Suppose a simple random sample of size $n=81$ is obtained from a population that is skewed right with $\mu=84$ and $\sigma=27$.
(a) Describe the sampling distribution of $\bar{x}$
(b) What is $P(\bar{x}> 8835)$ ?
(c) What is $P(x \leq 78)$ ?
(d) What is $P(82.5< \bar{x}< 91.5)$ ?
(a) Choose the correct description of the shape of the sampling distribution of $\bar{x}$.
A. The distribution is skewed left
B. The distribution is skewed right
C. The distribution is uniform
D. The distribution is approximately normal
E. The shape of the distribution is unknown

Find the mean and standard deviation of the sampling distribution of $\bar{x}$
\[
\begin{array}{l}
\mu_{x}^{-}=\square \\
\sigma_{x}=\square
\end{array}
\]
(Type integers or decimals. Do not round)
(b) $P(\bar{x}> 88.35)=\square$ (Round to four decimal places as needed)
(c) $P(x \leq 78)=\square$ (Round to four decimal places as needed )
(d) $P(82.5< \bar{x}< 91.5)=\square$ (Round to four decimal places as needed)

Step 1 ：The shape of the sampling distribution of the mean can be determined by the Central Limit Theorem, which states that if the sample size is large enough (usually n > 30), the sampling distribution of the mean will be approximately normal regardless of the shape of the population distribution. So, the distribution is approximately normal.

Step 2 ：The mean of the sampling distribution of the mean is equal to the population mean, and the standard deviation of the sampling distribution (also known as the standard error) is equal to the population standard deviation divided by the square root of the sample size. So, the mean of the sampling distribution of \(\bar{x}\) is \(\mu_{x}^{-}=84\) and the standard deviation is \(\sigma_{x}=3\).

Step 3 ：To find the probability that the sample mean is greater than 88.35, we need to standardize the value and find the corresponding z-score, then look up the probability in the standard normal distribution table. So, \(P(\bar{x}>88.35)=0.0735\) (rounded to four decimal places).

Step 4 ：The probability that a single observation is less than or equal to 78 is a question about the population distribution, not the sampling distribution. We can find this probability by standardizing the value and finding the corresponding z-score, then looking up the probability in the standard normal distribution table. So, \(P(x \leq 78)=0.4121\) (rounded to four decimal places).

Step 5 ：To find the probability that the sample mean is between 82.5 and 91.5, we need to standardize these values, find the corresponding z-scores, and find the probability between these z-scores in the standard normal distribution table. So, \(P(82.5<\bar{x}<91.5)=0.6853\) (rounded to four decimal places).

Step 6 ：Final Answer: \(\boxed{D}\), \(\boxed{\mu_{x}^{-}=84}\), \(\boxed{\sigma_{x}=3}\), \(\boxed{P(\bar{x}>88.35)=0.0735}\), \(\boxed{P(x \leq 78)=0.4121}\), \(\boxed{P(82.5<\bar{x}<91.5)=0.6853}\)