I need help with #1-4 questions. thank you 1) Human Resources Department of one

Question

I need help with #1-4 questions. thank you

1) Human Resources Department of one hotel reports that annual salries for its personnel have a mean of $40,000 and a standard deviation of $3500. If the salaries are symmetrically distributed, the interval of salaries representing two standard deviations on either side of the mean is

2) what proportion of the salaries is within $3500 of the mean?

3)what proportion of the salary is below $33000?

4)if the salaries are not symmetrical about the mean which proportion of the salariesare 5/2 standard deviation of the mean?

Explanation / Answer

1) Human Resources Department of one hotel reports that annual salaries for its personnel have a mean of $40,000 and a standard deviation of $3500. If the salaries are symmetrically distributed, the interval of salaries representing two standard deviations on either side of the mean is

Answer:

We are given

Mean = $40,000

SD = $3500

Mean – 2*SD = 40000 – 2*3500 = 40000 – 7000 = 33000

Mean + 2*SD = 40000 + 2*3500 = 40000 + 7000 = 47000

If the salaries are symmetrically distributed, the interval of salaries representing two standard deviations on either side of the mean is ($33,000, $47,000).

2) What proportion of the salaries is within $3500 of the mean?

We are given

Mean = $40,000

SD = $3500

Here, we have to find

P(40000 – 3500 < X < 40000 + 3500) = P(40000 – 1*3500 < X < 40000 + 1*3500) = P(36500 < X < 43500)

P(Mean – 1*SD < X < mean + 1*SD) = P(36500 < X < 43500)

We know that about 68% of the data falls within the one standard deviation from the mean. (Using empirical rule)

So, P(36500 < X < 43500) = 68% = 0.68

About 68% of the salaries is within $3500 of the mean.

3) What proportion of the salary is below $33000?

Here, we have to find P(X<33000)

Z = (X – mean) / SD

Z = (33000 – 40000) / 3500 = -7000/3500 = -2

P(Z<-2) = P(X<33000) = 0.02275 = 2.28% (by using z-table or excel)

About 2.28% of the salaries are below $33000.

4) If the salaries are not symmetrical about the mean which proportion of the salaries are 5/2 standard deviation of the mean?

Here we have to use Chebyshev’s theorem for finding the required proportion. We can use this theorem when data is not symmetrical about the mean. According to Chebyshev’s theorem,

Required proportion = (1 – (1/k^2)) where k>1

Here, we are given k = 5/2 = 2.5

Required proportion = (1 – (1/2.5^2))

Required proportion = (1 – (1/6.25))

Required proportion = (1 – 0.16)

Required proportion = 0.84

About 84% of the salaries are 5/2 standard deviation of the mean.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.