Question

On a test with a population mean of 75 and standard deviation equal to 16, if the scores are normally distributed, what percentage of scores fall between 70 and 80?

Answer 1

Answer:

Percentage of scores that fall between 70 and 80 = 24.34%

Step-by-step explanation:

We are given a test with a population mean of 75 and standard deviation equal to 16.

Let X = Percentage of scores

Since, X ~ N($\mu,\sigma^{2}$)

The z probability is given by;

           Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)    where, $\mu$ = 75  and  $\sigma$ = 16

So, P(70 < X < 80) = P(X < 80) - P(X <= 70)

P(X < 80) = P( $\frac{X-\mu}{\sigma}$ < $\frac{80-75}{16}$ ) = P(Z < 0.31) = 0.62172

P(X <= 70) = P( $\frac{X-\mu}{\sigma}$ < $\frac{70-75}{16}$ ) = P(Z < -0.31) = 1 - P(Z <= 0.31)

                                              = 1 - 0.62172 = 0.37828

Therefore, P(70 < X < 80) = 0.62172 - 0.37828 = 0.24344 or 24.34%