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

Step-by-step explanation:

Given : The test scores are normally distributed with

Mean : $\mu=\ 0$

Standard deviation :$\sigma= 1$

The formula to calculate the z-score :-

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

For x = -2.13

$z=\dfrac{-2.13-0}{1}=-2.13$

For x = 3.88

$z=\dfrac{3.88-0}{1}=3.88$

The p-value = $P(-2.13$

$0.9999477-0.0165858=0.9833619\approx0.9834$

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