Drums labeled 30 L are filled with a solution from a large vat. The amount of so

Question

Drums labeled 30 L are filled with a solution from a large vat. The amount of solution put into each drum is normally distributed with mean 30.01 L and standard deviation 0.1 L a) What is the probability that the total amount of solution contained in 50 drums is 3) more than 1501 L? (Hint: remember that the contents of each drum is added to the contents of the other drums.) If the total amount of solution in a vat is 2401 L, what is the probability that 80 drums can be filled without running out? How much solution should the vat contain sp that the probability is 0.9 that 80 drums can be filled without running out? b) c)

Explanation / Answer

a) expected amount in 50 drums =50*30.01 =1500.5

and std deviation =0.1*(50)1/2 =0.71

therefore probability that total solution exceeds 1501 L

b)

expected amount in 50 drums =80*30.01 =2400.8

and std deviation =0.1*(80)1/2 =0.89

therefore probability that amount in those drum should be at most 2401 L

c)

here for 90th percentile ; z =1.2816

tehrefore corresponding value =mean +z*std deviation =2400.8+1.2816*0.89 =2401.95

probability = P(X>1501) = P(Z>0.71)= 1-P(Z<0.71)= 1-0.7602= 0.2398

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