AP Statistics 2.3 Estimating Probabilities Using Simulation- Exam Style Questions - FRQs - New Syllabus

(a)

Step 1 — State the hypotheses:
\( H_0: \mu = 128 \text{ fl oz} \quad \text{vs.} \quad H_a: \mu < 128 \text{ fl oz} \)
where \(\mu\) is the true mean amount of milk in containers from this plant.
We test whether the mean is below 128 fl oz, since that would indicate non-compliance.
Step 2 — Identify the procedure and check conditions:
Use a one-sample \(t\)-test for a mean:
\( t = \frac{\bar{x} – \mu_0}{s/\sqrt{n}} \)
The problem states the filling process produces a normal distribution, so the normality condition is satisfied.
The containers were randomly sampled, so independence holds.
Step 3 — Compute the test statistic and p-value:
With \(\bar{x} = 127.2\), \(\mu_0 = 128\), \(s = 2.1\), \(n = 12\):
\( t = \frac{127.2 – 128}{2.1/\sqrt{12}} = \frac{-0.8}{0.6062} \approx -1.319 \)
Degrees of freedom: \(df = n – 1 = 11\).
For a one-tailed test with \(t = -1.319\) and \(df = 11\):
\( p\text{-value} = P(T_{11} < -1.319) \approx 0.107 \)
Step 4 — State the conclusion in context:
Since the p-value of \(0.107\) is greater than any reasonable significance level (e.g., \(\alpha = 0.05\)), we fail to reject \(H_0\). There is not sufficient evidence to conclude that the plant is out of compliance. The sample mean of 127.2 fl oz is below 128, but the difference is small enough that it could plausibly be due to random sampling variability alone.

(b)

Let \(X\) be the amount of milk in a randomly selected container, where \(X \sim N(128.0,\ 2.0)\). We want \(P(X \geq 125)\).
Standardize by converting to a \(z\)-score: \( z = \frac{125 – 128}{2} = \frac{-3}{2} = -1.5 \)
Using the standard normal table:
\( P(X \geq 125) = P(Z \geq -1.5) = 1 – P(Z < -1.5) = 1 – 0.0668 \)
\( \boxed{P(X \geq 125) = 0.9332} \)
So about 93.3% of containers from this machine will contain at least 125 fl oz.

(c)

Let \(X_{(1)} = \min(X_1, X_2, \ldots, X_{12})\) be the smallest value among 12 randomly selected containers.
For the minimum to be at least 125 fl oz, every single one of the 12 containers must contain at least 125 fl oz.
Since the containers are independent:
\( P(X_{(1)} \geq 125) = P(X_1 \geq 125) \times P(X_2 \geq 125) \times \cdots \times P(X_{12} \geq 125) \)
\( = [P(X \geq 125)]^{12} = (0.9332)^{12} \)
\( \boxed{P(X_{(1)} \geq 125) \approx 0.4362} \)
There is roughly a 43.6% chance that the smallest of 12 containers all meet the 125 fl oz threshold.
Even though each individual container has a 93.3% chance of passing, the probability that all 12 pass simultaneously drops considerably.

(d)

From the sorted list of 150 simulated minimums, we count how many are at least 125 fl oz. Scanning the table, the minimums first reach 125.00 at sample 85. From sample 85 through sample 150, that gives:
\( 150 – 85 + 1 = 66 \text{ minimums that are} \geq 125 \text{ fl oz} \)
The simulated probability estimate is therefore:
\( \hat{p} = \frac{66}{150} \approx \boxed{0.44} \)
Comparison with the theoretical value:
The theoretical probability from part (c) was \(0.4362\).
The simulation estimate of \(0.44\) is very close — the difference is only:
\( |0.44 – 0.4362| = 0.0038 \)
This is a tiny discrepancy, which is exactly what we expect from a simulation of this size.
The simulation does an excellent job of approximating the true theoretical probability, confirming that the independence-based calculation in part (c) is correct.