Automatic filler are used to fill cans of cooking oil, which have to meet a mini

Question

Automatic filler are used to fill cans of cooking oil, which have to meet a minimum specification of 10 oz. To ensure that every can meets this minimum, the company has set a target value for the process average at 11 oz. The standard deviation of the amount of oil in a can is known to be 0.20 oz. a- At the process avarage of 11 oz, what percentage of the cans will contain less than 10 oz of cooking oil? Assume that the amount of oil in cans is normally distributed. b- Assuming that virtually all calues of a normal distribution fall within a distance of 3.5 of the standard deviation from the mean, find the minimum value to which the proess avarage can be lowered so that virtually no can will have less than 10 oz?

Explanation / Answer

Mean = 11oz

Stdev = .20oz

a. P(X<10) = P(Z< 11-10/.2) = P(Z<5) = 1.00

b. P(X<10 ) = 0

So, 10-X / .2 = -3.5

10-X = -3.5*.2

X = 10+3.5*.2 = 10.7

So, the  minimum value to which the proess avarage can be lowered is 10.7

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