Normal Probability Distribution Group 2 The weekly incomes of a large group of s

Question

Normal Probability Distribution

Group 2 The weekly incomes of a large group of sales clerks are normally distributed with a mean of $1000 and a standard deviation of $125. Use the normal distribution to calculate the following:

5. Find the income that represents the 90th percentile.

6. 5% of the incomes are below what value?

7. The top 5% of the incomes are above what value?

8. Between which two values will the middle 50% of the data lie?

9. A. What percent of income lies within $250 of the population mean?

B. 68% of observations lie within which two values?

C. What percent of income lies within $375 of the population mean?

D. Discuss how your answers in A-C compare to the Empirical Rule.

10. Consider a sample with a mean of 30 and a standard deviation of 5. Use Chebyshev’s Theorem to find the percent of data that lies within the data range of 15 and 45?

Explanation / Answer

5.

Params of normal distribution is as follows:

Mean = 1000

Stdev = 125

P(represents 90th percentile) = ?

Z = 1.28 for 90th percentile
Therefore, 90th percentile is Xbar +/- Z*Sigma = $1000 +1.28*$125 = $1160

6. 5% of the income are below what value?
Z = -1.645
Therefore, 5% of the income is below Xbar +/- Z*Sigma = 1000-1.645*125 = 794.375

7. Top 5% are above what value?
Z = 1.645
Therefore, 5% of the income is above Xbar + Z*Sigma = 1000+1.645*125 = 1205.625

8. So, we find this by doing a Z value of .675
Z = .675
Therefore, 50% middle is 1000-.675*125 to 1000+.675*125
=915.625 to 1084.375

9.
A. P(X<250) = P(Z< 250-1000/125) = P(Z< -6) = 9.87E-10
B. 68% of obs, by empirical rule in normal distribution is +/- 1 deviation around mean
So, 1000+/- 1*125 = $875 to $1125
C. P(X<375) = P(Z< 375-1000/125) = P(Z< -5) = 2.87E-07
D. A and C have similar answers to the applications of empirical rules. Empirical rules states that a very high Z ( beyond 4) gives a area under curve of 0
A and C answers tend to 0

10. Sample mean = 30
Stdev = 5
15 and 45 are 5 deviations below and above mean. By applying Chebeshev' theorm we have:
P(15<X<45) = 1- 1/k^2, k is no. of deviation away from mean
So, P(15<X<45) = 1- (1/3^2) = 88.9%

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