Question

1. a sample of size 50 will be drawn from a population with mean 76 and standard deviation 14. find the 69th percentile of \\( \bar{x} \\).

Explanation:

Step1: Recall Sampling Distribution

For a sample of size \( n \), the sampling distribution of the sample mean \( \bar{x} \) has mean \( \mu_{\bar{x}}=\mu \) and standard deviation \( \sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}} \), where \( \mu \) is the population mean and \( \sigma \) is the population standard deviation. Here, \( \mu = 76 \), \( \sigma = 14 \), \( n = 50 \). So \( \mu_{\bar{x}} = 76 \), \( \sigma_{\bar{x}}=\frac{14}{\sqrt{50}}\approx1.9798 \).

Step2: Find Z - score for 69th Percentile

We need to find the \( z \)-score \( z_{0.69} \) such that \( P(Z\leq z_{0.69}) = 0.69 \). Using a standard normal table or calculator, \( z_{0.69}\approx0.4959 \) (since \( \Phi(0.4959)\approx0.69 \), where \( \Phi \) is the CDF of standard normal).

Step3: Use Formula for Percentile of \( \bar{x} \)

The formula for the \( p \)-th percentile of \( \bar{x} \) is \( \bar{x}_p=\mu_{\bar{x}}+z_p\times\sigma_{\bar{x}} \). Substitute \( \mu_{\bar{x}} = 76 \), \( z_p = 0.4959 \), \( \sigma_{\bar{x}}\approx1.9798 \):

\( \bar{x}_{0.69}=76 + 0.4959\times1.9798\approx76+0.981\approx76.981 \)

Answer:

\(\approx 76.98\) (or more precisely \(76.981\))