Answer:
Part B)
97.5%
Part C)
81.5%
Step-by-step explanation:
Part B)
ACT math scores for a particular year are approximately normally distributed
with a mean of 28 and a standard deviation of 2.4.
We want to find the Probability that a randomly selected score is less than 32.8.
First, we determine the z-score of 32.8 using:
![z = \frac{x - \bar x}{ \sigma}](https://tex.z-dn.net/?f=z%20%3D%20%20%5Cfrac%7Bx%20-%20%20%5Cbar%20x%7D%7B%20%5Csigma%7D%20)
![z = \frac{32.8 - 28}{2.4} = 2](https://tex.z-dn.net/?f=z%20%3D%20%20%5Cfrac%7B32.8%20-%2028%7D%7B2.4%7D%20%20%3D%202)
According to the Empirial rule, 95% of the distribution under the normal distribution curve is within 2 standard deviations of the mean, (-2 to 2).
This means from (0 to 2) corresponds to 47.5%.
Therefore the area less than 2, will be equal to:
50%+47.5%=97.5%
Part C)
We want to find the Probability that a randomly selected score is between 25.6 and 32.8.
That is
![P(25.6\:](https://tex.z-dn.net/?f=P%2825.6%5C%3A%3C%5C%3AX%5C%3A%20%3C%20%20%5C%3A%2032.8%29%3C%2Fp%3E%3Cp%3E)
We convert to z-scores to get:
![P( \frac{25.6 - 28}{2.4} \:](https://tex.z-dn.net/?f=P%28%20%5Cfrac%7B25.6%20-%2028%7D%7B2.4%7D%20%5C%3A%3C%5C%3Az%5C%3A%20%3C%20%20%5C%3A%20%20%5Cfrac%7B32.8%20-%2028%7D%7B2.4%7D%20%29)
This means that;
![P( - 1\:](https://tex.z-dn.net/?f=P%28%20%20-%201%5C%3A%3C%5C%3Az%5C%3A%20%3C%20%20%5C%3A%202%20%29)
Using the empirical rule again, 68% is within (-1 to 1), therefore 34% is within (-1 to 0).
And we know (-1 to 2)=(-1 to 0)+(0,2).
Therefore we have 34% +47.5%=81.5%
The required probability is 81.5%