Answer:
a. 4.05 b. 3.84 c. 1.2475 and 1.1344 d. 1.1169 and 1.0651 e. We can say that the overall job satistaction of senior executives and middle managers is about 4; however, there is more variability in the job satisfaction for senior executives than in the job satisfaction for middle managers.
Step-by-step explanation:
a. (1)(0.05)+(2)(0.09)+(3)(0.03)+(4)(0.42)+(5)(0.41) = 4.05
b. (1)(0.04)+(2)(0.1)+(3)(0.12)+(4)(0.46)+(5)(0.28) = 3.84
c. We compute the variances as follow:
= 1.2475 and
= 1.1344
d. The standard deviation is the squared root of the variance, therefore, we have
and
e. The expected value of the job satisfaction score for senior executives is very similar to the job satisfaction score for middle managers. We can say that the overall job satistaction of senior executives and middle managers is about 4; however, there is more variability in the job satisfaction for senior executives than in the job satisfaction for middle managers.
The answer is 11 my friend :)
Part (a): All are defective
Only one way of selecting the 5 defective transistors:
Number of ways of selections available = 6C5 = 16!/[5!*(15-5)!] = 4368
Probability they are all defective = Number of ways of selecting 5 defectives/Total number of ways possible = 1/4368 ≈ 0.000229
Part (b): None are defective
Total number of non defectives = 16 -5 = 11
Number of ways of selecting 5 non defective = 11C5 = 462 ways
Total number of ways possible = 16C5 = 4368
Probability of selecting 5 non defectives = 462/4368 = 11/104 ≈ 0.1058
Answer:
11.047---68.168---89.51
Step-by-step explanation:
just switch 11 and 89