Question

The mean weight of an adult is 76 kg with a variance of 100. If 142 adults are randomly selected what is the probability that the sample mean will be greater than 77.4 kg? Round your answer to four decimal places

Answer 1

The Solution:

Given:

$\begin{gathered} \mu=76kg \\ \sigma=\sqrt{100}=10 \\ n=142\text{ adults} \\ \bar{x}=77.4kg \end{gathered}$

Required:

Find the probability that the sample mean is greater than 77.4kg.

Using the z-statistic formula:

$Z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$

Substitute:

$Z=\frac{77.4-76}{\frac{10}{\sqrt{142}}}=\frac{1.4\sqrt{142}}{10}=1.6683$

From the z score tables, the probability that the sample mean is greater than 77.4kg is:

$P(Z\ge77.4)=0.047628\approx0.0476$

Answer:

0.0476