Answer to Question #336819 in Statistics and Probability for Rival

Question

A normal population has a mean of 50 and standard distribution of 10. Selecting 16 sample compute probability of sample mean

a) greater than 60

b) less than 45

c) between 45 and 60

Answer 1

We have a normal distribution, $μ=50,σ=10,n=16.$

Let's convert it to the standard normal distribution.

$\bar{z}=\cfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}.$

$\text{a) } \bar{z}_1=\cfrac{60-50}{10/\sqrt{16}}=4,\\ P(\bar{X}>60)=\\ =P(\bar{Z}>4)=\\ =1-P(\bar{Z}<4)=\\ =1-1=0\text{ (from z-table).}$

$\text{b) } \bar{z}_2=\cfrac{45-50}{10/\sqrt{16}}=-2,\\ P(\bar{X}<45)=\\ =P(\bar{Z}<-2)=\\ =0.0228\text{ (from z-table).}$

$\text{c) } P(45<\bar{X}<60)=\\ =P(-2<\bar{Z}<4)=\\ =P(\bar{Z}<4)-P(\bar{Z}<-2)=\\ =1-0.0228=0.9772.$