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([tex]\mu,\sigma^{2}[/tex])
The z probability is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1) where, [tex]\mu[/tex] = 75 and [tex]\sigma[/tex] = 16
So, P(70 < X < 80) = P(X < 80) - P(X <= 70)
P(X < 80) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{80-75}{16}[/tex] ) = P(Z < 0.31) = 0.62172
P(X <= 70) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{70-75}{16}[/tex] ) = 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%
