Question

ACT math scores for a particular year are approximately normally distributed

with a mean of 28 and a standard deviation of 2.4.​

Answer 1

Part B)

97.5%

Part C)

81.5%

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 = (x - \bar x)/( \sigma)`

`z = (32.8 - 28)/(2.4) = 2`

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\:<\:X\: < \: 32.8)</p><p>`

We convert to z-scores to get:

`P( (25.6 - 28)/(2.4) \:<\:z\: < \: (32.8 - 28)/(2.4) )`

This means that;

`P( - 1\:<\:z\: < \: 2 )`

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%