Suppose a meat pie manufacturer maintains a mean pie weight

Suppose a meat pie manufacturer maintains a mean pie weight of 850 grams with a standard deviation weight of all pies of 59 grams. Assume the weight of all pies is Normally Distributed. Find the weight of the pie such that only 0.2% of pies have a larger weight.

Solution

Mean = u = 850

SD = sigma = 59

P(larger weight) = 0.2% = 0.002

So, P(lower weight) = 1 - 0.002 = 0.998

Using below link, we can find z for each P
http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf

When P = 0.998, we see that z = 2.88

z = (x - u) / SD

2.88 = (x - 850) / 59

Crossmultiplying :

2.88(59) = x - 850

169.92 = x - 850

x = 850 + 169.92

x = 1019.92

So, the weight of the pie such that only 0.2% have a larger weight than it is : 1019.92 grams

Answer ---> 1019.92 grams