Suppose that in order to satisfy legal requirements a meat p

Suppose that in order to satisfy legal requirements, a meat pie manufacturer may only produce 0.2% of pies below a weight of 750 grams. Assume the manufacturer precisely meets this requirement and that the weight of all pies is Normally distributed. Given the Standard Deviation weight of all pies is 50 grams, what is the mean weight of all pies?

Now suppose that the standard deviation accidentally increases from 50 to 100 grams. What percent of pies now fall below 750 grams?

Solution

P(x < 750) = 0.2% = 0.002

x = 750

SD = sigma = 50

From the link ---> http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf,

When P = 0.002, z = -2.88

We know that z = (x - u) / SD

-2.88 = (750 - u) / 50

-144 = 750 - u

u = 750 + 144

u = 894 --> FIRST ANSWER

So, the mean weight of all the pies is : 894 grams

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Now, SD = 100

Mean, u = 750

z = (x - u) / SD

z = (750 - 894) / 100

z = -1.44

From the same link, P(z < -1.44) = 0.0749 = 7.49%

So, 7.49% fall below 750 grams ---> SECOND ANSWER