Sampling Distribution of the Sample Proportion
Model how p̂ changes across random samples, including its center p, standard error, success-failure condition, finite-population check, and normal probabilities.
Sampling Distribution Of The Sample Proportion: direct answer
Model how p̂ changes across random samples, including its center p, standard error, success-failure condition, finite-population check, and normal probabilities.
This page uses sampling distribution of the sample proportion 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 Proportion
| Center | μp̂=p |
|---|---|
| Standard error | √[p(1−p)/n] |
| Shape check | Expected successes and failures must be large enough for a normal approximation. |
| Without replacement | Use the finite-population/10% independence check when applicable. |
| Parameter vs statistic | p is fixed; p̂ varies from sample to sample. |
Concept mastery: Sampling Distribution of the Sample Proportion
p is fixed while p-hat varies
The population proportion p is a parameter. A sample proportion p̂ is a statistic that changes from sample to sample. Its repeated-sampling center is p under the usual random-sampling framework.
The standard error reflects binary variability
For independent Bernoulli observations, SD(p̂)=√[p(1−p)/n]. Variability is largest near p=.50 and smaller near 0 or 1, all else equal. Increasing n shrinks the standard error at the familiar 1/√n rate.
Success-failure counts support normal approximation
A normal model for p̂ requires enough expected successes and failures. The exact threshold depends on the course convention, but the logic is stable: when either side is too sparse, the distribution can be too discrete and skewed for a normal approximation.
Sampling without replacement needs a population-size check
The 10% condition is used when the sample is selected without replacement from a finite population. It limits dependence between selections so the standard-error formula based on approximate independence remains defensible.
Standardize using the sampling-distribution standard error
A probability involving p̂ is converted to z using the center p and SE=√[p(1−p)/n]. Substituting p̂ into this probability model changes the question; p̂-based standard errors belong to estimation procedures such as confidence intervals.
Worked analysis for Sampling Distribution of the Sample Proportion
Sample-proportion case 1: Urban Transit Riders
Let the population proportion be p=0.35 and take random samples of size 100. The sampling distribution of p̂ is centered at 0.35 with standard error √[p(1−p)/n]≈0.0477. The expected success and failure counts are np≈35.0 and n(1−p)≈65.0; both are comfortably above the usual normal-approximation threshold in this case.
For P(p̂>0.376), standardize with z=(0.376−0.35)/0.0477≈0.55. The parameter p belongs to the population, while p̂ varies from sample to sample. Confusing those roles would turn a sampling-distribution question into an incorrect statement about a fixed parameter.
Sample-proportion case 2: Regional Hospital Visits
Let the population proportion be p=0.42 and take random samples of size 120. The sampling distribution of p̂ is centered at 0.42 with standard error √[p(1−p)/n]≈0.0451. The expected success and failure counts are np≈50.4 and n(1−p)≈69.6; both are comfortably above the usual normal-approximation threshold in this case.
For P(p̂>0.452), standardize with z=(0.452−0.42)/0.0451≈0.70. The parameter p belongs to the population, while p̂ varies from sample to sample. Confusing those roles would turn a sampling-distribution question into an incorrect statement about a fixed parameter.
Sample-proportion case 3: Community College Placement Scores
Let the population proportion be p=0.49 and take random samples of size 150. The sampling distribution of p̂ is centered at 0.49 with standard error √[p(1−p)/n]≈0.0408. The expected success and failure counts are np≈73.5 and n(1−p)≈76.5; both are comfortably above the usual normal-approximation threshold in this case.
For P(p̂>0.525), standardize with z=(0.525−0.49)/0.0408≈0.85. The parameter p belongs to the population, while p̂ varies from sample to sample. Confusing those roles would turn a sampling-distribution question into an incorrect statement about a fixed parameter.
Sample-proportion case 4: Warehouse Order Times
Let the population proportion be p=0.56 and take random samples of size 200. The sampling distribution of p̂ is centered at 0.56 with standard error √[p(1−p)/n]≈0.0351. The expected success and failure counts are np≈112.0 and n(1−p)≈88.0; both are comfortably above the usual normal-approximation threshold in this case.
For P(p̂>0.574), standardize with z=(0.574−0.56)/0.0351≈0.40. The parameter p belongs to the population, while p̂ varies from sample to sample. Confusing those roles would turn a sampling-distribution question into an incorrect statement about a fixed parameter.
Sample-proportion case 5: Public-Library Checkouts
Let the population proportion be p=0.63 and take random samples of size 250. The sampling distribution of p̂ is centered at 0.63 with standard error √[p(1−p)/n]≈0.0305. The expected success and failure counts are np≈157.5 and n(1−p)≈92.5; both are comfortably above the usual normal-approximation threshold in this case.
For P(p̂>0.647), standardize with z=(0.647−0.63)/0.0305≈0.55. The parameter p belongs to the population, while p̂ varies from sample to sample. Confusing those roles would turn a sampling-distribution question into an incorrect statement about a fixed parameter.
Sample-proportion case 6: Solar-Panel Output Readings
Let the population proportion be p=0.29 and take random samples of size 80. The sampling distribution of p̂ is centered at 0.29 with standard error √[p(1−p)/n]≈0.0507. The expected success and failure counts are np≈23.2 and n(1−p)≈56.8; both are comfortably above the usual normal-approximation threshold in this case.
For P(p̂>0.326), standardize with z=(0.326−0.29)/0.0507≈0.70. The parameter p belongs to the population, while p̂ varies from sample to sample. Confusing those roles would turn a sampling-distribution question into an incorrect statement about a fixed parameter.
Sample-proportion case 7: School Attendance Rates
Let the population proportion be p=0.36 and take random samples of size 100. The sampling distribution of p̂ is centered at 0.36 with standard error √[p(1−p)/n]≈0.0480. The expected success and failure counts are np≈36.0 and n(1−p)≈64.0; both are comfortably above the usual normal-approximation threshold in this case.
For P(p̂>0.401), standardize with z=(0.401−0.36)/0.0480≈0.85. The parameter p belongs to the population, while p̂ varies from sample to sample. Confusing those roles would turn a sampling-distribution question into an incorrect statement about a fixed parameter.
Sample-proportion case 8: Restaurant Service Times
Let the population proportion be p=0.43 and take random samples of size 120. The sampling distribution of p̂ is centered at 0.43 with standard error √[p(1−p)/n]≈0.0452. The expected success and failure counts are np≈51.6 and n(1−p)≈68.4; both are comfortably above the usual normal-approximation threshold in this case.
For P(p̂>0.448), standardize with z=(0.448−0.43)/0.0452≈0.40. The parameter p belongs to the population, while p̂ varies from sample to sample. Confusing those roles would turn a sampling-distribution question into an incorrect statement about a fixed parameter.
Sample-proportion case 9: Wildlife Tag Measurements
Let the population proportion be p=0.50 and take random samples of size 150. The sampling distribution of p̂ is centered at 0.50 with standard error √[p(1−p)/n]≈0.0408. The expected success and failure counts are np≈75.0 and n(1−p)≈75.0; both are comfortably above the usual normal-approximation threshold in this case.
For P(p̂>0.522), standardize with z=(0.522−0.50)/0.0408≈0.55. The parameter p belongs to the population, while p̂ varies from sample to sample. Confusing those roles would turn a sampling-distribution question into an incorrect statement about a fixed parameter.
Sample-proportion case 10: Municipal Water-Use Records
Let the population proportion be p=0.57 and take random samples of size 200. The sampling distribution of p̂ is centered at 0.57 with standard error √[p(1−p)/n]≈0.0350. The expected success and failure counts are np≈114.0 and n(1−p)≈86.0; both are comfortably above the usual normal-approximation threshold in this case.
For P(p̂>0.595), standardize with z=(0.595−0.57)/0.0350≈0.70. The parameter p belongs to the population, while p̂ varies from sample to sample. Confusing those roles would turn a sampling-distribution question into an incorrect statement about a fixed parameter.
Sample-proportion case 11: Online Course Completion Times
Let the population proportion be p=0.64 and take random samples of size 250. The sampling distribution of p̂ is centered at 0.64 with standard error √[p(1−p)/n]≈0.0304. The expected success and failure counts are np≈160.0 and n(1−p)≈90.0; both are comfortably above the usual normal-approximation threshold in this case.
For P(p̂>0.666), standardize with z=(0.666−0.64)/0.0304≈0.85. The parameter p belongs to the population, while p̂ varies from sample to sample. Confusing those roles would turn a sampling-distribution question into an incorrect statement about a fixed parameter.
Sample-proportion case 12: Farm Yield Measurements
Let the population proportion be p=0.30 and take random samples of size 80. The sampling distribution of p̂ is centered at 0.30 with standard error √[p(1−p)/n]≈0.0512. The expected success and failure counts are np≈24.0 and n(1−p)≈56.0; both are comfortably above the usual normal-approximation threshold in this case.
For P(p̂>0.320), standardize with z=(0.320−0.30)/0.0512≈0.40. The parameter p belongs to the population, while p̂ varies from sample to sample. Confusing those roles would turn a sampling-distribution question into an incorrect statement about a fixed parameter.
Sample-proportion case 13: Clinic Appointment Waits
Let the population proportion be p=0.37 and take random samples of size 100. The sampling distribution of p̂ is centered at 0.37 with standard error √[p(1−p)/n]≈0.0483. The expected success and failure counts are np≈37.0 and n(1−p)≈63.0; both are comfortably above the usual normal-approximation threshold in this case.
For P(p̂>0.397), standardize with z=(0.397−0.37)/0.0483≈0.55. The parameter p belongs to the population, while p̂ varies from sample to sample. Confusing those roles would turn a sampling-distribution question into an incorrect statement about a fixed parameter.
Sample-proportion case 14: Manufacturing Fill Weights
Let the population proportion be p=0.44 and take random samples of size 120. The sampling distribution of p̂ is centered at 0.44 with standard error √[p(1−p)/n]≈0.0453. The expected success and failure counts are np≈52.8 and n(1−p)≈67.2; both are comfortably above the usual normal-approximation threshold in this case.
For P(p̂>0.472), standardize with z=(0.472−0.44)/0.0453≈0.70. The parameter p belongs to the population, while p̂ varies from sample to sample. Confusing those roles would turn a sampling-distribution question into an incorrect statement about a fixed parameter.
Sample-proportion case 15: County Commute Times
Let the population proportion be p=0.51 and take random samples of size 150. The sampling distribution of p̂ is centered at 0.51 with standard error √[p(1−p)/n]≈0.0408. The expected success and failure counts are np≈76.5 and n(1−p)≈73.5; both are comfortably above the usual normal-approximation threshold in this case.
For P(p̂>0.545), standardize with z=(0.545−0.51)/0.0408≈0.85. The parameter p belongs to the population, while p̂ varies from sample to sample. Confusing those roles would turn a sampling-distribution question into an incorrect statement about a fixed parameter.
Sample-proportion case 16: Energy Meter Readings
Let the population proportion be p=0.58 and take random samples of size 200. The sampling distribution of p̂ is centered at 0.58 with standard error √[p(1−p)/n]≈0.0349. The expected success and failure counts are np≈116.0 and n(1−p)≈84.0; both are comfortably above the usual normal-approximation threshold in this case.
For P(p̂>0.594), standardize with z=(0.594−0.58)/0.0349≈0.40. The parameter p belongs to the population, while p̂ varies from sample to sample. Confusing those roles would turn a sampling-distribution question into an incorrect statement about a fixed parameter.
Sample-proportion case 17: University Advising Durations
Let the population proportion be p=0.65 and take random samples of size 250. The sampling distribution of p̂ is centered at 0.65 with standard error √[p(1−p)/n]≈0.0302. The expected success and failure counts are np≈162.5 and n(1−p)≈87.5; both are comfortably above the usual normal-approximation threshold in this case.
For P(p̂>0.667), standardize with z=(0.667−0.65)/0.0302≈0.55. The parameter p belongs to the population, while p̂ varies from sample to sample. Confusing those roles would turn a sampling-distribution question into an incorrect statement about a fixed parameter.
Sample-proportion case 18: Sports Training Measurements
Let the population proportion be p=0.31 and take random samples of size 80. The sampling distribution of p̂ is centered at 0.31 with standard error √[p(1−p)/n]≈0.0517. The expected success and failure counts are np≈24.8 and n(1−p)≈55.2; both are comfortably above the usual normal-approximation threshold in this case.
For P(p̂>0.346), standardize with z=(0.346−0.31)/0.0517≈0.70. The parameter p belongs to the population, while p̂ varies from sample to sample. Confusing those roles would turn a sampling-distribution question into an incorrect statement about a fixed parameter.
Sample-proportion case 19: Call-Center Resolution Times
Let the population proportion be p=0.38 and take random samples of size 100. The sampling distribution of p̂ is centered at 0.38 with standard error √[p(1−p)/n]≈0.0485. The expected success and failure counts are np≈38.0 and n(1−p)≈62.0; both are comfortably above the usual normal-approximation threshold in this case.
For P(p̂>0.421), standardize with z=(0.421−0.38)/0.0485≈0.85. The parameter p belongs to the population, while p̂ varies from sample to sample. Confusing those roles would turn a sampling-distribution question into an incorrect statement about a fixed parameter.
Sample-proportion case 20: Retail Basket Totals
Let the population proportion be p=0.45 and take random samples of size 120. The sampling distribution of p̂ is centered at 0.45 with standard error √[p(1−p)/n]≈0.0454. The expected success and failure counts are np≈54.0 and n(1−p)≈66.0; both are comfortably above the usual normal-approximation threshold in this case.
For P(p̂>0.468), standardize with z=(0.468−0.45)/0.0454≈0.40. The parameter p belongs to the population, while p̂ varies from sample to sample. Confusing those roles would turn a sampling-distribution question into an incorrect statement about a fixed parameter.
Sample-proportion case 21: Lab Assay Values
Let the population proportion be p=0.52 and take random samples of size 150. The sampling distribution of p̂ is centered at 0.52 with standard error √[p(1−p)/n]≈0.0408. The expected success and failure counts are np≈78.0 and n(1−p)≈72.0; both are comfortably above the usual normal-approximation threshold in this case.
For P(p̂>0.542), standardize with z=(0.542−0.52)/0.0408≈0.55. The parameter p belongs to the population, while p̂ varies from sample to sample. Confusing those roles would turn a sampling-distribution question into an incorrect statement about a fixed parameter.
Sample-proportion case 22: Shipping Transit Times
Let the population proportion be p=0.59 and take random samples of size 200. The sampling distribution of p̂ is centered at 0.59 with standard error √[p(1−p)/n]≈0.0348. The expected success and failure counts are np≈118.0 and n(1−p)≈82.0; both are comfortably above the usual normal-approximation threshold in this case.
For P(p̂>0.614), standardize with z=(0.614−0.59)/0.0348≈0.70. The parameter p belongs to the population, while p̂ varies from sample to sample. Confusing those roles would turn a sampling-distribution question into an incorrect statement about a fixed parameter.
Sample-proportion case 23: District Test Scores
Let the population proportion be p=0.66 and take random samples of size 250. The sampling distribution of p̂ is centered at 0.66 with standard error √[p(1−p)/n]≈0.0300. The expected success and failure counts are np≈165.0 and n(1−p)≈85.0; both are comfortably above the usual normal-approximation threshold in this case.
For P(p̂>0.685), standardize with z=(0.685−0.66)/0.0300≈0.85. The parameter p belongs to the population, while p̂ varies from sample to sample. Confusing those roles would turn a sampling-distribution question into an incorrect statement about a fixed parameter.
Sample-proportion case 24: Park Visitor Durations
Let the population proportion be p=0.32 and take random samples of size 80. The sampling distribution of p̂ is centered at 0.32 with standard error √[p(1−p)/n]≈0.0522. The expected success and failure counts are np≈25.6 and n(1−p)≈54.4; both are comfortably above the usual normal-approximation threshold in this case.
For P(p̂>0.341), standardize with z=(0.341−0.32)/0.0522≈0.40. The parameter p belongs to the population, while p̂ varies from sample to sample. Confusing those roles would turn a sampling-distribution question into an incorrect statement about a fixed parameter.
Sample-proportion case 25: Airport Security Waits
Let the population proportion be p=0.39 and take random samples of size 100. The sampling distribution of p̂ is centered at 0.39 with standard error √[p(1−p)/n]≈0.0488. The expected success and failure counts are np≈39.0 and n(1−p)≈61.0; both are comfortably above the usual normal-approximation threshold in this case.
For P(p̂>0.417), standardize with z=(0.417−0.39)/0.0488≈0.55. The parameter p belongs to the population, while p̂ varies from sample to sample. Confusing those roles would turn a sampling-distribution question into an incorrect statement about a fixed parameter.
Sample-proportion case 26: Pharmacy Prescription Times
Let the population proportion be p=0.46 and take random samples of size 120. The sampling distribution of p̂ is centered at 0.46 with standard error √[p(1−p)/n]≈0.0455. The expected success and failure counts are np≈55.2 and n(1−p)≈64.8; both are comfortably above the usual normal-approximation threshold in this case.
For P(p̂>0.492), standardize with z=(0.492−0.46)/0.0455≈0.70. The parameter p belongs to the population, while p̂ varies from sample to sample. Confusing those roles would turn a sampling-distribution question into an incorrect statement about a fixed parameter.
Sample-proportion case 27: Bicycle Commute Distances
Let the population proportion be p=0.53 and take random samples of size 150. The sampling distribution of p̂ is centered at 0.53 with standard error √[p(1−p)/n]≈0.0408. The expected success and failure counts are np≈79.5 and n(1−p)≈70.5; both are comfortably above the usual normal-approximation threshold in this case.
For P(p̂>0.565), standardize with z=(0.565−0.53)/0.0408≈0.85. The parameter p belongs to the population, while p̂ varies from sample to sample. Confusing those roles would turn a sampling-distribution question into an incorrect statement about a fixed parameter.
Sample-proportion case 28: Community Garden Yields
Let the population proportion be p=0.60 and take random samples of size 200. The sampling distribution of p̂ is centered at 0.60 with standard error √[p(1−p)/n]≈0.0346. The expected success and failure counts are np≈120.0 and n(1−p)≈80.0; both are comfortably above the usual normal-approximation threshold in this case.
For P(p̂>0.614), standardize with z=(0.614−0.60)/0.0346≈0.40. The parameter p belongs to the population, while p̂ varies from sample to sample. Confusing those roles would turn a sampling-distribution question into an incorrect statement about a fixed parameter.
Sample-proportion case 29: Emergency Response Durations
Let the population proportion be p=0.67 and take random samples of size 250. The sampling distribution of p̂ is centered at 0.67 with standard error √[p(1−p)/n]≈0.0297. The expected success and failure counts are np≈167.5 and n(1−p)≈82.5; both are comfortably above the usual normal-approximation threshold in this case.
For P(p̂>0.686), standardize with z=(0.686−0.67)/0.0297≈0.55. The parameter p belongs to the population, while p̂ varies from sample to sample. Confusing those roles would turn a sampling-distribution question into an incorrect statement about a fixed parameter.
Sample-proportion case 30: College Credit Loads
Let the population proportion be p=0.33 and take random samples of size 80. The sampling distribution of p̂ is centered at 0.33 with standard error √[p(1−p)/n]≈0.0526. The expected success and failure counts are np≈26.4 and n(1−p)≈53.6; both are comfortably above the usual normal-approximation threshold in this case.
For P(p̂>0.367), standardize with z=(0.367−0.33)/0.0526≈0.70. The parameter p belongs to the population, while p̂ varies from sample to sample. Confusing those roles would turn a sampling-distribution question into an incorrect statement about a fixed parameter.
Sample-proportion case 31: Household Electricity Use
Let the population proportion be p=0.40 and take random samples of size 100. The sampling distribution of p̂ is centered at 0.40 with standard error √[p(1−p)/n]≈0.0490. The expected success and failure counts are np≈40.0 and n(1−p)≈60.0; both are comfortably above the usual normal-approximation threshold in this case.
For P(p̂>0.442), standardize with z=(0.442−0.40)/0.0490≈0.85. The parameter p belongs to the population, while p̂ varies from sample to sample. Confusing those roles would turn a sampling-distribution question into an incorrect statement about a fixed parameter.
Sample-proportion case 32: River Flow Measurements
Let the population proportion be p=0.47 and take random samples of size 120. The sampling distribution of p̂ is centered at 0.47 with standard error √[p(1−p)/n]≈0.0456. The expected success and failure counts are np≈56.4 and n(1−p)≈63.6; both are comfortably above the usual normal-approximation threshold in this case.
For P(p̂>0.488), standardize with z=(0.488−0.47)/0.0456≈0.40. The parameter p belongs to the population, while p̂ varies from sample to sample. Confusing those roles would turn a sampling-distribution question into an incorrect statement about a fixed parameter.
Sampling Distribution of the Sample Proportion: multiple-choice practice
Question 2. Sampling Distribution of a Sample Proportion
For a population proportion p=0.585, an SRS of n=400 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.607).
Answer: A
μp̂=p=0.585. σp̂=√p(1−p)n=√0.585(0.415)400=0.0246. z=(0.607−0.585)0.0246≈0.893, so the upper-tail probability is 0.1859.
Question 3. Sampling Distribution of a Sample Proportion
For a population proportion p=0.348, an SRS of n=144 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.384).
Answer: B
μp̂=p=0.348. σp̂=√p(1−p)n=√0.348(0.652)144=0.0397. z=(0.384−0.348)0.0397≈0.907, so the upper-tail probability is 0.1822.
Question 4. Sampling Distribution of a Sample Proportion
For a population proportion p=0.412, an SRS of n=100 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.471).
Answer: C
μp̂=p=0.412. σp̂=√p(1−p)n=√0.412(0.588)100=0.0492. z=(0.471−0.412)0.0492≈1.199, so the upper-tail probability is 0.1153.
Question 5. Sampling Distribution of a Sample Proportion
For a population proportion p=0.58, an SRS of n=144 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.617).
Answer: D
μp̂=p=0.58. σp̂=√p(1−p)n=√0.58(0.42)144=0.0411. z=(0.617−0.58)0.0411≈0.9, so the upper-tail probability is 0.1842.
Question 6. Sampling Distribution of a Sample Proportion
For a population proportion p=0.599, an SRS of n=196 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.62).
Answer: B
μp̂=p=0.599. σp̂=√p(1−p)n=√0.599(0.401)196=0.035. z=(0.62−0.599)0.035≈0.6, so the upper-tail probability is 0.2743.
Question 7. Sampling Distribution of a Sample Proportion
For a population proportion p=0.698, an SRS of n=196 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.728).
Answer: D
μp̂=p=0.698. σp̂=√p(1−p)n=√0.698(0.302)196=0.0328. z=(0.728−0.698)0.0328≈0.915, so the upper-tail probability is 0.1802.
Question 8. Sampling Distribution of a Sample Proportion
For a population proportion p=0.6, an SRS of n=196 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.642).
Answer: C
μp̂=p=0.6. σp̂=√p(1−p)n=√0.6(0.4)196=0.035. z=(0.642−0.6)0.035≈1.2, so the upper-tail probability is 0.115.
Question 9. Sampling Distribution of a Sample Proportion
For a population proportion p=0.259, an SRS of n=256 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.284).
Answer: D
μp̂=p=0.259. σp̂=√p(1−p)n=√0.259(0.741)256=0.0274. z=(0.284−0.259)0.0274≈0.913, so the upper-tail probability is 0.1806.
Question 10. Sampling Distribution of a Sample Proportion
For a population proportion p=0.69, an SRS of n=144 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.725).
Answer: A
μp̂=p=0.69. σp̂=√p(1−p)n=√0.69(0.31)144=0.0385. z=(0.725−0.69)0.0385≈0.908, so the upper-tail probability is 0.1819.
Question 11. Sampling Distribution of a Sample Proportion
For a population proportion p=0.299, an SRS of n=196 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.328).
Answer: A
μp̂=p=0.299. σp̂=√p(1−p)n=√0.299(0.701)196=0.0327. z=(0.328−0.299)0.0327≈0.887, so the upper-tail probability is 0.1876.
Question 12. Sampling Distribution of a Sample Proportion
For a population proportion p=0.57, an SRS of n=196 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.602).
Answer: C
μp̂=p=0.57. σp̂=√p(1−p)n=√0.57(0.43)196=0.0354. z=(0.602−0.57)0.0354≈0.905, so the upper-tail probability is 0.1828.
Question 13. Sampling Distribution of a Sample Proportion
For a population proportion p=0.431, an SRS of n=256 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.459).
Answer: A
μp̂=p=0.431. σp̂=√p(1−p)n=√0.431(0.569)256=0.031. z=(0.459−0.431)0.031≈0.905, so the upper-tail probability is 0.1828.
Question 14. Sampling Distribution of a Sample Proportion
For a population proportion p=0.421, an SRS of n=196 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.463).
Answer: A
μp̂=p=0.421. σp̂=√p(1−p)n=√0.421(0.579)196=0.0353. z=(0.463−0.421)0.0353≈1.191, so the upper-tail probability is 0.1168.
Question 15. Sampling Distribution of a Sample Proportion
For a population proportion p=0.574, an SRS of n=225 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.604).
Answer: D
μp̂=p=0.574. σp̂=√p(1−p)n=√0.574(0.426)225=0.033. z=(0.604−0.574)0.033≈0.91, so the upper-tail probability is 0.1814.
Question 16. Sampling Distribution of a Sample Proportion
For a population proportion p=0.331, an SRS of n=100 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.387).
Answer: D
μp̂=p=0.331. σp̂=√p(1−p)n=√0.331(0.669)100=0.0471. z=(0.387−0.331)0.0471≈1.19, so the upper-tail probability is 0.117.
Question 17. Sampling Distribution of a Sample Proportion
For a population proportion p=0.346, an SRS of n=225 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.375).
Answer: A
μp̂=p=0.346. σp̂=√p(1−p)n=√0.346(0.654)225=0.0317. z=(0.375−0.346)0.0317≈0.914, so the upper-tail probability is 0.1802.
Question 18. Sampling Distribution of a Sample Proportion
For a population proportion p=0.284, an SRS of n=256 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.301).
Answer: B
μp̂=p=0.284. σp̂=√p(1−p)n=√0.284(0.716)256=0.0282. z=(0.301−0.284)0.0282≈0.603, so the upper-tail probability is 0.2732.
Question 19. Sampling Distribution of a Sample Proportion
For a population proportion p=0.639, an SRS of n=100 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.682).
Answer: A
μp̂=p=0.639. σp̂=√p(1−p)n=√0.639(0.361)100=0.048. z=(0.682−0.639)0.048≈0.895, so the upper-tail probability is 0.1853.
Question 20. Sampling Distribution of a Sample Proportion
For a population proportion p=0.287, an SRS of n=196 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.306).
Answer: B
μp̂=p=0.287. σp̂=√p(1−p)n=√0.287(0.713)196=0.0323. z=(0.306−0.287)0.0323≈0.588, so the upper-tail probability is 0.2783.
Question 21. Sampling Distribution of a Sample Proportion
For a population proportion p=0.387, an SRS of n=400 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.409).
Answer: B
μp̂=p=0.387. σp̂=√p(1−p)n=√0.387(0.613)400=0.0244. z=(0.409−0.387)0.0244≈0.903, so the upper-tail probability is 0.1832.
Question 22. Sampling Distribution of a Sample Proportion
For a population proportion p=0.634, an SRS of n=225 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.663).
Answer: A
μp̂=p=0.634. σp̂=√p(1−p)n=√0.634(0.366)225=0.0321. z=(0.663−0.634)0.0321≈0.903, so the upper-tail probability is 0.1833.
Question 23. Sampling Distribution of a Sample Proportion
For a population proportion p=0.556, an SRS of n=144 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.581).
Answer: C
μp̂=p=0.556. σp̂=√p(1−p)n=√0.556(0.444)144=0.0414. z=(0.581−0.556)0.0414≈0.604, so the upper-tail probability is 0.273.
Question 24. Sampling Distribution of a Sample Proportion
For a population proportion p=0.616, an SRS of n=100 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.645).
Answer: B
μp̂=p=0.616. σp̂=√p(1−p)n=√0.616(0.384)100=0.0486. z=(0.645−0.616)0.0486≈0.596, so the upper-tail probability is 0.2755.
Question 25. Sampling Distribution of a Sample Proportion
For a population proportion p=0.566, an SRS of n=225 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.606).
Answer: B
μp̂=p=0.566. σp̂=√p(1−p)n=√0.566(0.434)225=0.033. z=(0.606−0.566)0.033≈1.211, so the upper-tail probability is 0.113.
Question 26. Sampling Distribution of a Sample Proportion
For a population proportion p=0.338, an SRS of n=400 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.366).
Answer: C
μp̂=p=0.338. σp̂=√p(1−p)n=√0.338(0.662)400=0.0237. z=(0.366−0.338)0.0237≈1.184, so the upper-tail probability is 0.1182.
Question 27. Sampling Distribution of a Sample Proportion
For a population proportion p=0.642, an SRS of n=400 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.656).
Answer: C
μp̂=p=0.642. σp̂=√p(1−p)n=√0.642(0.358)400=0.024. z=(0.656−0.642)0.024≈0.584, so the upper-tail probability is 0.2796.
Question 28. Sampling Distribution of a Sample Proportion
For a population proportion p=0.486, an SRS of n=196 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.518).
Answer: B
μp̂=p=0.486. σp̂=√p(1−p)n=√0.486(0.514)196=0.0357. z=(0.518−0.486)0.0357≈0.896, so the upper-tail probability is 0.185.
Question 29. Sampling Distribution of a Sample Proportion
For a population proportion p=0.253, an SRS of n=144 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.296).
Answer: C
μp̂=p=0.253. σp̂=√p(1−p)n=√0.253(0.747)144=0.0362. z=(0.296−0.253)0.0362≈1.187, so the upper-tail probability is 0.1176.
Question 30. Sampling Distribution of a Sample Proportion
For a population proportion p=0.544, an SRS of n=100 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.589).
Answer: A
μp̂=p=0.544. σp̂=√p(1−p)n=√0.544(0.456)100=0.0498. z=(0.589−0.544)0.0498≈0.904, so the upper-tail probability is 0.1831.
Question 31. Sampling Distribution of a Sample Proportion
For a population proportion p=0.269, an SRS of n=196 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.298).
Answer: C
μp̂=p=0.269. σp̂=√p(1−p)n=√0.269(0.731)196=0.0317. z=(0.298−0.269)0.0317≈0.916, so the upper-tail probability is 0.1799.
Question 32. Sampling Distribution of a Sample Proportion
For a population proportion p=0.33, an SRS of n=100 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.386).
Answer: C
μp̂=p=0.33. σp̂=√p(1−p)n=√0.33(0.67)100=0.047. z=(0.386−0.33)0.047≈1.191, so the upper-tail probability is 0.1168.
Question 33. Sampling Distribution of a Sample Proportion
For a population proportion p=0.452, an SRS of n=225 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.482).
Answer: B
μp̂=p=0.452. σp̂=√p(1−p)n=√0.452(0.548)225=0.0332. z=(0.482−0.452)0.0332≈0.904, so the upper-tail probability is 0.183.
Sampling Distribution of the Sample Proportion: free-response practice
FRQ set 2: Sampling Distribution of a Sample Proportion
Scenario. For a population proportion p=0.303, an SRS of n=100 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.358).
- 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
μp̂=p=0.303. σp̂=√p(1−p)n=√0.303(0.697)100=0.046. z=(0.358−0.303)0.046≈1.197, so the upper-tail probability is 0.1157.
FRQ set 3: Sampling Distribution of a Sample Proportion
Scenario. For a population proportion p=0.311, an SRS of n=144 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.346).
- 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
μp̂=p=0.311. σp̂=√p(1−p)n=√0.311(0.689)144=0.0386. z=(0.346−0.311)0.0386≈0.907, so the upper-tail probability is 0.1821.
FRQ set 4: Sampling Distribution of a Sample Proportion
Scenario. For a population proportion p=0.636, an SRS of n=100 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.665).
- 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
μp̂=p=0.636. σp̂=√p(1−p)n=√0.636(0.364)100=0.0481. z=(0.665−0.636)0.0481≈0.603, so the upper-tail probability is 0.2733.
FRQ set 5: Sampling Distribution of a Sample Proportion
Scenario. For a population proportion p=0.29, an SRS of n=256 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.316).
- 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
μp̂=p=0.29. σp̂=√p(1−p)n=√0.29(0.71)256=0.0284. z=(0.316−0.29)0.0284≈0.917, so the upper-tail probability is 0.1796.
FRQ set 6: Sampling Distribution of a Sample Proportion
Scenario. For a population proportion p=0.39, an SRS of n=100 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.434).
- 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
μp̂=p=0.39. σp̂=√p(1−p)n=√0.39(0.61)100=0.0488. z=(0.434−0.39)0.0488≈0.902, so the upper-tail probability is 0.1835.
FRQ set 7: Sampling Distribution of a Sample Proportion
Scenario. For a population proportion p=0.405, an SRS of n=225 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.444).
- 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
μp̂=p=0.405. σp̂=√p(1−p)n=√0.405(0.595)225=0.0327. z=(0.444−0.405)0.0327≈1.192, so the upper-tail probability is 0.1167.
FRQ set 8: Sampling Distribution of a Sample Proportion
Scenario. For a population proportion p=0.327, an SRS of n=144 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.374).
- 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
μp̂=p=0.327. σp̂=√p(1−p)n=√0.327(0.673)144=0.0391. z=(0.374−0.327)0.0391≈1.202, so the upper-tail probability is 0.1146.
FRQ set 9: Sampling Distribution of a Sample Proportion
Scenario. For a population proportion p=0.416, an SRS of n=256 is selected. Find the mean and SD of p̂ and approximate P(p̂>0.444).
- 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
μp̂=p=0.416. σp̂=√p(1−p)n=√0.416(0.584)256=0.0308. z=(0.444−0.416)0.0308≈0.909, so the upper-tail probability is 0.1817.
