Question

The mean salary at a local industrial plant is $27,700 with a standard deviation of $4400. The median salary is $27,100 and the 58th percentile is $28,800.

1. True or false: Approximately 58% of the salaries are above $28,800.
2. True or false: Joe's salary of $32,540 is 1.10 standard deviations above the mean
3. True or false: The percentile rank of $27,300 is 50
4. True or false: Approximately 8% of the salaries are between $27,100 and $28,800
5. If Tom's salary has a z-score of 0.6, how much does he earn (in dollars)?

Answer 1

Solution:

Given:

The mean salary at a local industrial plant is $27,700 with a standard deviation of $4400.

The median salary is $27,100 and the 58th percentile is $28,800.

Part 1) Approximately 58% of the salaries are above $28,800.

Since the 58th percentile is $28,800, that means 58% of the salaries are below $28,800.

Thus given statement is False.

Part 2)  Joe's salary of $32,540 is 1.10 standard deviations above the mean

Since z score is 1.10 and it is above 0, thus given statement is True.

Part 3) The percentile rank of $27,300 is 50

Since median salary is $27,100 , the percentile rank for $27,100 is 50.

Thus the percentile rank of $27,300 is not 50, thus given statement is False.

Part 4) Approximately 8% of the salaries are between $27,100 and $28,800

We have median salary = $27,100 , that is 50th percentile is $27,100

and  the 58th percentile is $28,800.

Thus 58-50= 8%

that is: approximately 8% of the salaries are between $27,100 and $28,800

thus given statement is True.

Part 5) If Tom's salary has a z-score of 0.6, how much does he earn (in dollars)?

z = 0.6