Question

Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. Find the probability that a given score is between negative 2.13 and 3.88 and draw a sketch of the region.

Answer 1

Answer: 0.9834

Explanation:

Given : The test scores are normally distributed with

Mean :
`\mu=\ 0`

Standard deviation :
`\sigma= 1`

The formula to calculate the z-score :-

`z=(x-\mu)/(\sigma)`

For x = -2.13

`z=(-2.13-0)/(1)=-2.13`

For x = 3.88

`z=(3.88-0)/(1)=3.88`

The p-value =
`P(-2.13<z<3.88)=P(z<3.88)-P(z<-2.13)`

`0.9999477-0.0165858=0.9833619\approx0.9834`

Hence, the probability that a given score is between negative 2.13 and 3.88 = 0.9834