Complete parts(a) and (b) using the probability distribution below. The number o
ID: 3294971 • Letter: C
Question
Complete parts(a) and (b) using the probability distribution below. The number of overtime is one week per employee (a) Find the mean, variance, and standard deviation of the probability distribution. Find the mean of the probability distribution. mu = (Round to one decimal place as needed.) Find the variance of the probability distribution. sigma^2 = (Round to one decimal place as needed.) Find the standard deviation of the probability distribution. sigma = (Round to one decimal place as needed.) (b) Interpret the results in the context of the real-life situation. A. An employee works an average of 3.4 overtime hours per week with a standard deviation of approximately 1.4 hours. B. An employee works an average 3.4 of overtime hours per week with a standard deviation of approximately 6 hours. C. An employee works an average of 1.4 overtime hours per week with a standard deviation of approximately 3.4 hours. D. An employee works an average of 2.0 overtime hours per week with a standard deviation of approximately 1.4 hours.Explanation / Answer
Solution:
(a) From the given data
X
P(X=x)
xP(X=x)
x2*P(X=x)
0
0.031
0
0
1
0.063
0.063
0.063
2
0.130
0.260
0.520
3
0.307
0.921
2.763
4
0.240
0.96
3.84
5
0.146
0.73
3.65
6
0.083
0.498
2.988
Total:
3.432
13.824
Mean : E(X) = xP(X=x) = 3.432
Now
E(X2) = x2P(X = x) = 13.824
2 = Var(X) = E(x2) - [E(X)]2
= 13.824 - (3.432)2 = 2.04538
= SD(X) = 2.04538 = 1.43017
(b) D. An employee works an average 2.0 overtime hours per week with a standard deviation of approximately 1.4 hours.
X
P(X=x)
xP(X=x)
x2*P(X=x)
0
0.031
0
0
1
0.063
0.063
0.063
2
0.130
0.260
0.520
3
0.307
0.921
2.763
4
0.240
0.96
3.84
5
0.146
0.73
3.65
6
0.083
0.498
2.988
Total:
3.432
13.824
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