AP Statistics 1.10 The Normal Distribution- FRQs - Exam Style Questions
Question
(i) Calculate the probability that the sample mean amount of gold applied to a random sample of \(n=2\) necklaces will be greater than \(303\) mg.
(ii) Suppose Cleo took a random sample of \(n=2\) necklaces that resulted in a sample mean amount of gold applied of \(303\) mg. Would that result indicate that the population mean amount of gold being applied by the machine is different from \(300\) mg? Justify your answer without performing an inference procedure.
(i) Describe the sampling distribution of the sample range for random samples of size \(n=2\) from a normal distribution with standard deviation \(\sigma=5\), as shown in Graph I.
(ii) Describe how the sampling distribution of the sample range for samples of size \(n=2\) changes as the value of the population standard deviation increases.
(i) Consider Cleo’s range of \(10\) mg from the sample of size \(n=2\). If the machine is working properly with a standard deviation of \(5\) mg, is a sample range of \(10\) mg unusual? Justify your answer.
(ii) Do Cleo’s sample mean of \(303\) mg and range of \(10\) mg indicate that the machine is not working properly? Explain your answer.
Most-appropriate topic codes (CED):
• TOPIC 5.7 — Sampling Distributions for Sample Means: (b)(i), (b)(ii), (d)(ii)
• TOPIC 5.1 — Introducing Statistics: Why Is My Sample Not Like Yours?: (c)(i), (c)(ii), (d)(i), (d)(ii)
▶️ Answer/Explanation
(a)
We are looking for \(P(296 < X < 304)\) for a normal distribution with \(\mu=300\) and \(\sigma=5\).
– Z-score for \(296\): \(z = \frac{296-300}{5} = -0.8\)
– Z-score for \(304\): \(z = \frac{304-300}{5} = 0.8\)
\(P(-0.8 < Z < 0.8) = P(Z < 0.8) – P(Z < -0.8) \approx 0.7881 – 0.2119 = 0.5762\).
\(\boxed{P \approx 0.576}\)
(b)
(i) The sampling distribution of \(\bar{x}\) for \(n=2\) is normal with \(\mu_{\bar{x}}=300\) and \(\sigma_{\bar{x}} = \frac{5}{\sqrt{2}} \approx 3.536\).
We need \(P(\bar{x} > 303)\).
\(z = \frac{303-300}{3.536} \approx 0.848\).
\(P(Z > 0.848) \approx 0.198\).
(ii) No, this result would not provide convincing evidence. A sample mean of \(303\) mg is not unusual because the probability of observing a sample mean this far or farther from \(300\) mg (\(P(\bar{x} \ge 303)\) or \(P(\bar{x} \le 297)\)) is large (\(2 \times 0.198 = 0.396\)).
(c)
(i) The sampling distribution of the sample range shown in Graph I is skewed to the right. The center is approximately \(6\) mg, and the values are spread from \(0\) mg to about \(25\) mg.
(ii) As the population standard deviation (\(\sigma\)) increases, the sampling distribution of the sample range becomes more spread out and its center (mean) increases.
(d)
(i) No, a sample range of \(10\) mg is not unusual. According to Graph I (\(\sigma=5\)), there is a notable proportion of the distribution at or above a sample range of \(10\) mg (approximately \(20\%\)), so this value occurs frequently by chance.
(ii) No, Cleo’s results do not indicate the machine is not working properly. As shown in part (b), a sample mean of \(303\) mg is not unusual. As shown in part (d-i), a sample range of \(10\) mg is also not unusual. Since neither the sample mean nor the sample range is an unusual result, there is no convincing evidence that the machine is not working properly.
