Question

A psychologist has designed a questionnaire to measure individuals' aggressiveness. Suppose that the scores on the questionnaire are normally distributed with a standard deviation of 80. Suppose also that exactly 15% of the scores exceed 700. Find the mean of the distribution of scores. Carry your intermediate computations to at least four decimal places. Round your answer to at least one decimal place.

Answer 1

Answer:

The mean score is of 617.4.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 p-value, we get the probability that the value of the measure is greater than X.

Suppose that the scores on the questionnaire are normally distributed with a standard deviation of 80.

This means that $\sigma = 80$

Suppose also that exactly 15% of the scores exceed 700.

This means that when X = 700, Z has a pvalue of 0.85. So X when X = 700, Z = 1.033. We use this to find $\mu$

$Z = \frac{X - \mu}{\sigma}$

$1.033 = \frac{700 - \mu}{80}$

$700 - \mu = 80*1.033$

$\mu = 700 - 80*1.033$

$\mu = 617.4$

The mean score is of 617.4.