Early Solutions to the 2026 AP Statistics FRQs | Step by Step

Part A: Find the five-number summary for Breed H

Given data. The ordered Breed H weights, in pounds, are:

• 48
• 48
• 55
• 56
• 56
• 57
• 62
• 66
• 72
• 72
• 72
• 73
• 80
• 80

There are \(n=14\) observations, so the median is the average of the seventh and eighth values.

1. Identify the endpoints. The minimum is \(48\) and the maximum is \(80\).
2. Calculate the median.
   \[\operatorname{Median}=\frac{62+66}{2}=64.\]
3. Find the first quartile. The lower half is \(48,48,55,56,56,57,62\). Its middle value is \(56\), so \(Q_1=56\).
4. Find the third quartile. The upper half is \(66,72,72,72,73,80,80\). Its middle value is \(72\), so \(Q_3=72\).

Part B: Compare center and variability

Read Breed J from the boxplot. Its approximate five-number summary is \((48,56,64,80,88)\) pounds.

1. Compare center. Both breeds have a median of \(64\) pounds, so their typical weights are the same when center is measured by the median.
2. Compare middle-spread using the IQR.
   \[\operatorname{IQR}_H=72-56=16,\qquad \operatorname{IQR}_J=80-56=24.\]
   Breed J's middle 50% covers a wider interval.
3. Compare overall spread using the range.
   \[\operatorname{Range}_H=80-48=32,\qquad \operatorname{Range}_J=88-48=40.\]
   Breed J also has the larger overall range.

Part C: Explain what the stem-and-leaf plot reveals

Part A: Identify the elements of the experiment

Part B: Give a valid random-assignment procedure

1. Label the rosebushes \(1,2,\ldots,30\).
2. Use a random-number generator to select 15 distinct labels from 1 through 30.
3. Assign the selected bushes to receive one-half cup of coffee grounds each week.
4. Assign the remaining 15 bushes to the no-coffee-grounds treatment.

This creates two groups of exactly 15 and gives every bush a fair random chance of receiving either treatment.

Part C: Interpret statistical significance at \(\alpha=0.05\)

The null hypothesis states that adding coffee grounds has no effect on the mean number of roses. Statistical significance at the \(0.05\) level means the randomization-based p-value is less than \(0.05\).

Part A: Find the probability a performance exceeds 120 seconds

1. Standardize 120 seconds.
   \[z=\frac{120-109}{16}=\frac{11}{16}=0.6875.\]
2. Use the upper tail of the standard normal distribution.
   \[P(T>120)=P(Z>0.6875)=1-\Phi(0.6875)\approx0.2458839.\]

The normal model and standardization process are reviewed in the verified normal distribution guide.

Part B: Find the probability of at least 3 long performances among 10

Define a success as a performance lasting longer than 120 seconds. From Part A, \(p=0.2458839\).

The binomial conditions are satisfied: there are 10 fixed trials, two outcomes on each trial, independent performances, and a constant success probability. Therefore,

Using the complement is shorter than adding eight separate probabilities:

Part C: Games until the first long performance

Let \(Y\) be the number of independent games attended up to and including the first game in which the performance exceeds 120 seconds. Then

Part D: Interpret the standard deviation

A correct interpretation names the random variable, includes units, and describes a typical distance from the mean. It should not describe the standard deviation as a probability.

Complete two-sample \(t\) test

Define \(\mu_C\) and \(\mu_N\) as the population mean numbers of oranges for similar trees treated with Brand C and Brand N, respectively.

1. State the hypotheses. The question asks whether the means differ, so use a two-sided alternative:
   \[H_0:\mu_C-\mu_N=0,\qquad H_a:\mu_C-\mu_N\ne0.\]
2. Check the conditions.
   • Random assignment: 58 trees were randomly assigned to each fertilizer, supporting a cause-and-effect conclusion for comparable trees.
   • Independent groups: Different trees received Brand C and Brand N.
   • Approximately normal sampling distribution: Both sample sizes are large, \(n_C=n_N=58>30\), so the Central Limit Theorem applies to \(\bar{X}_C-\bar{X}_N\).
3. Calculate the test statistic.
   \[ \begin{aligned} \bar{x}_C-\bar{x}_N&=141-148=-7,\\ SE&=\sqrt{\frac{s_C^2}{n_C}+\frac{s_N^2}{n_N}} =\sqrt{\frac{15^2}{58}+\frac{19^2}{58}} =\sqrt{\frac{586}{58}}\approx3.1786,\\ t&=\frac{-7-0}{3.1786}\approx-2.202. \end{aligned} \]
   Welch's approximation gives approximately \(108\) degrees of freedom.
4. Find the two-sided p-value.
   \[p=2P(T_{108}\le-2.202)\approx0.0298.\]
5. Make the decision. Since \(0.0298<0.05\), reject \(H_0\).

The athlete age-by-sport table

Part A: Calculate probabilities from the table

Part B: Interpret the mosaic plot

Part B(i): Meaning of \(b\)

The horizontal width of the football column is the proportion of all athletes who play football:

The vertical height is \(h=P(25\le A<30\mid F)\).

Part B(ii): Meaning of \(x=b\cdot h\)

Area in a mosaic plot represents a joint probability:

Part C: Determine whether two events are mutually exclusive or independent

Let \(B\) denote "Baseball" and \(G\) denote "age at least 35."

Part C(i): Mutually exclusive?

No. The table contains 61 athletes who are both baseball players and at least 35 years old:

Part C(ii): Independent?

Compare the conditional probability with the marginal probability:

Because these are not equal, knowing that an athlete plays baseball changes the probability that the athlete is at least 35. The equivalent product check also fails:

Part D: Decide whether a chi-square test for independence is appropriate

A chi-square inference test needs data from a random sample or randomized process when the purpose is to generalize beyond the observed individuals. This table is a census of all professional athletes in the three sports for the stated year, not a random sample.

The expected-count condition would be satisfied. The smallest expected count is

However, meeting the expected-count condition does not repair the failed random-sample condition.

Part A: Describe the overall relationship and make a prediction

Part A(i): Describe the relationship

The association is positive, roughly linear, and moderately strong to strong. Teams with more hits tend to score more runs. The points show noticeable but not excessive scatter around an increasing line, with no clear curve and no point completely isolated from the pattern.

A few teams near 1,400 hits and 650-700 runs fall below the main trend. The high-hits, high-runs point has leverage but remains consistent with the positive direction.

Part A(ii): Predict runs for a team with 1,250 hits

The regression equation is \(\hat{y}=-372.2+0.823x\). Substitute \(x=1250\):

Part B: Compare the salary classifications

Part B(i): Compare point A with its salary group

Point A is a square, so it represents a below-median-salary team. It is near 1,440 hits and 850 runs. Within the below-median group, it has the greatest run total and one of the greatest hit totals. It also lies above most square points and above the group's central pattern, so it produced unusually many runs for its number of hits compared with most lower-salary teams.

Part B(ii): Compare the strength of the two relationships

The above-median-salary dots follow a more tightly clustered increasing linear pattern. The below-median-salary squares show more vertical scatter at similar hit totals; around 1,350-1,450 hits, for example, their run totals range from roughly the mid-600s to about 850.

Part C: Construct confidence and prediction intervals at 1,250 hits

The sample size is \(n=30\), so the regression intervals use \(df=n-2=28\). The point estimate is \(\hat{y}=656.55\) runs.

Part C(i): Critical value

Part C(ii): 95% confidence interval for the mean response

The standard error for the mean response is \(17.48\) runs:

Part C(iii): 95% prediction interval for one team

The prediction standard error is \(56.78\) runs:

Part D: Explain why the prediction interval is wider

Part D(i): Means vary less than individuals

A distribution of sample means has less variability than a distribution of individual observations because averaging smooths individual-to-individual variation. In the simplest setting,

Part D(ii): Compare the standard-error formulas

For the mean response at \(x_0\),

For one future response,

The prediction formula contains an additional \(1\) under the square root. That term represents the natural residual variation of one team's run total around the regression line. A mean-response interval includes uncertainty in estimating the mean at \(x_0\); a prediction interval includes that uncertainty plus individual-team variation.