Question

without doing any computation, put the following in order from least to greatest, assuming the population is normally distributed with \\( \mu = 300 \\) and \\( \sigma = 25 \\).

(a) \\( p(280 \leq \overline{x} \leq 320) \\) for a random sample of size \\( n = 40 \\)
(b) \\( p(280 \leq \overline{x} \leq 320) \\) for a random sample of size \\( n = 50 \\)
(c) \\( p(280 \leq x \leq 320) \\)

Explanation:

• For a normal distribution, the standard deviation of the sample mean is \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\). As \(n\) increases, \(\sigma_{\bar{x}}\) decreases. A smaller standard deviation for the sample - mean distribution means that the values of \(\bar{x}\) are more concentrated around the mean \(\mu\).
• The probability \(P(280\leq x\leq320)\) for an individual observation (\(n = 1\)) has the largest standard deviation (\(\sigma=25\)).
• Between the two sample - mean probabilities, since \(n = 50>n = 40\), \(\sigma_{\bar{x}}\) for \(n = 50\) (\(\sigma_{\bar{x}}=\frac{25}{\sqrt{50}}\)) is smaller than \(\sigma_{\bar{x}}\) for \(n = 40\) (\(\sigma_{\bar{x}}=\frac{25}{\sqrt{40}}\)). A smaller standard deviation for the sample - mean distribution makes the probability \(P(280\leq\bar{x}\leq320)\) larger (because the distribution of \(\bar{x}\) is more concentrated around \(\mu = 300\)).

Answer:

\(P(280\leq x\leq320)<P(280\leq\bar{x}\leq320,n = 40)<P(280\leq\bar{x}\leq320,n = 50)\)