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?

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

Step-by-step 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

3x + 4y -25 =0 represents the amount of change sianna should recieve