Question

The mean salary at a local industrial plant is $27,100 with a standard deviation of $5400. The median salary is $⁢26,800 and the 61^st percentile is $28,300.

Step 1 of 5:

Based on the given information, determine if the following statement is true or false.

Approximately 61% of the salaries are less than or equal to $28,300.

Step 2 of 5:

Based on the given information, determine if the following statement is true or false.

Joe's salary of $⁢36,820 is 1.80 standard deviations above the mean.

Step 3 of 5:

Based on the given information, determine if the following statement is true or false.

The percentile rank of $27,000 is 50.

Step 4 of 5:

Based on the given information, determine if the following statement is true or false.

Approximately 11% of the salaries are between $⁢26,800 and $⁢28,300.

Answer 1

Step 1 of 5 Approximately 61% of the salaries is less than or equal to $28,300. True

Step 2 of 5:

x=μ+z(σ)

=$27100+(1.80×$5400)=$36820

The statement “Joe’s salary of $36,820 is 1.80 standard deviations above the mean” is True.

Step 3 of 5:

We have to check the percentile rank of $27000 is 50.

We have already given the median which is 50th percentile is $26800

Hence given statement is False

Step 4 of 5:

We know 50th percentile is $26800 and 61st percentile is $28300. If we take difference then

61-50 = 11

So 11% of the salaries are between $26800 and $28300

Hence given statement is True.

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