Suppose \( x \) has a distribution with \( \mu=16 \) and \( \sigma=15 \).
\( \Omega \) USE SALT
(a) If a random sample of size \( n=36 \) is drawn, find the following.
Find \( \mu_{\bar{x}}^{-} \)
\[
\mu_{x}^{-}=
\]
\( \square \)
Find \( \sigma_{\bar{x}}^{-} \)
\[
\sigma_{\bar{x}}=
\]
\( \square \)
Find \( P(16 \leq \bar{x} \leq 18) \). (Round your answer to four decimal places.)
\[
P(16 \leq \bar{x} \leq 18)=
\]
\( \square \)
(b) If a random sample of size \( n=64 \) is drawn, find the following.
Find \( \mu_{x^{\prime}}^{-} \)
\[
\mu_{x}^{-}=
\]
\( \square \)
Find \( \sigma_{\bar{x}}^{-} \)
\[
\sigma_{\bar{x}}=
\]
\( \square \)
Find \( P(16 \leq \bar{x} \leq 18) \). (Round your answer to four decimal places.)
\[
P(16 \leq \bar{x} \leq 18)=
\]
\( \square \)
(c) Why should you expect the probability of part (b) to be higher than that of part (a)? (Hint: Consider the standard deviations in parts (a) and (b).) The standard deviation of part (b) is: -- Select.\( - v \) part (a) because of the -- Select \( -v \) sample size. Therefore, the distribution about \( \mu_{\bar{x}} \) is - Select- \( -v \).