Question

A study of the impact of executive networking on firm performance measured firm performance as annual return on equity​ (ROE), recorded as a percentage. The mean ROE for the firms studied was 14.93​% and the standard deviation was 21.74​%. Assume that these values represent mu and sigma for the population ROE distribution and that this distribution is normal. What value of ROE will be exceeded by 78​% of the​ firms?
The value of ROE that will be exceeded by 78​% of the firms is blank %.
​(Round to two decimal places as​ needed.)

Answer 1

Answer:

The value of ROE that will be exceeded by 78​% of the firms is -1.77%.

Step-by-step explanation:

Problems of normally distributed samples can be 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:

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

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

In this problem, we have that:

The mean ROE for the firms studied was 14.93​% and the standard deviation was 21.74​%. This means that $\mu = 0.1493, \sigma = 0.2174$

What value of ROE will be exceeded by 78​% of the​ firms?

This is the value of X when Z has a pvalue of 1-0.78 = 0.22.

This is $Z = -0.77$

So:

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

$-0.77 = \frac{X - 0.1493}{0.2174}$

$X - 0.1493 = -0.77*0.2174$

$X = -0.0177$

The value of ROE that will be exceeded by 78​% of the firms is -1.77%.