Suppose a simple random sample of size $n=81$ is obtained from a population that is skewed right with $\mu=86$ and $\sigma=36$.
(a) Describe the sampling distribution of $\bar{x}$.
(b) What is $P(x> 90.4)$ ?
(c) What is $P(x \leq 77.6)$ ?
(d) What is $P(81.4< \bar{x}< 92.6)$ ?
Answer
\(\boxed{\text{Final Answer: The sampling distribution of } \bar{x} \text{ is approximately normally distributed with a mean of 86 and a standard deviation of 4. In mathematical notation, this can be written as } \bar{x} \sim N(86, 4^2)}\)
Step 1 ：Given a simple random sample of size \(n=81\) from a population that is skewed right with \(\mu=86\) and \(\sigma=36\).
Step 2 ：According to the Central Limit Theorem (CLT), if the sample size is sufficiently large, the sampling distribution of the sample mean will be approximately normally distributed, regardless of the shape of the population distribution.
Step 3 ：In this case, the sample size is \(n=81\), which is large enough for the CLT to apply.
Step 4 ：The mean of the sampling distribution of \(\bar{x}\) is equal to the population mean \(\mu\), and the standard deviation (also known as the standard error) is equal to the population standard deviation \(\sigma\) divided by the square root of the sample size \(n\).
Step 5 ：Calculating these values, we get \(\mu = 86\), \(\sigma = 36\), \(n = 81\), \(\bar{x}_{\mu} = 86\), and \(\bar{x}_{\sigma} = 4.0\).
Step 6 ：This means that the sampling distribution of \(\bar{x}\) is approximately normally distributed with a mean of 86 and a standard deviation of 4.
Step 7 ：\(\boxed{\text{Final Answer: The sampling distribution of } \bar{x} \text{ is approximately normally distributed with a mean of 86 and a standard deviation of 4. In mathematical notation, this can be written as } \bar{x} \sim N(86, 4^2)}\)
