Question

The average starting salary of this year's vocational school graduates is $35,000 with a standard deviation of $5,000. Furthermore, it is known that the starting salaries are normally distributed. What are the minimum and the maximum starting salaries of the middle 95% of the graduates

Answer 1

Answer:

Minimum: $25,200

Maximum: $44,800

Step-by-step explanation:

When the distribution is normal, we use 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 question:

$\mu = 35000, \sigma = 5000$

What are the minimum and the maximum starting salaries of the middle 95% of the graduates

Minimum: 50 - (95/2) = 2.5th percentile.

Maximum: 50 + (95/2) = 97.5th percentile

2.5th percentile:

X when Z has a pvalue of 0.025. So X when Z = -1.96.

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

$-1.96 = \frac{X - 35000}{5000}$

$X - 35000 = -1.96*5000$

$X = 25200$

The minimum is $25,200

97.5th percentile:

X when Z has a pvalue of 0.975. So X when Z = 1.96.

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

$1.96 = \frac{X - 35000}{5000}$

$X - 35000 = 1.96*5000$

$X = 44800$

The maximum is $44,800