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

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 = (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 = (X - \mu)/(\sigma)`

`-1.96 = (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 = (X - \mu)/(\sigma)`

`1.96 = (X - 35000)/(5000)`

`X - 35000 = 1.96*5000`

`X = 44800`

The maximum is $44,800