What Percent of Executives Earn $925 or Less?

A

z = (x-u)/sigma = 925 - 1000/100 = -0.75. From z-table z = 0.75 is 0.2734. So 1.0 - 0.7734 = 0.2266.

To determine the percentage of executives with an income of $925 or less, we need to calculate the z-score and use the standard normal distribution table.

The z-score measures the number of standard deviations an observation is from the mean. We can calculate the z-score using the formula:

z = (x - μ) / σ

Where: x = the value we are interested in (in this case, $925) μ = the mean income ($1,000) σ = the standard deviation ($100)

Plugging in the values, we have:

z = (925 - 1000) / 100 z = -0.75

Now, we need to find the area under the standard normal distribution curve for a z-score of -0.75. We can consult the standard normal distribution table or use a statistical calculator to find this value.

Looking up the z-score of -0.75 in the standard normal distribution table, we find that the area to the left of -0.75 is approximately 0.2266.

This means that approximately 22.66% of the executives have an income of $925 or less.

Since none of the given answer choices exactly match this result, we can conclude that the correct answer is "E. None of these answers."