Normal Distribution - statistic question?

Normal Distribution - statistic question?

I really need help on this case - I was able to get the answer to part A - 0.514

I need help explaining how to get the answer for the rest

thanks in advance

here is the case

Suppose the starting salaries for a business graduate in Virginia is adequately approximated by a normal distribution with mean of $42,000 and standard deviation $3,500.

a. You received a job offer of $43,800. What is the Z value for the offer?

b. What percentile of starting salaries is the offer? That is, what percentage of starting salaries are lower than the one offered you?

c. Since you finished in the top 20% of your class you feel your starting salary should be in the top 20% of starting salaries. What is the lowest salary you could get and be in the top 20%?

d. You read that starting salaries in Maryland were higher than Virginia, but couldn't find exact figures. You did find that the top 10% of salaries were $47,000 and above, while the bottom 10% were below $38,500. What are the mean and standard deviation of salaries in Maryland? (Assume salaries follow the normal distribution.)

e. Suppose a couple is considering searching for jobs in Virginia. One is a business major and the other a Computer Science major. Starting salaries for Computer Science majors in Virginia are adequately approximated by a normal distribution with mean of $52,000 and standard deviation $4,750. Starting salaries for business majors in Virginia are as shown in the opening paragraph. What is the mean and standard deviation of the couple's combined salaries? (Use the properties of Expected Values and Variances from the Text 8.3)

f. Using the Excel Random Number Generation function simulate the situations in part e and compare the results to the results you got using the equations. Use at least 1,000 iterations and show the Descriptive Statistics for the simulation results.

g. Create a histogram for your simulation results in part f. Describe the distribution of the combined salaries.



μ = 42000

σ = 3500

standardize x to z = (x - μ) / σ

P(x < 43800) = P( z < (43800-42000) / 3500)

= P(z < 0.5143) = .695

(From Normal probability table)


Find z from the normal probability table such that P( z > ? )=.20

(area above what value is 20%)

P( z > .84) = .20

z = (x - μ) / σ

.20 = (x- 42000)/3500

solve for x

x= (3500)(.20) + 42,000 =42,700


The z-value that corresponds to the top 10% is 1.28 and bottom 10% is -1.28

P( x > 47,000) = .10

z = (x - μ) / σ

(47000- μ) / σ =1.28

1.28σ + μ = 47,000 ---------(1)

P( x < 38,500) =-1.28

z = (x - μ) / σ

(38500- μ) / σ =-1.28

-1.28σ + μ = 38,500 ---------(2)

Solve (1) and (2)


2 μ = 85500

μ = 42750

plug this μ into equation (1) and solve for σ

1.28σ + μ = 47000

1.28σ +42750 = 47000

1.28 σ =4250

σ = 3320.31

Oct 25 at 12:22