part d please A simple random sample of size \( n = 49 \) is obtained from a population that is skewed right with \( \mu = 89 \) and \( \sigma = 21 \).(a) Describe the sampling distribution of \( \bar{x} \).- **Options**: - A. The distribution is skewed right. - B. The distribution is uniform. - C. The distribution is approximately normal. (Correct Answer) - D. The distribution is skewed left. - E. The shape of the distribution is unknown.Find the mean and standard deviation of the sampling distribution of \( \bar{x} \).- \( \mu_{\bar{x}} = 89 \)- \( \sigma_{\bar{x}} = 3 \) (Type integers or decimals. Do not round.)(b) \( P\left( \bar{x} > 94.7 \right) = 0.0287 \) (Round to four decimal places as needed.)(c) \( P\left( \bar{x} \leq 82.25 \right) = 0.0122 \) (Round to four decimal places as needed.)(d) \( P\left( 85.85 < \bar{x} < 95.15 \right) = \) (Round to four decimal places as needed.)

Suppose the variable of interest is x.It is given that; x follows a right skewed distribution mean µx = 89 and the standard deviation of σx = 21. The sample mean is denoted as x̄. The sample size is n = 49. The population is not normally distributed. However, the sample size is large (> 30). Thus, according to the central limit theorem, for large sample size, irrespective of the population distribution, the mean of the sampling distribution of the sample mean is, µx̄ = µ = 89. According to the central limit theorem, the standard deviation of the sampling distribution of the sample mean is, σx̄ = σx/√n = 21/√49 = 3. In this case, x̄ follows a normal distribution with mean µx̄ = 89 standard deviation σx̄ = 3. The z score formula is, z = (x̄ – µx̄)/σx̄. The value of P(85.85 < x̄ < 95.15) is, [or from standard normal table corresponding to row of 2.0 and column of .05, it is obtained that, P(z < 2.05) = 0.97982. The excel output is, From standard normal table corresponding to row of -1.0 and column of .05, it is obtained that, P(z < -1.05) = 0.14686. The excel output is, ] Therefore, P(85.85 < x̄ < 95.15) = 0.8330.