Question

Scores on a test are normally distributed with a mean of 81 and a standard deviation of 8. Find the probability that a randomly chosen score will be greater than 93.

Answer 1

We are given the following information.

Mean = 81

Standard deviation = 8

We are asked to find the probability that a randomly chosen score will be greater than 93.

Recall that the z-score is given by

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

Where μ is the mean, σ is the standard deviation, and x is the raw score.

$\begin{gathered} z=\frac{93-81}{8} \\ z=1.5 \end{gathered}$

So, the z-score is 1.5

$\begin{gathered} P(x>93)=1-P(x<93) \\ P(x>93)=1-P(z<1.5) \end{gathered}$

From the z-table, the probability corresponding to the z = 1.5 is found to be 0.93319

$\begin{gathered} P(x>93)=1-0.93319 \\ P(x>93)=0.0668 \end{gathered}$

There is a 0.0668 probability that a randomly chosen score will be greater than 93.