SOLUTION: Complete parts​ (a) and​ (b) using the probability distribution below.
The number of overtime hours worked in one week per employee
Overtime hours---0---6 on the be
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-> SOLUTION: Complete parts​ (a) and​ (b) using the probability distribution below.
The number of overtime hours worked in one week per employee
Overtime hours---0---6 on the be
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Question 1120593: Complete parts (a) and (b) using the probability distribution below.
The number of overtime hours worked in one week per employee
Overtime hours---0---6 on the below numbers
Probability---0.015----0.096 on the below numbers
0----0.015
1----0.066
2---0.172
3----0.273
4----0.237
5-----0.141
6-----0.096
(a) Find the mean, variance, and standard deviation of the probability distribution.
Find the mean of the probability distribution._______ Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! The mean is the sum of all integers * their probability
0+0.066+0.344+0.819
0.948+0.705+0.576=3.458 mean
variance is the sum of the difference between the integer and the mean * probability
3.458^2*0.015=0.1794
2.458^2*0.066=0.3988
1.458^2*0.172=0.3656
0.458^2*0.273=0.0573
0.542^2*0.237=0.0696
1.542^2*0.141=0.3353
2.542^2*0.096=0.6203
sum is 2.0263 variance
sd is sqrt (2.0263)=1.423 standard deviation