Question

ACT math scores for a particular year are approximately normally distributed

Answer 1

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:

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

$z = \frac{32.8 - 28}{2.4} = 2$

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

$P(25.6\:$

We convert to z-scores to get:

$P( \frac{25.6 - 28}{2.4} \:$

This means that;

$P( - 1\:$

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%