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White raven [17]
3 years ago
7

Hello! Mind Helping Me?

Mathematics
2 answers:
scoundrel [369]3 years ago
6 0
The answer would be 2 Pounds
umka21 [38]3 years ago
3 0

Answer:

2 pounds

if you add .5 4 times with each other, you get 2

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The following probability distributions of job satisfaction scores for a sample of information
kodGreya [7K]

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)^2(0.05)+(2)^2(0.09)+(3)^2(0.03)+(4)^2(0.42)+(5)^2(0.41)] - 4.05^2 = 1.2475 and [(1)^2(0.04)+(2)^2(0.1)+(3)^2(0.12)+(4)^2(0.46)+(5)^2(0.28)]-3.84^2 = 1.1344

d. The standard deviation is the squared root of the variance, therefore, we have \sqrt{1.2475} = 1.1169 and \sqrt{1.1344} = 1.0651

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.

4 0
3 years ago
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