Question

the average starting salary for this year's graduates at a large university (lu) is $55,000 with a standard deviation of $6,000. furthermore, it is known that the starting salaries are normally distributed. individuals with starting salaries of less than $39,600 receive a low income tax break. what percentage of the graduates will receive the tax break?

Answer 1

0.84% of the graduates will receive the tax break

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

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

Z = x - μ / σ

Z-scores are used to measure how far a measure is from the mean. We find the p-value associated with this Z-score by looking at the z-score table after finding the Z-score. The p-value represents the probability that the measure is smaller than X, which is the percentile of X. The probability of the measure being greater than X is calculated by subtracting 1 from the pvalue.

μ = $55,000 σ = $6,000

This is the pvalue of Z when X = $39,600. So

z = 39600 - 55000 / 6000

z = -2.5

z = -2.5 has a o value of 0.0084

0.84% of the graduates will receive the tax break.

To learn more about similar tax break:

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