Question

The scores for a state's peace officer training test are normally distributed with a mean of 55 and a standard deviation of 12. The test scores range from 0 to 100. A state agency will only interview applicants with test scores of 43 or greater. About what percent of the people have test scores that make them eligible to be interviewed?

Answer 1

About 84.13% of the test takers will have a score of 43 or higher on the peace officer training test and be eligible to be interviewed, based on the normal distribution with a mean of 55 and a standard deviation of 12.

Step-by-step explanation:

The question is asking to find out what percent of people have test scores that make them eligible to be interviewed for a peace officer training test where the scores are normally distributed with a mean of 55 and a standard deviation of 12. To be eligible for an interview, applicants must have a score of 43 or higher. To calculate this, we need to determine the z-score for a test score of 43.

The z-score is calculated by subtracting the mean from the score, and then dividing by the standard deviation. For a score of 43, the z-score is:

Z = (43 - 55) / 12 = -12 / 12 = -1

Next, we would use a z-table or relevant software to find the proportion of scores that fall above a z-score of -1. The z-table tells us that approximately 84.13% of the distribution falls above a z-score of -1. Therefore, about 84.13% of test takers will score 43 or higher and be eligible for an interview.

Answer 2

About 84% of the people have test scores that make them eligible to be interviewed

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

`\mu = 55, \sigma = 12`

A state agency will only interview applicants with test scores of 43 or greater. About what percent of the people have test scores that make them eligible to be interviewed?

This is 1 subtracted by the pvalue of Z when X = 43. So

`Z = (X - \mu)/(\sigma)`

`Z = (43 - 55)/(12)`

`Z = -1`

`Z = -1` has a pvalue of 0.1587

1 - 0.1587 = 0.8413

Rounding to the nearest percent

About 84% of the people have test scores that make them eligible to be interviewed