The weight of a mini candy bar is normally distributed with

The weight of a \"mini\" candy bar is normally distributed with a standard deviation of 0.50 ounces. The production manager wants no more than 5% of the individual candy bars to weigh in excess of 5.1 ounces. What should he establish as the average weight in order for production to meet this specification?

Solution

P ( Z < X ) = 0.95
Value of z to the cumulative probability of 0.95 from normal table is 1.645
P( X-u/s.d < X - U /0.5 ) = 0.95
That is, ( 5.1 - U /0.5 ) = 1.6449
--> ( 5.1 - U ) = 1.6449*0.5
--> ( 5.1 - U /0.5 ) = -0.0627
--> U = -0.8224 + 5.1 = 4.27764