Question

what is the probability of seeing a sample mean for 21 observations less or equal to the sample mean that we observed

Answer 1

Answer:

$P(x \le 21) = 0.69146$

Step-by-step explanation:

The missing parameters are:

$n = 64$ --- population

$\mu = 20$ --- population mean

$\sigma = 16$ -- population standard deviation

Required

$P(x \le 21)$

First, calculate the sample standard deviation

$\sigma_x = \frac{\sigma}{\sqrt n}$

$\sigma_x = \frac{16}{\sqrt {64}}$

$\sigma_x = \frac{16}{8}$

$\sigma_x = 2$

Next, calculate the sample mean $\bar_x$

$\bar x = \mu$

So:

$\bar x = 20$

So, we have:

$\sigma_x = 2$

$\bar x = 20$

$x = 21$

Calculate the z score

$x = \frac{x - \mu}{\sigma}$

$x = \frac{21 - 20}{2}$

$x = \frac{1}{2}$

$x = 0.50$

So, we have:

$P(x \le 21) = P(z \le 0.50)$

From the z table

$P(z \le 0.50) = 0.69146$

So:

$P(x \le 21) = 0.69146$