manufacturing fill weights has mean 52 and standard deviation 12. Imagine taking many independent random samples of size 36 and recording each sample mean. The center of that sampling distribution is 52, while its standard error is 12/√36≈2.000. The population SD describes individual variation; the standard error describes sample-to-sample variation in x̄.

For a cutoff 53.62, standardization gives z=(53.62−52)/2.000≈0.81. The probability question is about the statistic across repeated samples, not about a randomly selected individual. This distinction is the main interpretive checkpoint in case 14.

Sampling-distribution case 15: County Commute Times

A population of county commute times has mean 56 and standard deviation 8. Imagine taking many independent random samples of size 49 and recording each sample mean. The center of that sampling distribution is 56, while its standard error is 8/√49≈1.143. The population SD describes individual variation; the standard error describes sample-to-sample variation in x̄.

For a cutoff 57.13, standardization gives z=(57.13−56)/1.143≈0.99. The probability question is about the statistic across repeated samples, not about a randomly selected individual. This distinction is the main interpretive checkpoint in case 15.

Sampling-distribution case 16: Energy Meter Readings

A population of energy meter readings has mean 60 and standard deviation 9. Imagine taking many independent random samples of size 64 and recording each sample mean. The center of that sampling distribution is 60, while its standard error is 9/√64≈1.125. The population SD describes individual variation; the standard error describes sample-to-sample variation in x̄.

For a cutoff 60.51, standardization gives z=(60.51−60)/1.125≈0.45. The probability question is about the statistic across repeated samples, not about a randomly selected individual. This distinction is the main interpretive checkpoint in case 16.

Sampling-distribution case 17: University Advising Durations

A population of university advising durations has mean 64 and standard deviation 10. Imagine taking many independent random samples of size 81 and recording each sample mean. The center of that sampling distribution is 64, while its standard error is 10/√81≈1.111. The population SD describes individual variation; the standard error describes sample-to-sample variation in x̄.

For a cutoff 64.70, standardization gives z=(64.70−64)/1.111≈0.63. The probability question is about the statistic across repeated samples, not about a randomly selected individual. This distinction is the main interpretive checkpoint in case 17.

Sampling-distribution case 18: Sports Training Measurements

A population of sports training measurements has mean 68 and standard deviation 11. Imagine taking many independent random samples of size 16 and recording each sample mean. The center of that sampling distribution is 68, while its standard error is 11/√16≈2.750. The population SD describes individual variation; the standard error describes sample-to-sample variation in x̄.

For a cutoff 70.23, standardization gives z=(70.23−68)/2.750≈0.81. The probability question is about the statistic across repeated samples, not about a randomly selected individual. This distinction is the main interpretive checkpoint in case 18.

Sampling-distribution case 19: Call-Center Resolution Times

A population of call-center resolution times has mean 72 and standard deviation 12. Imagine taking many independent random samples of size 25 and recording each sample mean. The center of that sampling distribution is 72, while its standard error is 12/√25≈2.400. The population SD describes individual variation; the standard error describes sample-to-sample variation in x̄.

For a cutoff 74.38, standardization gives z=(74.38−72)/2.400≈0.99. The probability question is about the statistic across repeated samples, not about a randomly selected individual. This distinction is the main interpretive checkpoint in case 19.

Sampling-distribution case 20: Retail Basket Totals

A population of retail basket totals has mean 76 and standard deviation 8. Imagine taking many independent random samples of size 36 and recording each sample mean. The center of that sampling distribution is 76, while its standard error is 8/√36≈1.333. The population SD describes individual variation; the standard error describes sample-to-sample variation in x̄.

For a cutoff 76.60, standardization gives z=(76.60−76)/1.333≈0.45. The probability question is about the statistic across repeated samples, not about a randomly selected individual. This distinction is the main interpretive checkpoint in case 20.

Sampling-distribution case 21: Lab Assay Values

A population of lab assay values has mean 52 and standard deviation 9. Imagine taking many independent random samples of size 49 and recording each sample mean. The center of that sampling distribution is 52, while its standard error is 9/√49≈1.286. The population SD describes individual variation; the standard error describes sample-to-sample variation in x̄.

For a cutoff 52.81, standardization gives z=(52.81−52)/1.286≈0.63. The probability question is about the statistic across repeated samples, not about a randomly selected individual. This distinction is the main interpretive checkpoint in case 21.

Sampling-distribution case 22: Shipping Transit Times

A population of shipping transit times has mean 56 and standard deviation 10. Imagine taking many independent random samples of size 64 and recording each sample mean. The center of that sampling distribution is 56, while its standard error is 10/√64≈1.250. The population SD describes individual variation; the standard error describes sample-to-sample variation in x̄.

For a cutoff 57.01, standardization gives z=(57.01−56)/1.250≈0.81. The probability question is about the statistic across repeated samples, not about a randomly selected individual. This distinction is the main interpretive checkpoint in case 22.

Sampling-distribution case 23: District Test Scores

A population of district test scores has mean 60 and standard deviation 11. Imagine taking many independent random samples of size 81 and recording each sample mean. The center of that sampling distribution is 60, while its standard error is 11/√81≈1.222. The population SD describes individual variation; the standard error describes sample-to-sample variation in x̄.

For a cutoff 61.21, standardization gives z=(61.21−60)/1.222≈0.99. The probability question is about the statistic across repeated samples, not about a randomly selected individual. This distinction is the main interpretive checkpoint in case 23.

Sampling-distribution case 24: Park Visitor Durations

A population of park visitor durations has mean 64 and standard deviation 12. Imagine taking many independent random samples of size 16 and recording each sample mean. The center of that sampling distribution is 64, while its standard error is 12/√16≈3.000. The population SD describes individual variation; the standard error describes sample-to-sample variation in x̄.

For a cutoff 65.35, standardization gives z=(65.35−64)/3.000≈0.45. The probability question is about the statistic across repeated samples, not about a randomly selected individual. This distinction is the main interpretive checkpoint in case 24.

Sampling-distribution case 25: Airport Security Waits

A population of airport security waits has mean 68 and standard deviation 8. Imagine taking many independent random samples of size 25 and recording each sample mean. The center of that sampling distribution is 68, while its standard error is 8/√25≈1.600. The population SD describes individual variation; the standard error describes sample-to-sample variation in x̄.

For a cutoff 69.01, standardization gives z=(69.01−68)/1.600≈0.63. The probability question is about the statistic across repeated samples, not about a randomly selected individual. This distinction is the main interpretive checkpoint in case 25.

Sampling-distribution case 26: Pharmacy Prescription Times

A population of pharmacy prescription times has mean 72 and standard deviation 9. Imagine taking many independent random samples of size 36 and recording each sample mean. The center of that sampling distribution is 72, while its standard error is 9/√36≈1.500. The population SD describes individual variation; the standard error describes sample-to-sample variation in x̄.

For a cutoff 73.22, standardization gives z=(73.22−72)/1.500≈0.81. The probability question is about the statistic across repeated samples, not about a randomly selected individual. This distinction is the main interpretive checkpoint in case 26.

Sampling-distribution case 27: Bicycle Commute Distances

A population of bicycle commute distances has mean 76 and standard deviation 10. Imagine taking many independent random samples of size 49 and recording each sample mean. The center of that sampling distribution is 76, while its standard error is 10/√49≈1.429. The population SD describes individual variation; the standard error describes sample-to-sample variation in x̄.

For a cutoff 77.41, standardization gives z=(77.41−76)/1.429≈0.99. The probability question is about the statistic across repeated samples, not about a randomly selected individual. This distinction is the main interpretive checkpoint in case 27.

Sampling-distribution case 28: Community Garden Yields

A population of community garden yields has mean 52 and standard deviation 11. Imagine taking many independent random samples of size 64 and recording each sample mean. The center of that sampling distribution is 52, while its standard error is 11/√64≈1.375. The population SD describes individual variation; the standard error describes sample-to-sample variation in x̄.

For a cutoff 52.62, standardization gives z=(52.62−52)/1.375≈0.45. The probability question is about the statistic across repeated samples, not about a randomly selected individual. This distinction is the main interpretive checkpoint in case 28.

Sampling-distribution case 29: Emergency Response Durations

A population of emergency response durations has mean 56 and standard deviation 12. Imagine taking many independent random samples of size 81 and recording each sample mean. The center of that sampling distribution is 56, while its standard error is 12/√81≈1.333. The population SD describes individual variation; the standard error describes sample-to-sample variation in x̄.

For a cutoff 56.84, standardization gives z=(56.84−56)/1.333≈0.63. The probability question is about the statistic across repeated samples, not about a randomly selected individual. This distinction is the main interpretive checkpoint in case 29.

Sampling-distribution case 30: College Credit Loads

A population of college credit loads has mean 60 and standard deviation 8. Imagine taking many independent random samples of size 16 and recording each sample mean. The center of that sampling distribution is 60, while its standard error is 8/√16≈2.000. The population SD describes individual variation; the standard error describes sample-to-sample variation in x̄.

For a cutoff 61.62, standardization gives z=(61.62−60)/2.000≈0.81. The probability question is about the statistic across repeated samples, not about a randomly selected individual. This distinction is the main interpretive checkpoint in case 30.

Sampling-distribution case 31: Household Electricity Use

A population of household electricity use has mean 64 and standard deviation 9. Imagine taking many independent random samples of size 25 and recording each sample mean. The center of that sampling distribution is 64, while its standard error is 9/√25≈1.800. The population SD describes individual variation; the standard error describes sample-to-sample variation in x̄.

For a cutoff 65.78, standardization gives z=(65.78−64)/1.800≈0.99. The probability question is about the statistic across repeated samples, not about a randomly selected individual. This distinction is the main interpretive checkpoint in case 31.

Sampling-distribution case 32: River Flow Measurements

A population of river flow measurements has mean 68 and standard deviation 10. Imagine taking many independent random samples of size 36 and recording each sample mean. The center of that sampling distribution is 68, while its standard error is 10/√36≈1.667. The population SD describes individual variation; the standard error describes sample-to-sample variation in x̄.

For a cutoff 68.75, standardization gives z=(68.75−68)/1.667≈0.45. The probability question is about the statistic across repeated samples, not about a randomly selected individual. This distinction is the main interpretive checkpoint in case 32.

Sampling-distribution case 33: Mobile Data Usage

A population of mobile data usage has mean 72 and standard deviation 11. Imagine taking many independent random samples of size 49 and recording each sample mean. The center of that sampling distribution is 72, while its standard error is 11/√49≈1.571. The population SD describes individual variation; the standard error describes sample-to-sample variation in x̄.

For a cutoff 72.99, standardization gives z=(72.99−72)/1.571≈0.63. The probability question is about the statistic across repeated samples, not about a randomly selected individual. This distinction is the main interpretive checkpoint in case 33.

Sampling-distribution case 34: Food Delivery Times

A population of food delivery times has mean 76 and standard deviation 12. Imagine taking many independent random samples of size 64 and recording each sample mean. The center of that sampling distribution is 76, while its standard error is 12/√64≈1.500. The population SD describes individual variation; the standard error describes sample-to-sample variation in x̄.

For a cutoff 77.22, standardization gives z=(77.22−76)/1.500≈0.81. The probability question is about the statistic across repeated samples, not about a randomly selected individual. This distinction is the main interpretive checkpoint in case 34.

Sampling Distributions: Center, Spread, and Shape: multiple-choice practice

Question 1. Sampling Distributions

A population at a regional airport authority in Lakeside district during a semester-long cohort study has mean 109 and SD 23. For random samples of size 25, find the mean and SD of x̄ and P(x̄>114.8) under a normal/CLT approximation.

A. Mean 109, SD 23, because averaging does not change spread.
B. Mean 109, SD 0.92, probability 0.8944.
C. Mean 4.36, SD 23, probability 0.1056.
D. Mean 109, SD 4.6, probability 0.1056.

Answer: D

μ_x̄=μ=109. σ_x̄=σ√n=23√25=4.6. The cutoff has z=(114.8−109)4.6=1.25, so P(x̄>cutoff)=0.1056.

Question 2. Sampling Distributions

A population at a digital learning platform in Pine Ridge during a semester-long cohort study has mean 75 and SD 9. For random samples of size 100, find the mean and SD of x̄ and P(x̄>75.72) under a normal/CLT approximation.

A. Mean 75, SD 0.09, probability 0.7881.
B. Mean 75, SD 9, because averaging does not change spread.
C. Mean 75, SD 0.9, probability 0.2119.
D. Mean 0.75, SD 9, probability 0.2119.

Answer: C

μ_x̄=μ=75. σ_x̄=σ√n=9√100=0.9. The cutoff has z=(75.72−75)0.9=0.8, so P(x̄>cutoff)=0.2119.

Question 3. Sampling Distributions

A population at a university advising center in Westview during a two-month observation window has mean 56 and SD 23. For random samples of size 49, find the mean and SD of x̄ and P(x̄>60.11) under a normal/CLT approximation.

A. Mean 56, SD 23, because averaging does not change spread.
B. Mean 56, SD 0.469, probability 0.8945.
C. Mean 56, SD 3.286, probability 0.1055.
D. Mean 1.143, SD 23, probability 0.1055.

Answer: C

μ_x̄=μ=56. σ_x̄=σ√n=23√49=3.286. The cutoff has z=(60.11−56)3.286=1.251, so P(x̄>cutoff)=0.1055.

Question 4. Sampling Distributions

A population at a city recreation department in Midwest consortium during a spring 2027 pilot has mean 54 and SD 21. For random samples of size 100, find the mean and SD of x̄ and P(x̄>56.62) under a normal/CLT approximation.

A. Mean 54, SD 21, because averaging does not change spread.
B. Mean 54, SD 0.21, probability 0.8939.
C. Mean 54, SD 2.1, probability 0.1061.
D. Mean 0.54, SD 21, probability 0.1061.

Answer: C

μ_x̄=μ=54. σ_x̄=σ√n=21√100=2.1. The cutoff has z=(56.62−54)2.1=1.248, so P(x̄>cutoff)=0.1061.

Question 5. Sampling Distributions

A population at a municipal emergency dispatch center in Pacific Northwest during a semester-long cohort study has mean 48 and SD 8. For random samples of size 49, find the mean and SD of x̄ and P(x̄>49.14) under a normal/CLT approximation.

A. Mean 48, SD 0.163, probability 0.8407.
B. Mean 48, SD 1.143, probability 0.1593.
C. Mean 0.98, SD 8, probability 0.1593.
D. Mean 48, SD 8, because averaging does not change spread.

Answer: B

μ_x̄=μ=48. σ_x̄=σ√n=8√49=1.143. The cutoff has z=(49.14−48)1.143=0.998, so P(x̄>cutoff)=0.1593.

Question 6. Sampling Distributions

A population at a housing authority in New England network during a monthly quality review has mean 99 and SD 19. For random samples of size 64, find the mean and SD of x̄ and P(x̄>101.4) under a normal/CLT approximation.

A. Mean 99, SD 2.375, probability 0.1581.
B. Mean 1.547, SD 19, probability 0.1581.
C. Mean 99, SD 19, because averaging does not change spread.
D. Mean 99, SD 0.297, probability 0.8419.

Answer: A

μ_x̄=μ=99. σ_x̄=σ√n=19√64=2.375. The cutoff has z=(101.4−99)2.375=1.002, so P(x̄>cutoff)=0.1581.

Question 7. Sampling Distributions

A population at a school district in Mountain Region during a quarterly performance study has mean 48 and SD 15. For random samples of size 36, find the mean and SD of x̄ and P(x̄>50) under a normal/CLT approximation.

A. Mean 48, SD 0.417, probability 0.7881.
B. Mean 1.333, SD 15, probability 0.2119.
C. Mean 48, SD 2.5, probability 0.2119.
D. Mean 48, SD 15, because averaging does not change spread.

Answer: C

μ_x̄=μ=48. σ_x̄=σ√n=15√36=2.5. The cutoff has z=(50−48)2.5=0.8, so P(x̄>cutoff)=0.2119.

Question 8. Sampling Distributions

A population at a municipal water office in New England network during a pre-exam training cycle has mean 55 and SD 24. For random samples of size 49, find the mean and SD of x̄ and P(x̄>58.43) under a normal/CLT approximation.

A. Mean 55, SD 3.429, probability 0.1586.
B. Mean 55, SD 24, because averaging does not change spread.
C. Mean 1.122, SD 24, probability 0.1586.
D. Mean 55, SD 0.49, probability 0.8414.

Answer: A

μ_x̄=μ=55. σ_x̄=σ√n=24√49=3.429. The cutoff has z=(58.43−55)3.429=1, so P(x̄>cutoff)=0.1586.

Question 9. Sampling Distributions

A population at a state park in Metro East during a school-year data collection has mean 84 and SD 22. For random samples of size 25, find the mean and SD of x̄ and P(x̄>86.2) under a normal/CLT approximation.

A. Mean 84, SD 0.88, probability 0.6915.
B. Mean 84, SD 22, because averaging does not change spread.
C. Mean 84, SD 4.4, probability 0.3085.
D. Mean 3.36, SD 22, probability 0.3085.

Answer: C

μ_x̄=μ=84. σ_x̄=σ√n=22√25=4.4. The cutoff has z=(86.2−84)4.4=0.5, so P(x̄>cutoff)=0.3085.

Question 10. Sampling Distributions

A population at a regional airport authority in Pine Ridge during a fall 2026 audit has mean 55 and SD 15. For random samples of size 36, find the mean and SD of x̄ and P(x̄>58.12) under a normal/CLT approximation.

A. Mean 55, SD 15, because averaging does not change spread.
B. Mean 55, SD 2.5, probability 0.106.
C. Mean 1.528, SD 15, probability 0.106.
D. Mean 55, SD 0.417, probability 0.894.

Answer: B

μ_x̄=μ=55. σ_x̄=σ√n=15√36=2.5. The cutoff has z=(58.12−55)2.5=1.248, so P(x̄>cutoff)=0.106.

Question 11. Sampling Distributions

A population at a city recreation department in Lakeside district during a pre-exam training cycle has mean 94 and SD 24. For random samples of size 25, find the mean and SD of x̄ and P(x̄>96.4) under a normal/CLT approximation.

A. Mean 3.76, SD 24, probability 0.3085.
B. Mean 94, SD 0.96, probability 0.6915.
C. Mean 94, SD 24, because averaging does not change spread.
D. Mean 94, SD 4.8, probability 0.3085.

Answer: D

μ_x̄=μ=94. σ_x̄=σ√n=24√25=4.8. The cutoff has z=(96.4−94)4.8=0.5, so P(x̄>cutoff)=0.3085.

Question 12. Sampling Distributions

A population at a university advising center in Riverbend during a community outreach cycle has mean 84 and SD 21. For random samples of size 100, find the mean and SD of x̄ and P(x̄>85.68) under a normal/CLT approximation.

A. Mean 84, SD 21, because averaging does not change spread.
B. Mean 0.84, SD 21, probability 0.2119.
C. Mean 84, SD 2.1, probability 0.2119.
D. Mean 84, SD 0.21, probability 0.7881.

Answer: C

μ_x̄=μ=84. σ_x̄=σ√n=21√100=2.1. The cutoff has z=(85.68−84)2.1=0.8, so P(x̄>cutoff)=0.2119.

Question 13. Sampling Distributions

A population at a city recreation department in Desert County during a two-month observation window has mean 74 and SD 10. For random samples of size 64, find the mean and SD of x̄ and P(x̄>75) under a normal/CLT approximation.

A. Mean 74, SD 10, because averaging does not change spread.
B. Mean 74, SD 0.156, probability 0.7881.
C. Mean 74, SD 1.25, probability 0.2119.
D. Mean 1.156, SD 10, probability 0.2119.

Answer: C

μ_x̄=μ=74. σ_x̄=σ√n=10√64=1.25. The cutoff has z=(75−74)1.25=0.8, so P(x̄>cutoff)=0.2119.

Question 14. Sampling Distributions

A population at a county election office in Atlantic Corridor during a summer implementation review has mean 72 and SD 21. For random samples of size 100, find the mean and SD of x̄ and P(x̄>74.62) under a normal/CLT approximation.

A. Mean 0.72, SD 21, probability 0.1061.
B. Mean 72, SD 2.1, probability 0.1061.
C. Mean 72, SD 0.21, probability 0.8939.
D. Mean 72, SD 21, because averaging does not change spread.

Answer: B

μ_x̄=μ=72. σ_x̄=σ√n=21√100=2.1. The cutoff has z=(74.62−72)2.1=1.248, so P(x̄>cutoff)=0.1061.

Question 15. Sampling Distributions

A population at a grocery cooperative in Mountain Region during a fall 2026 audit has mean 57 and SD 21. For random samples of size 100, find the mean and SD of x̄ and P(x̄>59.1) under a normal/CLT approximation.

A. Mean 57, SD 2.1, probability 0.1587.
B. Mean 0.57, SD 21, probability 0.1587.
C. Mean 57, SD 21, because averaging does not change spread.
D. Mean 57, SD 0.21, probability 0.8413.

Answer: A

μ_x̄=μ=57. σ_x̄=σ√n=21√100=2.1. The cutoff has z=(59.1−57)2.1=1, so P(x̄>cutoff)=0.1587.

Question 16. Sampling Distributions

A population at a regional airport authority in New England network during a pre-exam training cycle has mean 86 and SD 19. For random samples of size 36, find the mean and SD of x̄ and P(x̄>87.58) under a normal/CLT approximation.

A. Mean 86, SD 19, because averaging does not change spread.
B. Mean 86, SD 3.167, probability 0.3089.
C. Mean 2.389, SD 19, probability 0.3089.
D. Mean 86, SD 0.528, probability 0.6911.

Answer: B

μ_x̄=μ=86. σ_x̄=σ√n=19√36=3.167. The cutoff has z=(87.58−86)3.167=0.499, so P(x̄>cutoff)=0.3089.

Question 17. Sampling Distributions

A population at a university advising center in Coastal Plains during a yearly program evaluation has mean 98 and SD 15. For random samples of size 64, find the mean and SD of x̄ and P(x̄>99.88) under a normal/CLT approximation.

A. Mean 98, SD 1.875, probability 0.158.
B. Mean 1.531, SD 15, probability 0.158.
C. Mean 98, SD 15, because averaging does not change spread.
D. Mean 98, SD 0.234, probability 0.842.

Answer: A

μ_x̄=μ=98. σ_x̄=σ√n=15√64=1.875. The cutoff has z=(99.88−98)1.875=1.003, so P(x̄>cutoff)=0.158.

Question 18. Sampling Distributions

A population at a housing authority in Coastal Plains during a follow-up evaluation period has mean 94 and SD 12. For random samples of size 36, find the mean and SD of x̄ and P(x̄>95.6) under a normal/CLT approximation.

A. Mean 94, SD 12, because averaging does not change spread.
B. Mean 2.611, SD 12, probability 0.2119.
C. Mean 94, SD 0.333, probability 0.7881.
D. Mean 94, SD 2, probability 0.2119.

Answer: D

μ_x̄=μ=94. σ_x̄=σ√n=12√36=2. The cutoff has z=(95.6−94)2=0.8, so P(x̄>cutoff)=0.2119.

Question 19. Sampling Distributions

A population at a university advising center in Westview during a monthly quality review has mean 63 and SD 13. For random samples of size 25, find the mean and SD of x̄ and P(x̄>65.08) under a normal/CLT approximation.

A. Mean 2.52, SD 13, probability 0.2119.
B. Mean 63, SD 13, because averaging does not change spread.
C. Mean 63, SD 0.52, probability 0.7881.
D. Mean 63, SD 2.6, probability 0.2119.

Answer: D

μ_x̄=μ=63. σ_x̄=σ√n=13√25=2.6. The cutoff has z=(65.08−63)2.6=0.8, so P(x̄>cutoff)=0.2119.

Question 20. Sampling Distributions

A population at a digital learning platform in Sunbelt district during a randomized pilot period has mean 108 and SD 13. For random samples of size 49, find the mean and SD of x̄ and P(x̄>108.9) under a normal/CLT approximation.

A. Mean 108, SD 0.265, probability 0.6917.
B. Mean 2.204, SD 13, probability 0.3083.
C. Mean 108, SD 1.857, probability 0.3083.
D. Mean 108, SD 13, because averaging does not change spread.

Answer: C

μ_x̄=μ=108. σ_x̄=σ√n=13√49=1.857. The cutoff has z=(108.9−108)1.857=0.501, so P(x̄>cutoff)=0.3083.

Question 21. Sampling Distributions

A population at a farm cooperative in Prairie District during a quarterly performance study has mean 45 and SD 12. For random samples of size 36, find the mean and SD of x̄ and P(x̄>47) under a normal/CLT approximation.

A. Mean 45, SD 2, probability 0.1587.
B. Mean 45, SD 12, because averaging does not change spread.
C. Mean 45, SD 0.333, probability 0.8413.
D. Mean 1.25, SD 12, probability 0.1587.

Answer: A

μ_x̄=μ=45. σ_x̄=σ√n=12√36=2. The cutoff has z=(47−45)2=1, so P(x̄>cutoff)=0.1587.

Question 22. Sampling Distributions

A population at a school district in Lakeside district during a school-year data collection has mean 60 and SD 13. For random samples of size 64, find the mean and SD of x̄ and P(x̄>60.81) under a normal/CLT approximation.

A. Mean 0.938, SD 13, probability 0.3091.
B. Mean 60, SD 0.203, probability 0.6909.
C. Mean 60, SD 13, because averaging does not change spread.
D. Mean 60, SD 1.625, probability 0.3091.

Answer: D

μ_x̄=μ=60. σ_x̄=σ√n=13√64=1.625. The cutoff has z=(60.81−60)1.625=0.498, so P(x̄>cutoff)=0.3091.

Question 23. Sampling Distributions

A population at a county election office in Atlantic Corridor during a summer implementation review has mean 73 and SD 8. For random samples of size 100, find the mean and SD of x̄ and P(x̄>74) under a normal/CLT approximation.

A. Mean 73, SD 0.8, probability 0.1056.
B. Mean 0.73, SD 8, probability 0.1056.
C. Mean 73, SD 8, because averaging does not change spread.
D. Mean 73, SD 0.08, probability 0.8944.

Answer: A

μ_x̄=μ=73. σ_x̄=σ√n=8√100=0.8. The cutoff has z=(74−73)0.8=1.25, so P(x̄>cutoff)=0.1056.

Question 24. Sampling Distributions

A population at a municipal water office in Lakeside district during a summer implementation review has mean 72 and SD 18. For random samples of size 25, find the mean and SD of x̄ and P(x̄>75.6) under a normal/CLT approximation.

A. Mean 72, SD 3.6, probability 0.1587.
B. Mean 2.88, SD 18, probability 0.1587.
C. Mean 72, SD 0.72, probability 0.8413.
D. Mean 72, SD 18, because averaging does not change spread.

Answer: A

μ_x̄=μ=72. σ_x̄=σ√n=18√25=3.6. The cutoff has z=(75.6−72)3.6=1, so P(x̄>cutoff)=0.1587.

Question 25. Sampling Distributions

A population at a municipal emergency dispatch center in Great Lakes during a pre-exam training cycle has mean 73 and SD 8. For random samples of size 49, find the mean and SD of x̄ and P(x̄>73.91) under a normal/CLT approximation.

A. Mean 73, SD 1.143, probability 0.2129.
B. Mean 73, SD 8, because averaging does not change spread.
C. Mean 1.49, SD 8, probability 0.2129.
D. Mean 73, SD 0.163, probability 0.7871.

Answer: A

μ_x̄=μ=73. σ_x̄=σ√n=8√49=1.143. The cutoff has z=(73.91−73)1.143=0.796, so P(x̄>cutoff)=0.2129.

Question 26. Sampling Distributions

A population at a regional manufacturer in Atlantic Corridor during a summer implementation review has mean 75 and SD 23. For random samples of size 49, find the mean and SD of x̄ and P(x̄>77.63) under a normal/CLT approximation.

A. Mean 75, SD 3.286, probability 0.2117.
B. Mean 1.531, SD 23, probability 0.2117.
C. Mean 75, SD 23, because averaging does not change spread.
D. Mean 75, SD 0.469, probability 0.7883.

Answer: A

μ_x̄=μ=75. σ_x̄=σ√n=23√49=3.286. The cutoff has z=(77.63−75)3.286=0.8, so P(x̄>cutoff)=0.2117.

Question 27. Sampling Distributions

A population at a grocery cooperative in Atlantic Corridor during a community outreach cycle has mean 56 and SD 24. For random samples of size 25, find the mean and SD of x̄ and P(x̄>59.84) under a normal/CLT approximation.

A. Mean 56, SD 24, because averaging does not change spread.
B. Mean 56, SD 0.96, probability 0.7881.
C. Mean 2.24, SD 24, probability 0.2119.
D. Mean 56, SD 4.8, probability 0.2119.

Answer: D

μ_x̄=μ=56. σ_x̄=σ√n=24√25=4.8. The cutoff has z=(59.84−56)4.8=0.8, so P(x̄>cutoff)=0.2119.

Question 28. Sampling Distributions

A population at a municipal emergency dispatch center in Riverbend during a baseline measurement week has mean 104 and SD 18. For random samples of size 25, find the mean and SD of x̄ and P(x̄>107.6) under a normal/CLT approximation.

A. Mean 4.16, SD 18, probability 0.1587.
B. Mean 104, SD 0.72, probability 0.8413.
C. Mean 104, SD 3.6, probability 0.1587.
D. Mean 104, SD 18, because averaging does not change spread.

Answer: C

μ_x̄=μ=104. σ_x̄=σ√n=18√25=3.6. The cutoff has z=(107.6−104)3.6=1, so P(x̄>cutoff)=0.1587.

Question 29. Sampling Distributions

A population at a state park in Mountain Region during a service-improvement study has mean 109 and SD 15. For random samples of size 25, find the mean and SD of x̄ and P(x̄>111.4) under a normal/CLT approximation.

A. Mean 109, SD 3, probability 0.2119.
B. Mean 109, SD 0.6, probability 0.7881.
C. Mean 4.36, SD 15, probability 0.2119.
D. Mean 109, SD 15, because averaging does not change spread.

Answer: A

μ_x̄=μ=109. σ_x̄=σ√n=15√25=3. The cutoff has z=(111.4−109)3=0.8, so P(x̄>cutoff)=0.2119.

Question 30. Sampling Distributions

A population at a municipal water office in Lakeside district during a two-month observation window has mean 95 and SD 11. For random samples of size 36, find the mean and SD of x̄ and P(x̄>96.47) under a normal/CLT approximation.

A. Mean 95, SD 11, because averaging does not change spread.
B. Mean 95, SD 1.833, probability 0.2113.
C. Mean 2.639, SD 11, probability 0.2113.
D. Mean 95, SD 0.306, probability 0.7887.

Answer: B

μ_x̄=μ=95. σ_x̄=σ√n=11√36=1.833. The cutoff has z=(96.47−95)1.833=0.802, so P(x̄>cutoff)=0.2113.

Sampling Distributions: Center, Spread, and Shape: free-response practice

FRQ set 1: Sampling Distributions

Scenario. A population at a school district in Lakeside district during a yearly program evaluation has mean 71 and SD 9. For random samples of size 36, find the mean and SD of x̄ and P(x̄>72.2) under a normal/CLT approximation.

Identify the statistic and describe its sampling distribution target.
Verify independence and the normal/large-count or central-limit condition.
Calculate the sampling-distribution center, standard deviation, standardized value, and requested probability.
Interpret the probability across repeated random samples of the stated size.

Model response

μ_x̄=μ=71. σ_x̄=σ√n=9√36=1.5. The cutoff has z=(72.2−71)1.5=0.8, so P(x̄>cutoff)=0.2119.

FRQ set 2: Sampling Distributions

Scenario. A population at a wildlife clinic in New England network during a two-month observation window has mean 88 and SD 9. For random samples of size 49, find the mean and SD of x̄ and P(x̄>89.29) under a normal/CLT approximation.

Identify the statistic and describe its sampling distribution target.
Verify independence and the normal/large-count or central-limit condition.
Calculate the sampling-distribution center, standard deviation, standardized value, and requested probability.
Interpret the probability across repeated random samples of the stated size.

Model response

μ_x̄=μ=88. σ_x̄=σ√n=9√49=1.286. The cutoff has z=(89.29−88)1.286=1.003, so P(x̄>cutoff)=0.1579.

FRQ set 3: Sampling Distributions

Scenario. A population at a school district in Metro East during a pre-exam training cycle has mean 102 and SD 21. For random samples of size 49, find the mean and SD of x̄ and P(x̄>104.4) under a normal/CLT approximation.

Identify the statistic and describe its sampling distribution target.
Verify independence and the normal/large-count or central-limit condition.
Calculate the sampling-distribution center, standard deviation, standardized value, and requested probability.
Interpret the probability across repeated random samples of the stated size.

Model response

μ_x̄=μ=102. σ_x̄=σ√n=21√49=3. The cutoff has z=(104.4−102)3=0.8, so P(x̄>cutoff)=0.2119.

FRQ set 4: Sampling Distributions

Scenario. A population at a public health department in Midwest consortium during a service-improvement study has mean 45 and SD 21. For random samples of size 25, find the mean and SD of x̄ and P(x̄>48.36) under a normal/CLT approximation.

Identify the statistic and describe its sampling distribution target.
Verify independence and the normal/large-count or central-limit condition.
Calculate the sampling-distribution center, standard deviation, standardized value, and requested probability.
Interpret the probability across repeated random samples of the stated size.

Model response

μ_x̄=μ=45. σ_x̄=σ√n=21√25=4.2. The cutoff has z=(48.36−45)4.2=0.8, so P(x̄>cutoff)=0.2119.

FRQ set 5: Sampling Distributions

Scenario. A population at a county library in New England network during a pre-exam training cycle has mean 53 and SD 21. For random samples of size 64, find the mean and SD of x̄ and P(x̄>56.28) under a normal/CLT approximation.

Identify the statistic and describe its sampling distribution target.
Verify independence and the normal/large-count or central-limit condition.
Calculate the sampling-distribution center, standard deviation, standardized value, and requested probability.
Interpret the probability across repeated random samples of the stated size.

Model response

μ_x̄=μ=53. σ_x̄=σ√n=21√64=2.625. The cutoff has z=(56.28−53)2.625=1.25, so P(x̄>cutoff)=0.1057.

FRQ set 6: Sampling Distributions

Scenario. A population at a city transit agency in Great Lakes during a spring 2027 pilot has mean 67 and SD 20. For random samples of size 100, find the mean and SD of x̄ and P(x̄>68.6) under a normal/CLT approximation.

Identify the statistic and describe its sampling distribution target.
Verify independence and the normal/large-count or central-limit condition.
Calculate the sampling-distribution center, standard deviation, standardized value, and requested probability.
Interpret the probability across repeated random samples of the stated size.

Model response

μ_x̄=μ=67. σ_x̄=σ√n=20√100=2. The cutoff has z=(68.6−67)2=0.8, so P(x̄>cutoff)=0.2119.

FRQ set 7: Sampling Distributions

Scenario. A population at a county library in South Harbor during a fall 2026 audit has mean 59 and SD 20. For random samples of size 100, find the mean and SD of x̄ and P(x̄>61.5) under a normal/CLT approximation.

Identify the statistic and describe its sampling distribution target.
Verify independence and the normal/large-count or central-limit condition.
Calculate the sampling-distribution center, standard deviation, standardized value, and requested probability.
Interpret the probability across repeated random samples of the stated size.

Model response

μ_x̄=μ=59. σ_x̄=σ√n=20√100=2. The cutoff has z=(61.5−59)2=1.25, so P(x̄>cutoff)=0.1056.

FRQ set 8: Sampling Distributions

Scenario. A population at a municipal emergency dispatch center in South Harbor during a two-month observation window has mean 89 and SD 13. For random samples of size 36, find the mean and SD of x̄ and P(x̄>91.71) under a normal/CLT approximation.

Identify the statistic and describe its sampling distribution target.
Verify independence and the normal/large-count or central-limit condition.
Calculate the sampling-distribution center, standard deviation, standardized value, and requested probability.
Interpret the probability across repeated random samples of the stated size.

Model response

μ_x̄=μ=89. σ_x̄=σ√n=13√36=2.167. The cutoff has z=(91.71−89)2.167=1.251, so P(x̄>cutoff)=0.1055.