Question 1173293: 1:a production department has 35 similar milling machines.the number of breakdown on each machine average 0.06 per week.determine the probability of having (a):one
(B):less than three machines breaking down in any week
2;A company employing 60 people pays its employees an average wages of TZs 725 an hour with standard deviation of TZS 60.suppose the wages are approximately normally distributed.
1: determine the number of workers receiving wages between TZS 665 and TZS 770 an hour inclusive
2: determine the minimum hourly wages received by the highest 5% of the employees
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 1. a. one breakdown is 35*0.06*0.94^34=0.2562
b. for < 3, look at 0, which has prob. .94^35=0.1147 and 2, which would be 35C2*0.06^2*0.94^33=0.2780
That sum is 0.6489, when the prob. of 1 is included.
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2.z=(x-mean)/sd or (665-725)/60 which is -1 and z=(770-725)/60 or 0.75
That probability is 0.6147. For 60 people, that would be 37 to the nearest integer.
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the top 5% are at z=1.645
therefore, 1.645=(x-725)/60
or 98.7=x-725
x=TZS 823.69 which can be rounded if needed
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