Question

The following probability distributions of job satisfaction scores for a sample of information systems (IS) senior executives and middle managers range from a low of 1 (very dissatisfied) to a high of 5 (very satisfied).

Probability
Job Satisfaction Score IS Senior Executives IS Middle Managers
1 0.05 0.04
2 0.09 0.10
3 0.03 0.12
4 0.44 0.45
5 0.39 0.29
1. What is the expected value of the job satisfaction score for senior executives (to 2 decimals)?
2. What is the expected value of the job satisfaction score for middle managers (to 2 decimals)?
3. Compute the variance of job satisfaction scores for executives and middle managers (to 2 decimals).
4. Compute the standard deviation of job satisfaction scores for both probability distributions (to 2 decimals).
5. What comparison can you make about the job satisfaction of senior executives and middle managers?

Answer 1

1.)E₁ = 4,03

2) E₂ = 3,85

3) σ₁² = 1,23

4)σ₂² = 1,14

5)σ₁ = 1,11

6)σ₂ = 1,07

5 ) Distribution of senior executive and Middle Manager look pretty similar , diference in median 0,18 and difference in standard deviation 0,04

Explanation:

1.-Expected value of the job satisfaction score for senior executive

E₁, variance σ₁ and standard deviation σ₁ and expected value of the job satisfaction score for Middle Manager E₂ variance σ₂ and standard deviation σ₂

E(X) = ∑ ( xi *f(xi)

JSS IS senior executive IS Middle Managers

1 0,05 0,04

2 0,09 0,1

3 0,03 0,12

4 0,44 0,45

5 0,39 0,29

∑ xi*f(xi) = 1*0,05 + 2 *0,09 + 3*0,03 + 4 * 0,44 + 5 * 0,39

E₁ = 0,05 + 0,18 + 0,09 + 1,76 + 1,95

1.)E₁ = 4,03

2.) ∑ xi*f(xi)

E₂ = 1* 0,04 + 2 * 0,1 + 3*0,12 + 4*0,45 +5*0,29

E₂ = 0,04 + 0,2 + 0,36 + 1,8 + 1,45

E₂ = 3,85

2 ) E₂ = 3,85

3) Variance of job satisfaction scores for senior executives.

x² f(x) x²* f(x)

1 0,05 0,05

4 0,09 0,36

9 0,03 0,27

16 0,44 7,04

25 0,39 9,75

σ₁² = [ ∑ x²*f(x) ] - μ² ∑ x²*f(x) = 17,47

In this case μ = E μ² = ( 4,03)² = 16,24

For senior executives: σ₁² = 17,47 - 16,24 = 1,23

σ₁² = 1,23

And standard deviation σ₁ = 1,11

For Middle managers

x² f(x) x²* f(x)

1 0,04 0,04

4 0,1 0,4

9 0,12 1,08

16 0,45 7,2

25 0,29 7,25

∑ x²*f(x) = 15,97

σ₂² = 15,97 - (3,85)² = 15,97 - 14,82

σ₂² = 1,14

σ₂ = √1,14 = 1,07

5 ) Distribution of senior executive and Middle Manager look pretty similar , diference in median 0,18 and difference in standard deviation 0,04