Sampling Distribution of the Sample Mean
Analyze the repeated-sampling behavior of x̄: center at μ, standard error σ/√n, normality conditions, and probability statements about sample means.
Sampling Distribution Of The Sample Mean: direct answer
Analyze the repeated-sampling behavior of x̄: center at μ, standard error σ/√n, normality conditions, and probability statements about sample means.
This page uses sampling distribution of the sample mean 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 Distribution of the Sample Mean
| Center | μx̄=μ |
|---|---|
| Standard error | σx̄=σ/√n |
| Larger n | Reduces standard error at the 1/√n rate. |
| Without replacement | Check that the sample is small relative to the population for approximate independence. |
| Probability language | Describe x̄ from samples of the stated size, not individual observations. |
Concept mastery: Sampling Distribution of the Sample Mean
The sample mean targets a population mean
Across repeated samples of the same size, x̄ changes while μ remains fixed. The sampling distribution of x̄ is centered at μ, so x̄ is an unbiased estimator of μ under the standard random-sampling framework.
Averaging shrinks variability by √n
If observations are independent with population SD σ, then SD(x̄)=σ/√n. Doubling n does not halve the standard error; reducing it by half requires multiplying sample size by four.
Population shape and sampling-distribution shape are different
A nonnormal population can still produce an approximately normal distribution of x̄ for sufficiently large n. Conversely, a small sample from a strongly skewed or heavy-tailed population may not justify a normal approximation.
The 10% condition is about dependence
When sampling without replacement, observations are not exactly independent. Keeping the sample no more than about 10% of the population is a common condition used to make the independence approximation reasonable for standard-error calculations.
Probability statements must name x̄
After standardizing a cutoff, interpret the tail probability as the chance that a random sample of the specified size produces a sample mean beyond the cutoff. Do not describe it as the chance an individual observation exceeds that value.
Worked analysis for Sampling Distribution of the Sample Mean
Sample-mean case 1: Urban Transit Riders
Suppose the population mean is 63, population SD is 12, and samples of size 36 are selected independently. For x̄, μx̄=63 and σx̄=12/√36≈2.000. The sample size n=36 is large enough for the exercise’s stated normal/CLT approximation. The standard-error denominator is √n because averaging stabilizes the statistic as n grows.
To evaluate P(x̄>64.40), use z=(64.40−63)/2.000≈0.70 and then the appropriate normal tail. The wording of the conclusion must say “a sample mean from samples of size 36,” not “an observation,” because the random object here is x̄.
Sample-mean case 2: Regional Hospital Visits
Suppose the population mean is 66, population SD is 14, and samples of size 49 are selected independently. For x̄, μx̄=66 and σx̄=14/√49≈2.000. The sample size n=49 is large enough for the exercise’s stated normal/CLT approximation. The standard-error denominator is √n because averaging stabilizes the statistic as n grows.
To evaluate P(x̄>67.80), use z=(67.80−66)/2.000≈0.90 and then the appropriate normal tail. The wording of the conclusion must say “a sample mean from samples of size 49,” not “an observation,” because the random object here is x̄.
Sample-mean case 3: Community College Placement Scores
Suppose the population mean is 69, population SD is 16, and samples of size 64 are selected independently. For x̄, μx̄=69 and σx̄=16/√64≈2.000. The population is described as approximately normal. The standard-error denominator is √n because averaging stabilizes the statistic as n grows.
To evaluate P(x̄>71.20), use z=(71.20−69)/2.000≈1.10 and then the appropriate normal tail. The wording of the conclusion must say “a sample mean from samples of size 64,” not “an observation,” because the random object here is x̄.
Sample-mean case 4: Warehouse Order Times
Suppose the population mean is 72, population SD is 10, and samples of size 100 are selected independently. For x̄, μx̄=72 and σx̄=10/√100≈1.000. The sample size n=100 is large enough for the exercise’s stated normal/CLT approximation. The standard-error denominator is √n because averaging stabilizes the statistic as n grows.
To evaluate P(x̄>73.30), use z=(73.30−72)/1.000≈1.30 and then the appropriate normal tail. The wording of the conclusion must say “a sample mean from samples of size 100,” not “an observation,” because the random object here is x̄.
Sample-mean case 5: Public-Library Checkouts
Suppose the population mean is 75, population SD is 12, and samples of size 25 are selected independently. For x̄, μx̄=75 and σx̄=12/√25≈2.400. The sample size n=25 is large enough for the exercise’s stated normal/CLT approximation. The standard-error denominator is √n because averaging stabilizes the statistic as n grows.
To evaluate P(x̄>76.20), use z=(76.20−75)/2.400≈0.50 and then the appropriate normal tail. The wording of the conclusion must say “a sample mean from samples of size 25,” not “an observation,” because the random object here is x̄.
Sample-mean case 6: Solar-Panel Output Readings
Suppose the population mean is 60, population SD is 14, and samples of size 36 are selected independently. For x̄, μx̄=60 and σx̄=14/√36≈2.333. The population is described as approximately normal. The standard-error denominator is √n because averaging stabilizes the statistic as n grows.
To evaluate P(x̄>61.63), use z=(61.63−60)/2.333≈0.70 and then the appropriate normal tail. The wording of the conclusion must say “a sample mean from samples of size 36,” not “an observation,” because the random object here is x̄.
Sample-mean case 7: School Attendance Rates
Suppose the population mean is 63, population SD is 16, and samples of size 49 are selected independently. For x̄, μx̄=63 and σx̄=16/√49≈2.286. The sample size n=49 is large enough for the exercise’s stated normal/CLT approximation. The standard-error denominator is √n because averaging stabilizes the statistic as n grows.
To evaluate P(x̄>65.06), use z=(65.06−63)/2.286≈0.90 and then the appropriate normal tail. The wording of the conclusion must say “a sample mean from samples of size 49,” not “an observation,” because the random object here is x̄.
Sample-mean case 8: Restaurant Service Times
Suppose the population mean is 66, population SD is 10, and samples of size 64 are selected independently. For x̄, μx̄=66 and σx̄=10/√64≈1.250. The sample size n=64 is large enough for the exercise’s stated normal/CLT approximation. The standard-error denominator is √n because averaging stabilizes the statistic as n grows.
To evaluate P(x̄>67.38), use z=(67.38−66)/1.250≈1.10 and then the appropriate normal tail. The wording of the conclusion must say “a sample mean from samples of size 64,” not “an observation,” because the random object here is x̄.
Sample-mean case 9: Wildlife Tag Measurements
Suppose the population mean is 69, population SD is 12, and samples of size 100 are selected independently. For x̄, μx̄=69 and σx̄=12/√100≈1.200. The population is described as approximately normal. The standard-error denominator is √n because averaging stabilizes the statistic as n grows.
To evaluate P(x̄>70.56), use z=(70.56−69)/1.200≈1.30 and then the appropriate normal tail. The wording of the conclusion must say “a sample mean from samples of size 100,” not “an observation,” because the random object here is x̄.
Sample-mean case 10: Municipal Water-Use Records
Suppose the population mean is 72, population SD is 14, and samples of size 25 are selected independently. For x̄, μx̄=72 and σx̄=14/√25≈2.800. The sample size n=25 is large enough for the exercise’s stated normal/CLT approximation. The standard-error denominator is √n because averaging stabilizes the statistic as n grows.
To evaluate P(x̄>73.40), use z=(73.40−72)/2.800≈0.50 and then the appropriate normal tail. The wording of the conclusion must say “a sample mean from samples of size 25,” not “an observation,” because the random object here is x̄.
Sample-mean case 11: Online Course Completion Times
Suppose the population mean is 75, population SD is 16, and samples of size 36 are selected independently. For x̄, μx̄=75 and σx̄=16/√36≈2.667. The sample size n=36 is large enough for the exercise’s stated normal/CLT approximation. The standard-error denominator is √n because averaging stabilizes the statistic as n grows.
To evaluate P(x̄>76.87), use z=(76.87−75)/2.667≈0.70 and then the appropriate normal tail. The wording of the conclusion must say “a sample mean from samples of size 36,” not “an observation,” because the random object here is x̄.
Sample-mean case 12: Farm Yield Measurements
Suppose the population mean is 60, population SD is 10, and samples of size 49 are selected independently. For x̄, μx̄=60 and σx̄=10/√49≈1.429. The population is described as approximately normal. The standard-error denominator is √n because averaging stabilizes the statistic as n grows.
To evaluate P(x̄>61.29), use z=(61.29−60)/1.429≈0.90 and then the appropriate normal tail. The wording of the conclusion must say “a sample mean from samples of size 49,” not “an observation,” because the random object here is x̄.
Sample-mean case 13: Clinic Appointment Waits
Suppose the population mean is 63, population SD is 12, and samples of size 64 are selected independently. For x̄, μx̄=63 and σx̄=12/√64≈1.500. The sample size n=64 is large enough for the exercise’s stated normal/CLT approximation. The standard-error denominator is √n because averaging stabilizes the statistic as n grows.
To evaluate P(x̄>64.65), use z=(64.65−63)/1.500≈1.10 and then the appropriate normal tail. The wording of the conclusion must say “a sample mean from samples of size 64,” not “an observation,” because the random object here is x̄.
Sample-mean case 14: Manufacturing Fill Weights
Suppose the population mean is 66, population SD is 14, and samples of size 100 are selected independently. For x̄, μx̄=66 and σx̄=14/√100≈1.400. The sample size n=100 is large enough for the exercise’s stated normal/CLT approximation. The standard-error denominator is √n because averaging stabilizes the statistic as n grows.
To evaluate P(x̄>67.82), use z=(67.82−66)/1.400≈1.30 and then the appropriate normal tail. The wording of the conclusion must say “a sample mean from samples of size 100,” not “an observation,” because the random object here is x̄.
Sample-mean case 15: County Commute Times
Suppose the population mean is 69, population SD is 16, and samples of size 25 are selected independently. For x̄, μx̄=69 and σx̄=16/√25≈3.200. The population is described as approximately normal. The standard-error denominator is √n because averaging stabilizes the statistic as n grows.
To evaluate P(x̄>70.60), use z=(70.60−69)/3.200≈0.50 and then the appropriate normal tail. The wording of the conclusion must say “a sample mean from samples of size 25,” not “an observation,” because the random object here is x̄.
Sample-mean case 16: Energy Meter Readings
Suppose the population mean is 72, population SD is 10, and samples of size 36 are selected independently. For x̄, μx̄=72 and σx̄=10/√36≈1.667. The sample size n=36 is large enough for the exercise’s stated normal/CLT approximation. The standard-error denominator is √n because averaging stabilizes the statistic as n grows.
To evaluate P(x̄>73.17), use z=(73.17−72)/1.667≈0.70 and then the appropriate normal tail. The wording of the conclusion must say “a sample mean from samples of size 36,” not “an observation,” because the random object here is x̄.
Sample-mean case 17: University Advising Durations
Suppose the population mean is 75, population SD is 12, and samples of size 49 are selected independently. For x̄, μx̄=75 and σx̄=12/√49≈1.714. The sample size n=49 is large enough for the exercise’s stated normal/CLT approximation. The standard-error denominator is √n because averaging stabilizes the statistic as n grows.
To evaluate P(x̄>76.54), use z=(76.54−75)/1.714≈0.90 and then the appropriate normal tail. The wording of the conclusion must say “a sample mean from samples of size 49,” not “an observation,” because the random object here is x̄.
Sample-mean case 18: Sports Training Measurements
Suppose the population mean is 60, population SD is 14, and samples of size 64 are selected independently. For x̄, μx̄=60 and σx̄=14/√64≈1.750. The population is described as approximately normal. The standard-error denominator is √n because averaging stabilizes the statistic as n grows.
To evaluate P(x̄>61.92), use z=(61.92−60)/1.750≈1.10 and then the appropriate normal tail. The wording of the conclusion must say “a sample mean from samples of size 64,” not “an observation,” because the random object here is x̄.
Sample-mean case 19: Call-Center Resolution Times
Suppose the population mean is 63, population SD is 16, and samples of size 100 are selected independently. For x̄, μx̄=63 and σx̄=16/√100≈1.600. The sample size n=100 is large enough for the exercise’s stated normal/CLT approximation. The standard-error denominator is √n because averaging stabilizes the statistic as n grows.
To evaluate P(x̄>65.08), use z=(65.08−63)/1.600≈1.30 and then the appropriate normal tail. The wording of the conclusion must say “a sample mean from samples of size 100,” not “an observation,” because the random object here is x̄.
Sample-mean case 20: Retail Basket Totals
Suppose the population mean is 66, population SD is 10, and samples of size 25 are selected independently. For x̄, μx̄=66 and σx̄=10/√25≈2.000. The sample size n=25 is large enough for the exercise’s stated normal/CLT approximation. The standard-error denominator is √n because averaging stabilizes the statistic as n grows.
To evaluate P(x̄>67.00), use z=(67.00−66)/2.000≈0.50 and then the appropriate normal tail. The wording of the conclusion must say “a sample mean from samples of size 25,” not “an observation,” because the random object here is x̄.
Sample-mean case 21: Lab Assay Values
Suppose the population mean is 69, population SD is 12, and samples of size 36 are selected independently. For x̄, μx̄=69 and σx̄=12/√36≈2.000. The population is described as approximately normal. The standard-error denominator is √n because averaging stabilizes the statistic as n grows.
To evaluate P(x̄>70.40), use z=(70.40−69)/2.000≈0.70 and then the appropriate normal tail. The wording of the conclusion must say “a sample mean from samples of size 36,” not “an observation,” because the random object here is x̄.
Sample-mean case 22: Shipping Transit Times
Suppose the population mean is 72, population SD is 14, and samples of size 49 are selected independently. For x̄, μx̄=72 and σx̄=14/√49≈2.000. The sample size n=49 is large enough for the exercise’s stated normal/CLT approximation. The standard-error denominator is √n because averaging stabilizes the statistic as n grows.
To evaluate P(x̄>73.80), use z=(73.80−72)/2.000≈0.90 and then the appropriate normal tail. The wording of the conclusion must say “a sample mean from samples of size 49,” not “an observation,” because the random object here is x̄.
Sample-mean case 23: District Test Scores
Suppose the population mean is 75, population SD is 16, and samples of size 64 are selected independently. For x̄, μx̄=75 and σx̄=16/√64≈2.000. The sample size n=64 is large enough for the exercise’s stated normal/CLT approximation. The standard-error denominator is √n because averaging stabilizes the statistic as n grows.
To evaluate P(x̄>77.20), use z=(77.20−75)/2.000≈1.10 and then the appropriate normal tail. The wording of the conclusion must say “a sample mean from samples of size 64,” not “an observation,” because the random object here is x̄.
Sample-mean case 24: Park Visitor Durations
Suppose the population mean is 60, population SD is 10, and samples of size 100 are selected independently. For x̄, μx̄=60 and σx̄=10/√100≈1.000. The population is described as approximately normal. The standard-error denominator is √n because averaging stabilizes the statistic as n grows.
To evaluate P(x̄>61.30), use z=(61.30−60)/1.000≈1.30 and then the appropriate normal tail. The wording of the conclusion must say “a sample mean from samples of size 100,” not “an observation,” because the random object here is x̄.
Sample-mean case 25: Airport Security Waits
Suppose the population mean is 63, population SD is 12, and samples of size 25 are selected independently. For x̄, μx̄=63 and σx̄=12/√25≈2.400. The sample size n=25 is large enough for the exercise’s stated normal/CLT approximation. The standard-error denominator is √n because averaging stabilizes the statistic as n grows.
To evaluate P(x̄>64.20), use z=(64.20−63)/2.400≈0.50 and then the appropriate normal tail. The wording of the conclusion must say “a sample mean from samples of size 25,” not “an observation,” because the random object here is x̄.
Sample-mean case 26: Pharmacy Prescription Times
Suppose the population mean is 66, population SD is 14, and samples of size 36 are selected independently. For x̄, μx̄=66 and σx̄=14/√36≈2.333. The sample size n=36 is large enough for the exercise’s stated normal/CLT approximation. The standard-error denominator is √n because averaging stabilizes the statistic as n grows.
To evaluate P(x̄>67.63), use z=(67.63−66)/2.333≈0.70 and then the appropriate normal tail. The wording of the conclusion must say “a sample mean from samples of size 36,” not “an observation,” because the random object here is x̄.
Sample-mean case 27: Bicycle Commute Distances
Suppose the population mean is 69, population SD is 16, and samples of size 49 are selected independently. For x̄, μx̄=69 and σx̄=16/√49≈2.286. The population is described as approximately normal. The standard-error denominator is √n because averaging stabilizes the statistic as n grows.
To evaluate P(x̄>71.06), use z=(71.06−69)/2.286≈0.90 and then the appropriate normal tail. The wording of the conclusion must say “a sample mean from samples of size 49,” not “an observation,” because the random object here is x̄.
Sample-mean case 28: Community Garden Yields
Suppose the population mean is 72, population SD is 10, and samples of size 64 are selected independently. For x̄, μx̄=72 and σx̄=10/√64≈1.250. The sample size n=64 is large enough for the exercise’s stated normal/CLT approximation. The standard-error denominator is √n because averaging stabilizes the statistic as n grows.
To evaluate P(x̄>73.38), use z=(73.38−72)/1.250≈1.10 and then the appropriate normal tail. The wording of the conclusion must say “a sample mean from samples of size 64,” not “an observation,” because the random object here is x̄.
Sample-mean case 29: Emergency Response Durations
Suppose the population mean is 75, population SD is 12, and samples of size 100 are selected independently. For x̄, μx̄=75 and σx̄=12/√100≈1.200. The sample size n=100 is large enough for the exercise’s stated normal/CLT approximation. The standard-error denominator is √n because averaging stabilizes the statistic as n grows.
To evaluate P(x̄>76.56), use z=(76.56−75)/1.200≈1.30 and then the appropriate normal tail. The wording of the conclusion must say “a sample mean from samples of size 100,” not “an observation,” because the random object here is x̄.
Sample-mean case 30: College Credit Loads
Suppose the population mean is 60, population SD is 14, and samples of size 25 are selected independently. For x̄, μx̄=60 and σx̄=14/√25≈2.800. The population is described as approximately normal. The standard-error denominator is √n because averaging stabilizes the statistic as n grows.
To evaluate P(x̄>61.40), use z=(61.40−60)/2.800≈0.50 and then the appropriate normal tail. The wording of the conclusion must say “a sample mean from samples of size 25,” not “an observation,” because the random object here is x̄.
Sampling Distribution of the Sample Mean: multiple-choice practice
Question 1. Sampling Distribution of a Sample Mean
A population at a farm cooperative in Desert County during a semester-long cohort study has mean 100 and SD 15. For random samples of size 64, find the mean and SD of x̄ and P(x̄>100.9) under a normal/CLT approximation.
Answer: A
μx̄=μ=100. σx̄=σ√n=15√64=1.875. The cutoff has z=(100.9−100)1.875=0.501, so P(x̄>cutoff)=0.3081.
Question 2. Sampling Distribution of a Sample Mean
A population at a public health department in North Valley during a service-improvement study has mean 70 and SD 12. For random samples of size 49, find the mean and SD of x̄ and P(x̄>70.86) under a normal/CLT approximation.
Answer: B
μx̄=μ=70. σx̄=σ√n=12√49=1.714. The cutoff has z=(70.86−70)1.714=0.502, so P(x̄>cutoff)=0.308.
Question 3. Sampling Distribution of a Sample Mean
A population at a city recreation department in Pine Ridge during a follow-up evaluation period has mean 49 and SD 22. For random samples of size 36, find the mean and SD of x̄ and P(x̄>52.67) under a normal/CLT approximation.
Answer: A
μx̄=μ=49. σx̄=σ√n=22√36=3.667. The cutoff has z=(52.67−49)3.667=1.001, so P(x̄>cutoff)=0.1584.
Question 4. Sampling Distribution of a Sample Mean
A population at a digital learning platform in Sunbelt district during a two-month observation window has mean 105 and SD 12. For random samples of size 64, find the mean and SD of x̄ and P(x̄>106.2) under a normal/CLT approximation.
Answer: C
μx̄=μ=105. σx̄=σ√n=12√64=1.5. The cutoff has z=(106.2−105)1.5=0.8, so P(x̄>cutoff)=0.2119.
Question 5. Sampling Distribution of a Sample Mean
A population at a city recreation department in Desert County during a winter readiness review has mean 58 and SD 20. For random samples of size 25, find the mean and SD of x̄ and P(x̄>60) under a normal/CLT approximation.
Answer: C
μx̄=μ=58. σx̄=σ√n=20√25=4. The cutoff has z=(60−58)4=0.5, so P(x̄>cutoff)=0.3085.
Question 6. Sampling Distribution of a Sample Mean
A population at a food safety laboratory in Midwest consortium during a semester-long cohort study has mean 50 and SD 17. For random samples of size 49, find the mean and SD of x̄ and P(x̄>53.04) under a normal/CLT approximation.
Answer: A
μx̄=μ=50. σx̄=σ√n=17√49=2.429. The cutoff has z=(53.04−50)2.429=1.252, so P(x̄>cutoff)=0.1053.
Question 7. Sampling Distribution of a Sample Mean
A population at a municipal emergency dispatch center in Desert County during a yearly program evaluation has mean 92 and SD 10. For random samples of size 100, find the mean and SD of x̄ and P(x̄>93.25) under a normal/CLT approximation.
Answer: A
μx̄=μ=92. σx̄=σ√n=10√100=1. The cutoff has z=(93.25−92)1=1.25, so P(x̄>cutoff)=0.1056.
Question 8. Sampling Distribution of a Sample Mean
A population at a county library in North Valley during a quarterly performance study has mean 76 and SD 13. For random samples of size 64, find the mean and SD of x̄ and P(x̄>77.3) under a normal/CLT approximation.
Answer: B
μx̄=μ=76. σx̄=σ√n=13√64=1.625. The cutoff has z=(77.3−76)1.625=0.8, so P(x̄>cutoff)=0.2119.
Question 9. Sampling Distribution of a Sample Mean
A population at a wildlife clinic in Prairie District during a yearly program evaluation has mean 81 and SD 14. For random samples of size 36, find the mean and SD of x̄ and P(x̄>83.33) under a normal/CLT approximation.
Answer: C
μx̄=μ=81. σx̄=σ√n=14√36=2.333. The cutoff has z=(83.33−81)2.333=0.999, so P(x̄>cutoff)=0.159.
Question 10. Sampling Distribution of a Sample Mean
A population at a city transit agency in Central County during a weekday operations study has mean 107 and SD 10. For random samples of size 100, find the mean and SD of x̄ and P(x̄>107.5) under a normal/CLT approximation.
Answer: B
μx̄=μ=107. σx̄=σ√n=10√100=1. The cutoff has z=(107.5−107)1=0.5, so P(x̄>cutoff)=0.3085.
Question 11. Sampling Distribution of a Sample Mean
A population at a university advising center in North Valley during a monthly quality review has mean 61 and SD 10. For random samples of size 100, find the mean and SD of x̄ and P(x̄>61.8) under a normal/CLT approximation.
Answer: D
μx̄=μ=61. σx̄=σ√n=10√100=1. The cutoff has z=(61.8−61)1=0.8, so P(x̄>cutoff)=0.2119.
Question 12. Sampling Distribution of a Sample Mean
A population at a wildlife clinic in Atlantic Corridor during a monthly quality review has mean 46 and SD 8. For random samples of size 64, find the mean and SD of x̄ and P(x̄>47) under a normal/CLT approximation.
Answer: D
μx̄=μ=46. σx̄=σ√n=8√64=1. The cutoff has z=(47−46)1=1, so P(x̄>cutoff)=0.1587.
Question 13. Sampling Distribution of a Sample Mean
A population at a municipal emergency dispatch center in Pacific Northwest during a baseline measurement week has mean 74 and SD 23. For random samples of size 64, find the mean and SD of x̄ and P(x̄>76.3) under a normal/CLT approximation.
Answer: B
μx̄=μ=74. σx̄=σ√n=23√64=2.875. The cutoff has z=(76.3−74)2.875=0.8, so P(x̄>cutoff)=0.2119.
Question 14. Sampling Distribution of a Sample Mean
A population at a county library in Cedar Grove during a two-month observation window has mean 82 and SD 9. For random samples of size 36, find the mean and SD of x̄ and P(x̄>83.88) under a normal/CLT approximation.
Answer: B
μx̄=μ=82. σx̄=σ√n=9√36=1.5. The cutoff has z=(83.88−82)1.5=1.253, so P(x̄>cutoff)=0.105.
Question 15. Sampling Distribution of a Sample Mean
A population at a housing authority in Prairie District during a summer implementation review has mean 52 and SD 14. For random samples of size 64, find the mean and SD of x̄ and P(x̄>54.19) under a normal/CLT approximation.
Answer: C
μx̄=μ=52. σx̄=σ√n=14√64=1.75. The cutoff has z=(54.19−52)1.75=1.251, so P(x̄>cutoff)=0.1054.
Question 16. Sampling Distribution of a Sample Mean
A population at a regional hospital in Sunbelt district during a school-year data collection has mean 95 and SD 12. For random samples of size 49, find the mean and SD of x̄ and P(x̄>95.86) under a normal/CLT approximation.
Answer: A
μx̄=μ=95. σx̄=σ√n=12√49=1.714. The cutoff has z=(95.86−95)1.714=0.502, so P(x̄>cutoff)=0.308.
Question 17. Sampling Distribution of a Sample Mean
A population at a county election office in Westview during a yearly program evaluation has mean 55 and SD 21. For random samples of size 36, find the mean and SD of x̄ and P(x̄>59.38) under a normal/CLT approximation.
Answer: C
μx̄=μ=55. σx̄=σ√n=21√36=3.5. The cutoff has z=(59.38−55)3.5=1.251, so P(x̄>cutoff)=0.1054.
Question 18. Sampling Distribution of a Sample Mean
A population at a university advising center in Midwest consortium during a pre-exam training cycle has mean 95 and SD 22. For random samples of size 100, find the mean and SD of x̄ and P(x̄>97.75) under a normal/CLT approximation.
Answer: C
μx̄=μ=95. σx̄=σ√n=22√100=2.2. The cutoff has z=(97.75−95)2.2=1.25, so P(x̄>cutoff)=0.1056.
Question 19. Sampling Distribution of a Sample Mean
A population at a recycling program in Cedar Grove during a monthly quality review has mean 105 and SD 15. For random samples of size 36, find the mean and SD of x̄ and P(x̄>107.0) under a normal/CLT approximation.
Answer: B
μx̄=μ=105. σx̄=σ√n=15√36=2.5. The cutoff has z=(107.0−105)2.5=0.8, so P(x̄>cutoff)=0.2119.
Question 20. Sampling Distribution of a Sample Mean
A population at a wildlife clinic in Great Lakes during a two-month observation window has mean 81 and SD 19. For random samples of size 25, find the mean and SD of x̄ and P(x̄>85.75) under a normal/CLT approximation.
Answer: B
μx̄=μ=81. σx̄=σ√n=19√25=3.8. The cutoff has z=(85.75−81)3.8=1.25, so P(x̄>cutoff)=0.1056.
Question 21. Sampling Distribution of a Sample Mean
A population at a farm cooperative in South Harbor during a summer implementation review has mean 99 and SD 24. For random samples of size 49, find the mean and SD of x̄ and P(x̄>103.3) under a normal/CLT approximation.
Answer: D
μx̄=μ=99. σx̄=σ√n=24√49=3.429. The cutoff has z=(103.3−99)3.429=1.251, so P(x̄>cutoff)=0.1054.
Question 22. Sampling Distribution of a Sample Mean
A population at a food safety laboratory in Sunbelt district during a winter readiness review has mean 68 and SD 13. For random samples of size 36, find the mean and SD of x̄ and P(x̄>69.73) under a normal/CLT approximation.
Answer: C
μx̄=μ=68. σx̄=σ√n=13√36=2.167. The cutoff has z=(69.73−68)2.167=0.798, so P(x̄>cutoff)=0.2123.
Question 23. Sampling Distribution of a Sample Mean
A population at a county election office in Great Lakes during a follow-up evaluation period has mean 59 and SD 13. For random samples of size 100, find the mean and SD of x̄ and P(x̄>60.62) under a normal/CLT approximation.
Answer: B
μx̄=μ=59. σx̄=σ√n=13√100=1.3. The cutoff has z=(60.62−59)1.3=1.246, so P(x̄>cutoff)=0.1064.
Question 24. Sampling Distribution of a Sample Mean
A population at a wildlife clinic in Pine Ridge during a six-week field trial has mean 90 and SD 18. For random samples of size 64, find the mean and SD of x̄ and P(x̄>91.8) under a normal/CLT approximation.
Answer: C
μx̄=μ=90. σx̄=σ√n=18√64=2.25. The cutoff has z=(91.8−90)2.25=0.8, so P(x̄>cutoff)=0.2119.
Question 25. Sampling Distribution of a Sample Mean
A population at a county election office in Cedar Grove during a service-improvement study has mean 74 and SD 24. For random samples of size 25, find the mean and SD of x̄ and P(x̄>76.4) under a normal/CLT approximation.
Answer: B
μx̄=μ=74. σx̄=σ√n=24√25=4.8. The cutoff has z=(76.4−74)4.8=0.5, so P(x̄>cutoff)=0.3085.
Question 26. Sampling Distribution of a Sample Mean
A population at a digital learning platform in Cedar Grove during a pre-exam training cycle has mean 98 and SD 18. For random samples of size 36, find the mean and SD of x̄ and P(x̄>101.0) under a normal/CLT approximation.
Answer: B
μx̄=μ=98. σx̄=σ√n=18√36=3. The cutoff has z=(101.0−98)3=1, so P(x̄>cutoff)=0.1587.
Question 27. Sampling Distribution of a Sample Mean
A population at a city transit agency in Desert County during a weekday operations study has mean 89 and SD 17. For random samples of size 49, find the mean and SD of x̄ and P(x̄>90.94) under a normal/CLT approximation.
Answer: A
μx̄=μ=89. σx̄=σ√n=17√49=2.429. The cutoff has z=(90.94−89)2.429=0.799, so P(x̄>cutoff)=0.2122.
Question 28. Sampling Distribution of a Sample Mean
A population at a city recreation department in Great Lakes during a multiweek validation study has mean 70 and SD 16. For random samples of size 64, find the mean and SD of x̄ and P(x̄>72.5) under a normal/CLT approximation.
Answer: A
μx̄=μ=70. σx̄=σ√n=16√64=2. The cutoff has z=(72.5−70)2=1.25, so P(x̄>cutoff)=0.1056.
Question 29. Sampling Distribution of a Sample Mean
A population at a county library in Pacific Northwest during a six-week field trial has mean 104 and SD 15. For random samples of size 25, find the mean and SD of x̄ and P(x̄>107.8) under a normal/CLT approximation.
Answer: C
μx̄=μ=104. σx̄=σ√n=15√25=3. The cutoff has z=(107.8−104)3=1.25, so P(x̄>cutoff)=0.1056.
Question 30. Sampling Distribution of a Sample Mean
A population at a city recreation department in Prairie District during a monthly quality review has mean 51 and SD 14. For random samples of size 36, find the mean and SD of x̄ and P(x̄>53.92) under a normal/CLT approximation.
Answer: C
μx̄=μ=51. σx̄=σ√n=14√36=2.333. The cutoff has z=(53.92−51)2.333=1.251, so P(x̄>cutoff)=0.1054.
Question 31. Sampling Distribution of a Sample Mean
A population at a housing authority in Midwest consortium during a multiweek validation study has mean 97 and SD 14. For random samples of size 36, find the mean and SD of x̄ and P(x̄>99.33) under a normal/CLT approximation.
Answer: A
μx̄=μ=97. σx̄=σ√n=14√36=2.333. The cutoff has z=(99.33−97)2.333=0.999, so P(x̄>cutoff)=0.159.
Question 32. Sampling Distribution of a Sample Mean
A population at a public high school in Pacific Northwest during a regional benchmarking study has mean 98 and SD 12. For random samples of size 100, find the mean and SD of x̄ and P(x̄>99.5) under a normal/CLT approximation.
Answer: B
μx̄=μ=98. σx̄=σ√n=12√100=1.2. The cutoff has z=(99.5−98)1.2=1.25, so P(x̄>cutoff)=0.1056.
Sampling Distribution of the Sample Mean: free-response practice
FRQ set 1: Sampling Distribution of a Sample Mean
Scenario. A population at a grocery cooperative in Atlantic Corridor during a two-month observation window has mean 102 and SD 13. For random samples of size 25, find the mean and SD of x̄ and P(x̄>105.2) 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
μx̄=μ=102. σx̄=σ√n=13√25=2.6. The cutoff has z=(105.2−102)2.6=1.25, so P(x̄>cutoff)=0.1056.
FRQ set 2: Sampling Distribution of a Sample Mean
Scenario. A population at a county election office in Desert County during a winter readiness review has mean 100 and SD 14. For random samples of size 25, find the mean and SD of x̄ and P(x̄>102.2) 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
μx̄=μ=100. σx̄=σ√n=14√25=2.8. The cutoff has z=(102.2−100)2.8=0.8, so P(x̄>cutoff)=0.2119.
FRQ set 3: Sampling Distribution of a Sample Mean
Scenario. A population at a school district in Capital Region during a semester-long cohort study has mean 97 and SD 15. For random samples of size 49, find the mean and SD of x̄ and P(x̄>99.14) 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
μx̄=μ=97. σx̄=σ√n=15√49=2.143. The cutoff has z=(99.14−97)2.143=0.999, so P(x̄>cutoff)=0.159.
FRQ set 4: Sampling Distribution of a Sample Mean
Scenario. A population at a regional manufacturer in Central County during a fall 2026 audit has mean 105 and SD 20. For random samples of size 36, find the mean and SD of x̄ and P(x̄>109.2) 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
μx̄=μ=105. σx̄=σ√n=20√36=3.333. The cutoff has z=(109.2−105)3.333=1.251, so P(x̄>cutoff)=0.1055.
FRQ set 5: Sampling Distribution of a Sample Mean
Scenario. A population at a public health department in Great Lakes during a quarterly performance study has mean 107 and SD 10. For random samples of size 49, find the mean and SD of x̄ and P(x̄>108.1) 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
μx̄=μ=107. σx̄=σ√n=10√49=1.429. The cutoff has z=(108.1−107)1.429=0.798, so P(x̄>cutoff)=0.2124.
FRQ set 6: Sampling Distribution of a Sample Mean
Scenario. A population at a food safety laboratory in Riverbend during a multiweek validation study has mean 80 and SD 14. For random samples of size 64, find the mean and SD of x̄ and P(x̄>82.19) 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
μx̄=μ=80. σx̄=σ√n=14√64=1.75. The cutoff has z=(82.19−80)1.75=1.251, so P(x̄>cutoff)=0.1054.
FRQ set 7: Sampling Distribution of a Sample Mean
Scenario. A population at a farm cooperative in Central County during a quarterly performance study has mean 57 and SD 23. For random samples of size 25, find the mean and SD of x̄ and P(x̄>59.3) 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
μx̄=μ=57. σx̄=σ√n=23√25=4.6. The cutoff has z=(59.3−57)4.6=0.5, so P(x̄>cutoff)=0.3085.
FRQ set 8: Sampling Distribution of a Sample Mean
Scenario. A population at a solar installer in Pacific Northwest during a monthly quality review has mean 68 and SD 9. For random samples of size 36, find the mean and SD of x̄ and P(x̄>68.75) 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
μx̄=μ=68. σx̄=σ√n=9√36=1.5. The cutoff has z=(68.75−68)1.5=0.5, so P(x̄>cutoff)=0.3085.
