the income of a group of 10,000 health system employees was found to be normally distributed with a mean=750 pm and standard deviation=50. show that of this group about 95% had income exceeding 668 and only 5% had income exceeding 832

Z = (score-mean)/SD

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z scores.

10000

It is clear that because in normal distribution curve within 1 standard deviation (700-800)at 68.34% and within 2SD (650-850)at 95%

To show that about 95% of the group had income exceeding 668 and only 5% had income exceeding 832, we can use the concept of Z-scores.

A Z-score represents the number of standard deviations an individual's income falls from the mean. It can be calculated using the Z-score formula:

Z = (X - μ) / σ

Where:
- Z is the Z-score
- X is the individual's income
- μ is the mean income
- σ is the standard deviation of income

Let's calculate the Z-scores for the given income values:

For income exceeding 668:
Z1 = (668 - 750) / 50
Z1 = -1.64

For income exceeding 832:
Z2 = (832 - 750) / 50
Z2 = 1.64

Now, we need to find the area under the normal distribution curve corresponding to these Z-scores.

Using a Z-table or a statistical calculator, we can find that the area to the left of Z = -1.64 is approximately 0.0495 and the area to the left of Z = 1.64 is approximately 0.9505.

Since we want to find the proportion of employees above these income levels, we subtract the obtained areas from 1 to get the remaining areas.

For income exceeding 668:
Proportion = 1 - 0.0495
Proportion ≈ 0.9505 (95.05%)

For income exceeding 832:
Proportion = 1 - 0.9505
Proportion ≈ 0.0495 (4.95%)

Therefore, we can conclude that about 95% of the group had income exceeding 668 and only 5% had income exceeding 832.