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:

Step1: Recall the Central Limit Theorem and the formula for the standard error of the sample mean

The standard error of the sample mean is \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\). As \(n\) increases, \(\sigma_{\bar{x}}\) decreases. For a normal population, the probability \(P(a\leq\bar{x}\leq b)\) is related to the \(z -\)scores \(z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}\). A smaller standard error \(\sigma_{\bar{x}}\) will lead to larger \(|z|\) - values for the same \(\bar{x}\) - values (relative to \(\mu\)).

Step2: Analyze the population probability versus the sample - mean probabilities

For the population (\(n = 1\)), the standard deviation is \(\sigma=25\). For sample means:

• For \(n = 40\), \(\sigma_{\bar{x}_1}=\frac{25}{\sqrt{40}}\approx3.95\)
• For \(n = 50\), \(\sigma_{\bar{x}_2}=\frac{25}{\sqrt{50}}\approx3.54\)

The interval \(280\leq x\leq320\) for the population (\(n = 1\)) has \(z\) - scores \(z_1=\frac{280 - 300}{25}=- 0.8\) and \(z_2=\frac{320 - 300}{25}=0.8\).
For the sample - mean intervals:

• For \(n = 40\), \(z_{1a}=\frac{280 - 300}{\frac{25}{\sqrt{40}}}\approx\frac{-20}{3.95}\approx - 5.06\) and \(z_{2a}=\frac{320 - 300}{\frac{25}{\sqrt{40}}}\approx5.06\)
• For \(n = 50\), \(z_{1b}=\frac{280 - 300}{\frac{25}{\sqrt{50}}}\approx\frac{-20}{3.54}\approx - 5.65\) and \(z_{2b}=\frac{320 - 300}{\frac{25}{\sqrt{50}}}\approx5.65\)

The probability \(P(a\leq X\leq b)\) for a normal distribution \(X\sim N(\mu,\sigma)\) is given by \(P(a\leq X\leq b)=\Phi(\frac{b - \mu}{\sigma})-\Phi(\frac{a - \mu}{\sigma})\), where \(\Phi(z)\) is the cumulative distribution function of the standard normal distribution.
Since the tails of the normal distribution get thinner as we move further out from the mean, and the interval \(280\leq x\leq320\) is wider in terms of the number of standard deviations for the sample - mean distributions (because \(\sigma_{\bar{x}}\lt\sigma\)) compared to the population distribution. Also, as \(n\) increases, the standard error \(\sigma_{\bar{x}}\) decreases, and the probability \(P(280\leq\bar{x}\leq320)\) for the sample - mean distribution increases (because the sample - mean distribution is more concentrated around \(\mu\)).

Answer:

\((c)<(a)<(b)\)