📊AP Statistics Review
AP Stats FRQ Practice Prompt Samples & Feedback (Unit 5)
AP Stats FRQ Practice Prompt Samples & Feedback (Unit 5)
The FRQ is a great way to prep for the AP exam! Review FRQ practice writing samples from Unit 5 and corresponding feedback from Fiveable teacher Jerry Kosoff.
AP Statistics FRQ exam format
Section II has 6 free-response questions. Part A includes Questions 1–5 completed in 65 minutes; Part B is Question 6, the investigative task, completed in 25 minutes. On the exam, FRQs assess skills like selecting methods, data analysis, probability/simulation, and statistical argumentation. This Unit 5 practice problem is most aligned to probability and sampling distributions and to communicating statistical reasoning clearly.
The FRQ Practice Prompt
A researcher in Yellowstone National Park observed the “Old Faithful” geyser for several weeks. For each eruption of the geyser, the duration from start to end, in seconds, was recorded. The histogram below summarizes the results from 421 observations. The mean of the distribution is 210 seconds, with a standard deviation of 68 seconds.
a. Describe the sampling distribution of sample mean eruption length for random samples of 40 eruptions from the researcher’s observations.
b. What is the probability that the sample mean eruption length for a random sample of 40 eruptions is 200 seconds or less?
Clarifying note: For this exercise, treat the 421 recorded eruptions as the population being sampled from. Therefore, the independence condition can be checked with n < 0.10N: 40 < 0.10(421) = 42.1, so the 10% condition is satisfied. If a problem instead sampled from all eruptions in nature, students would need information about that larger population.
FRQ Writing Samples and Feedback
FRQ Practice Submission 1
a) The sampling distribution of sample mean eruption length for random samples of 40 eruptions has a mean of 210 seconds, and it is bimodal with peaks at 100-125 seconds and 250-275 seconds. The shape of the sampling distribution of sample mean eruption length seems to be roughly symmetrical, and the range of the sampling distribution of sample mean eruption length for random samples of 40 eruptions is no more than 250 seconds.
b) x = mean geyser eruption duration for a random sample of 40 eruptions
Conditions: Random - stated that there were random samples of 40 eruptions, 10% Rule for Independence - because samples are drawn from the 421 recorded observations in this prompt, 40 < 0.10(421) = 42.1, so the 10% condition is satisfied, Normal/Large Sample - satisfied since n =40 >= 30; therefore, the sampling distribution of sample mean eruption duration is approximately normal.
P(x<200) = P(z<-0.14) = .4443 <–from Table A using z-score of -0.14
z = (200-210)/68 = -0.14
[pretend i drew a picture of a normal distribution here with 210 as median, 200 slightly to left of it, and everything shaded below -0.14]
The probability that the sample mean eruption duration for a random sample of 40 eruptions is 200 seconds is less is 0.4443.
Teacher feedback
In part (a), you appear to misunderstand what you’re being asked to describe. You describe the distribution provided by the histogram. However, the histogram is really providing the distribution of the “population” in this scenario; we are being asked to describe what it would look like if we took repeated samples of 40 eruptions from the graph shown and create a new graph of x-bars. Since n = 40 is sufficiently large, the Central Limit Theorem tells us that the sampling distribution of x̄ is approximately normal, even though the original eruption-length distribution is not normal. We would describe the sampling distribution as approximately normal with mean 210 seconds and standard deviation 68/√40 ≈ 10.75 seconds.
The misconception in part (a) then extends to part (b) - you use the “original” standard deviation when we should instead use the standard error of 68/sqrt(40) = 10.751 seconds. This impacts the z-score you would get and ultimately your associated probability.
Another small thing: you had the right idea checking the 10% condition, but used the wrong numbers. Because samples are drawn from the 421 recorded observations in this prompt, 40 < 0.10(421) = 42.1, so the 10% condition is satisfied.
FRQ Practice Submission 2
a) The sampling distribution of the sample mean is approximately normal because n = 40 is at least 30, so the Central Limit Theorem applies. The center of the sampling distribution is 210 seconds. The variability is 10.752 seconds because 68/sqrt(40) = 10.752.
b) P(x̄<=200) = ?
Use the z = (x-bar - mu)/(standard deviation/sqrt(n)) equation
(200-210)/10.752 = -0.93
Using Table A, a z-score of -0.93 is a probability of 0.1762.
The probability that the sample mean eruption length for a random sample of 40 eruptions is 200 seconds or less is 0.1762.
Teacher feedback
In part (a), you correctly describe the shape, center, and spread of the sampling distribution, citing the Central Limit Theorem as the reason for the distribution being “approximately normal.” Nice correction to focus on the sampling distribution rather than the population distribution.
In part (b), you do a good job of communicating the probability you are asked to find, then carry out calculations correctly and answer in context. Nice job! On an AP Statistics FRQ, imprecise notation or terminology can cost communication credit or prevent earning a component point if it makes your statistical meaning unclear, so use x̄ and “sampling distribution” precisely.
FRQ Practice Submission 3
a.) The sampling distribution is approximately normal because according to the Central Limit Theorem, if the sampling size (40) is greater than 30, the shape is approximately normal. The mean of the sampling distribution is 210 seconds. The standard deviation of the sampling distribution is 68/sqrt(40)=10.75.
b.) P(x̄<200)= P(z<-0.93)= .1762
z= (200-210)/10.75 = -.93
There is a 17.62% chance that the sample mean eruption length for a random sample of 40 eruptions is 200 seconds or less.
Teacher feedback
Good on both parts! In part (a), you give correct descriptions for shape, center, and spread, and correctly invoke the Central Limit Theorem since n = 40 > 30. In part (b), you calculate the correct probability. Small note on notation: because the question is about a sample mean, use x̄ rather than a generic X. On an AP Statistics FRQ, imprecise notation or terminology can cost communication credit or prevent earning a component point if it makes your statistical meaning unclear, so use x̄ and “sampling distribution” precisely.
FRQ Practice Submission 4
A. The sampling distribution is approximately normal due to the Central Limit Theorem (40 is greater than 30). There seem to be no unusual features to note. For the center, the mean of the distribution is 210. And for the spread, the standard deviation is 10.75 ( 68/ square root of 40).
B. Required Assumptions:
Sampling: There is a random sample of 40 eruptions.
Approximately Normal: 40 is greater than 30 therefore it meets the Central Limit Theorem so we can assume the sampling distribution of x̄ is approximately normal.
Independence: 40 < 0.10(421) = 42.1, so the 10% condition is satisfied.
The mean is 210. Standard deviation is 10.75 ( 68/ square root of 40). I then proceeded to find the z-score: (200-210)/10.75=-.9302.
The probability statement is P(Z is less than or equal to -.9302).
Then using my calculator I did normalcdf(-1000,-.9302,0,1) and found the probability which is .1761.
To conclude, there’s a 17.61% chance that the sample mean of a random sample of 40 eruptions is 200 seconds or less.
Also I would have added a sketch to show the distribution
Teacher feedback
Well done! You’ve correctly invoked the CLT in part (a) to justify your shape being approximately normal, while giving correct measures of center and spread. Be careful - your first words should refer to “the sampling distribution” instead of “the sample” - there’s a big difference in those two things. On an AP Statistics FRQ, imprecise notation or terminology can cost communication credit or prevent earning a component point if it makes your statistical meaning unclear, so use x̄ and “sampling distribution” precisely.
Your calculations in part (b) are on the right track as well.
FRQ Practice Submission 5
- A. The sampling distribution of sample mean eruption length for random samples of 40 eruptions would be approximately normal due to the Central Limit Theorem — because the sample size is greater than 30, the sampling distribution will be approximately normal.
- B. The probability that the sample mean eruption length for a random sample of 40 eruptions is 200 seconds or less is about 44.04%. The Z-score would be about -0.15.
Teacher feedback
For part (a), you correctly identify the shape as “approximately normal” due to the CLT (and give the correct reason, n = 40 > 30). However, a description of a distribution should include measures of center and spread to go with shape. (Many teachers use “S.O.C.S.” or “C.U.S.S.” as acronyms to help students remember - Shape/Outliers/Center/Spread or Center/Unusual Features/Shape/Spread). In this case, you did not mention the mean of the sampling distribution (which would still be 210) or the standard error (which would be 68/sqrt(40) = 10.75). This then impacted your probability calculation in part (b).
FRQ Practice Submission 6
a) The sampling distribution of sample mean eruption length for random samples of 40 eruptions from the researcher’s observations is approximately normal(random samples of 40 eruptions > 30; Central Limit theorem). The distribution has a mean(center) of 210 sec and and standard deviation (spread) of 10.75 sec ( 68/ sqrt 40).
b) The probability that the sample mean eruption length for a random sample of 40 eruptions is 0.176.
Conditions: Random: Random sample of 40 eruptions was taken.
Independent: Random sample of 40 eruptions is less than 10% of the population; 40 < 0.10(421) = 42.1, so the 10% condition is satisfied.
Normal: 40 samples > 30 ; Central Limit Theorem is satisfied
Calculator: normalcdf[ Lower:0, Upper: 200, u: 210, st. dev. : 10.75] = 0.1761
Teacher feedback
Almost perfect. Part (a) is strong. In part (b), using normalcdf(-1E99, 200, 210, 10.75) is the most general setup for “200 or less.” Because eruption times are nonnegative, using 0 as the lower bound gives essentially the same probability here, so this is better treated as a notation/setup refinement rather than a meaningful numerical error.
FRQ Practice Submission 7
a) The distribution eruption length is approximately normal as n>30 with a range of 225 seconds, a mean of 210 seconds, and a standard deviation of ~10.7517 seconds.
Do I need to mention outliers and range here?
b) According to the central limit theorem, a sample of n>=30 so our sample of 40 tells us this is approximately normal. Also, 40 < 0.10(421) = 42.1, so the 10% condition is satisfied, and we are told the sample is random.
normCdf(lower=-1e99,upper=200,μ=210,σ=10.751744)=0.176164
Teacher feedback
part (a) has most of what is needed, though be careful to describe the sampling distribution of x̄, not the original distribution of eruption lengths. Center, shape, and spread are the big ideas here, and you should show where the 10.75 seconds calculation came from. Range is not really the focus for this description.
Part (b), you’ve done all appropriate calculations.
FRQ Practice Submission 8
a. The sampling distribution of 40 random samples of eruption would be approximately normal. The distribution of 40 random samples would be centered at the mean of 210. The shape of the distribution would be bell-shaped and approximately symmetrical. The sampling distribution would be spread with a standard deviation of 68/sqrt(40) = 10.752. The sampling distribution would not have any unusual features or gaps.
b. Assumptions:
-We have a random sample of geyser eruptions.
-Because samples are drawn from the 421 recorded observations in this prompt, 40 < 0.10(421) = 42.1, so the 10% condition is satisfied.
-Since the sample size is large enough (n>30) due to CLT, the sampling distribution is approximately normal
-The standard deviation of the eruption-length distribution is given as 68 seconds, so the standard deviation of x̄ is 68/√40.
Calculations
p(x_bar ≤ 200) = normalcdf(-1E99, 200, 210, 10.752) = 0.1762
Conclusion
The probability that the sample mean eruption length for a random sample of 40 eruptions is 200 seconds or less is 0.1762.
Teacher feedback
Solid work! The only possible issue: in part (a), you mention the shape as “approximately normal”, but don’t give the reason for that until part (b) (since n = 40 > 30, the CLT applies). It’s always safest to show that the CLT applies right when you describe the sampling distribution of x̄.
Also, note that “sigma known” is not a required condition to justify this probability calculation; the relevant justifications are random sampling, independence/10% condition, and the CLT/large sample condition.
FRQ Practice Submission 9
a. The sampling distribution of sample mean eruption length for random samples from the researchers observations is approximately normal 40>30 so CLT applies, it is centered at mu=210 and the samples were chosen randomly and the spread is 68/sqrt40=10.75 and 40 < 0.10(421) = 42.1, so the 10% condition is satisfied.
b. 200-210/10.75=.93
p=.1762
The probability that the sample mean eruption length for a random sample of 40 eruptions is 200 seconds or less is .1762.
Teacher feedback
Good work. Part (a) includes the key ideas of shape, center, and spread, though the independence check and random sampling details are more useful when justifying the probability calculation in part (b).
In part (b), be careful with your written work: the standardized value should be . Your final probability is correct, but it helps to show the negative sign and label the z-score clearly.
FRQ Practice Submission 10
a. The sampling distribution of the sample mean eruption length for random samples of 40 eruptions from the researcher’s observations can be described as approximately normal (by the Central Limit Theorem, as n = 40 >30 and is thus sufficiently large) with a mean of 210 seconds and a standard deviation of 68/sqrt(40) = 10.7517 seconds (N(210, 10.7517))).
b. Let x̄ represent the sample mean eruption length for a random sample of 40 eruptions. Then x̄ is approximately distributed as Normal with mean 210 and standard deviation 68/√40 ≈ 10.7517.
The probability that the sample mean eruption length for a random sample of 40 eruptions is 200 seconds or less is: P(x̄<200) = normcdf(lowerbound = -infinity, upperbound = 200, mu = 210, sigma = 10.7517) = 0.176164.
Teacher feedback:
Nicely done. You’ve correctly justified the sampling distribution with the CLT and then done appropriate calculations in part (b).
One notation note: using x̄ is the preferred symbol here because the quantity being modeled is the sample mean, not an individual eruption time X.
FRQ Practice Submission 11
a) The sampling distribution of sample mean eruption length for random samples of 40 eruptions has a mean of 210 seconds and a standard deviation of 10.75 (68/sqrt40). It is approximately normally distributed as the central limit theorem is applicable because n is greater than or equal to 30 (n=40). Because the sampling distribution is approximately normal, its shape can be described as symmetrical.
b) Conditions: There is a random sample of 40 eruptions. The sampling distribution is approximately normal because n>30, so the Central Limit Theorem applies.
We want P(x̄ ≤ 200).
z=(200-210)/(68/sqrt40)= -.93
P(x̄ ≤ 200) = P(Z ≤ -0.93) ≈ .176
The probability that the sample mean eruption length of a random sample of 40 eruptions is 200s or less is .176
Teacher feedback
Nice work overall. Your setup and probability statement are correct: P(x̄ ≤ 200) = P(Z ≤ -0.93) ≈ 0.176. One wording note: because the Central Limit Theorem is being used, describe the sampling distribution of x̄ as approximately normal rather than exactly normal. Also, be explicit that 68/√40 ≈ 10.75 seconds is the standard deviation of the sampling distribution of x̄ (the standard error).
FRQ Practice Submission 12
a. Since the sample size is large (n=40 above 30), we can approximate the sampling distribution of x̄ with a normal curve. Since the data come from random samples, the mean of the sampling distribution of x̄ is 210 seconds, the same as the mean of the eruption-length distribution described in the prompt. Because samples are drawn from the 421 recorded observations in this prompt, 40 < 0.10(421) = 42.1, so the 10% condition is satisfied. The standard deviation of x̄ (the standard error) is population standard deviation divided by the square root of n. We have 68/sqrt(40)=10.752.
b. We let x̄ be the mean duration of a random sample of 40 eruptions. We want to find P(x̄≤200), which is equivalent to P(z≤(200-210)/10.752=-0.93). We use normal cdf with a lower bound of -∞, an upper bound of -0.93, a mean of 0, and a standard deviation of 1, and we get 0.176. The probability that the sample mean eruption length for a random sample of 40 eruptions is 200 seconds or less is 0.176.
Teacher feedback
Your overall approach is good, and the probability calculation in part (b) is correct.
A couple of terms in part (a) needed tightening: it’s not correct to say “the population mean is the sample mean,” and 68/√40 is not a “sample standard deviation.” Better language is: For the sampling distribution of x̄, the mean is 210 seconds, the same as the mean of the eruption-length distribution described in the prompt. The standard deviation of x̄ (the standard error) is 68/√40 ≈ 10.752 seconds. On an AP Statistics FRQ, imprecise terminology can cost communication credit if it makes your meaning unclear.