The problem asks to find the z-score for different sample means ($\bar{X}$) given a population mean ($\mu = 18$) and a population standard deviation ($\sigma = 4$). We are given different sample sizes ($n$) for each case.

1. Problem Description

The problem asks to find the z-score for different sample means ($\bar{X}$) given a population mean ($\mu = 18$) and a population standard deviation ($\sigma = 4$). We are given different sample sizes ($n$) for each case.

2. Solution Steps

The formula for the z-score of a sample mean is:

$z = \frac{\bar{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$

where:

$\bar{X}$ is the sample mean

$\mu$ is the population mean

$\sigma$ is the population standard deviation

$n$ is the sample size

a. $\bar{X} = 10$, $n = 15$

$z = \frac{10 - 18}{\frac{4}{\sqrt{15}}} = \frac{-8}{\frac{4}{\sqrt{15}}} = \frac{-8\sqrt{15}}{4} = -2\sqrt{15} \approx -7.746$

b. $\bar{X} = 10$, $n = 30$

$z = \frac{10 - 18}{\frac{4}{\sqrt{30}}} = \frac{-8}{\frac{4}{\sqrt{30}}} = \frac{-8\sqrt{30}}{4} = -2\sqrt{30} \approx -10.954$

c. $\bar{X} = 20$, $n = 12$

$z = \frac{20 - 18}{\frac{4}{\sqrt{12}}} = \frac{2}{\frac{4}{\sqrt{12}}} = \frac{2\sqrt{12}}{4} = \frac{\sqrt{12}}{2} = \frac{2\sqrt{3}}{2} = \sqrt{3} \approx 1.732$

d. $\bar{X} = 20$, $n = 24$

$z = \frac{20 - 18}{\frac{4}{\sqrt{24}}} = \frac{2}{\frac{4}{\sqrt{24}}} = \frac{2\sqrt{24}}{4} = \frac{\sqrt{24}}{2} = \frac{2\sqrt{6}}{2} = \sqrt{6} \approx 2.449$

3. Final Answer

a. $z \approx -7.746$

b. $z \approx -10.954$

c. $z \approx 1.732$

d. $z \approx 2.449$