Question

A population has a mean of 200 and a standard deviation of 50. Suppose a sample of size 100 is selected. What is the probability that the sample mean will be within LaTeX: \pm±10 of the population mean?

Answer 1

Answer: 0.9544

Step-by-step explanation:

As per given have ,

Population mean : $\mu=200$

Standard deviation : $\sigma=50$

Sample size : n= 100

We assume that the population is normally distributed.

Let $\overline{x}$ be the sample mean.

Now , the probability that the sample mean will be within ±10 of the population mean will be :-

$P(200-10<\overlien{x}<200+10)=P(190<x<210)$

$P(\dfrac{190-200}{\dfrac{50}{\sqrt{100}}}<\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}<\dfrac{210-200}{\dfrac{50}{\sqrt{100}}})$

$=P(\dfrac{-10}{\dfrac{50}{10}}<z<\dfrac{10}{\dfrac{50}{10}})\ \ [\because\ z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}]$

$=P(\dfrac{-10}{5}<z<\dfrac{10}{5})=P(-2<z<2) \\\\=2P(z<2)-1\ \ [\because\ P(-z<Z<z)=2P(Z<z)-1]\\\\=2(0.9772)-1\ \ [\text{[By z-table]}]\\\\=0.9544$

Hence, the required probability is 0.9544.