Question

The mean weight of an adult is 60 kilograms with a variance of 100. If 118 adults are randomly selected, what is the probability that the sample mean would be greater than 62.2 kilograms? Round your answer to four decimal places.

Answer 1

Step 1

Given; The mean weight of an adult is 60 kilograms with a variance of 100. If 118 adults are randomly selected, what is the probability that the sample mean would be greater than 62.2 kilograms? Round your answer to four.

Step 2

In a set with mean and standard deviation, the z-score of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

The Central Limit Theorem establishes that, for a random variable X, with mean and standard deviation, the sample means with size n of at least 30 can be approximated to a normal distribution with mean and standard deviation;

$s=\frac{\sigma}{\sqrt{n}}$$\begin{gathered} \mu=60kg \\ s=\sqrt{variance}=\sqrt{100}=10 \\ n=118 \\ \end{gathered}$$\begin{gathered} s=\frac{10}{\sqrt{118}} \\ s=\frac{5\sqrt{118}}{59} \end{gathered}$

What is the probability that the sample mean would be greater than 62.2 kilograms?

$\begin{gathered} z=\frac{62.2-60}{\frac{5\sqrt{118}}{59}} \\ z=2.38981 \end{gathered}$$The\text{ p-value=0.0084285362}$

Thus;

$0.0084284970$

Answer;

$0.0084\text{ to 4 d.p}$