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inna [77]
3 years ago
10

The starting salaries of individuals with an MBA degree are normally distributed with a mean of $40,000 and a standard deviation

of $5,000. What percentage of MBA's will have starting salaries of $34,000 to $46,000?
a. 38.49%
b. 38.59%
c. 50%
d. 76.98%
Mathematics
1 answer:
liraira [26]3 years ago
7 0

Answer:

d. 76.98%

Step-by-step explanation:

Problems of normally distributed samples are 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:

\mu = 40000, \sigma = 5000

What percentage of MBA's will have starting salaries of $34,000 to $46,000?

This is the pvalue of Z when X = 46000 subtracted by the pvalue of Z when X = 34000. So

X = 46000

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

Z = \frac{46000 - 40000}{5000}

Z = 1.2

Z = 1.2 has a pvalue of 0.8849

X = 34000

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

Z = \frac{34000 - 40000}{5000}

Z = -1.2

Z = -1.2 has a pvalue of 0.1151

0.8849 - 0.1151 = 0.7698

So the correct answer is:

d. 76.98%

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Answer:

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Step-by-step explanation:

circular plate had a diameter of 10 cm. The rectangular table it sits on

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