Answer:
Kindly see explanation
Step-by-step explanation:
Given the data:
JOB satisfaction (x) : 1, 2, 3, 4, 5
IS Senior ; p(x) : 0.05, 0.09, 0.03, 0.42, 0.41
IS Middle ; p(x) : 0.04, 0.10, 0.12, 0.46, 0.28
Expected value of job satisfaction for senior executives :
E(x) = Σp(x) * x = [(1*0.05) + (2*0.09) + (3*0.03) + (4*0.42) + (5*0.41)]
= 4.05
B) Σp(x) * x = [(1*0.04) + (2*0.10) + (3*0.12) + (4*0.46) + (5*0.28)]
= 3.84
C.) Job variance for executives:
Σ(x - E(x))² * p(x) :
((1 - 4.05)^2 * 0.05) + ((2 - 4.05)^2 * 0.09) + ((3 - 4.05)^2 * 0.03) + ((4 - 4.05)^2 * 0.42) + ((5 - 4.05)^2 * 0.41)
= 1.2475 = 1.25 ( using calculator)
job variance for middle managers :
Σ(x - E(x))² * p(x) :
((1 - 3.84)^2 * 0.04) + ((2 - 3.84)^2 * 0.10) + ((3 - 3.84)^2 * 0.12) + ((4 - 3.84)^2 * 0.46) + ((5 - 3.84)^2 * 0.28)
= 1.1344 = 1.13 (using calculator)
Standard deviation(sd) for (senior managers) :
sd = √variance
sd = √1.25 = 1.118 = 1.12
Standard deviation(sd) for (middle managers) :
sd = √variance
sd = √1.13 = 1.063 = 1.06
The expected value (mean) for senior executives (4.05) is slightly higher Than obtained for middle managers (3.84). Similarly the measure of Variation in job satisfaction ; standard deviation and variance is also higher for senior managers than in middle managers.
