Boxes are labeled as containing 500 g of cereal. The machine filling the boxes produces weights that are normally distributed with standard deviation 10 g. (a) If the target weight is 500 g, what is the probability that the machine produces a box with less than 475 g of cereal? (Round your answer to three decimal places.) (b) Suppose a law states that no more than 4% of a manufacturer's cereal boxes can contain less than the stated weight of 500 g. At what target weight should the manufacturer set its filling machine? (Round your answer to the nearest gram.)

Here we want the area under the Normal curve of mean [tex]\bf \bar x[/tex] = 500 g and standard deviation s = 10 g to the left of 475 (the probability that the machine produces a box with less than 475 g of cereal).

With the help of a spreadsheet we found that value is 0.0062 or 0.62% (another way of seeing it is that 62 boxes out of 10,000 will have 475 g or less)

See picture

b)

Here we want to find a value of [tex]\bf \mu[/tex] such that the area of the Normal curve of mean [tex]\bf \mu[/tex] and standard deviation 10 to the left of 500 is 4% or 0.04

If we make the change

[tex]\bf z=\frac{500-\mu}{s}[/tex]

then this is equivalent to finding a value of z for which the area under the Normal curve N(0,1) (mean = 0 s = 1) to the left of z is 4% = 0.04

Either by using a table or spreadsheet, we find that value is z = -1.751

So,

[tex]\bf z=\frac{500-\mu}{s}\rightarrow -1.751=\frac{500-\mu}{10}\rightarrow \mu=500+10*1.751=517.51[/tex]

and the manufacturer must set its filling machine to the target weight of 518 g.