Question

A normal Distribution has a mean of 101 and a standard Deviation of 5. find the probability that a value selected at random is in the following interval. at least 91

Answer 1

STEP 1: Find the z scores of 91

`\begin{gathered} z=\frac{x-\mu}{s\mathrm{}d}\text{ where }\mu=\text{ mean and s.d = standard deviation} \\ z=(91-101)/(5) \\ z=-(10)/(2) \\ z=-2 \end{gathered}`

STEP 2

To find the probability that a value selected at random is in the following interval of at least 91​.

We get the probability that a value selected at random is in the following interval of at most 91​, and subtract it from 1.

Pr(at least 91)= 1 - Pr(at most 91)

Pr(at least 91)= 1 - probabilites less than a z - score of -2 from the graph

Pr(at least 91) = 1-(0.017+0.005+0.001)

=0.977

Therefore the probability of at least 91 =0.977

That is 97.7%