The weights of certain machine components are normally distributed with a mean of 8.5 g and a standard deviation of 0.09 g. Find the two weights that separate the top 3% and the bottom 3%. These weights could serve as limits used to identify which components should be rejected. Round to the nearest hundredth of a gram.

To solve this problem, lets say that

X = the weight of the machine components.

X is normally distributed with mean=8.5 and sd=0.09

We need to find x1 and x2 such that

P (Xx2) = 0.03

Standardizing:

P (Z< (x1 - 8.5) / 0.09) = 0.03

P (Z > (x2 - 8.5) / 0.09) = 0.03.

From the Z standard table, we can see that approximately P = 0.03 is achieved when Z equals to:

z = - 1.88 and z = 1.88

Therefore,

P (Z1.88) = 0.03

So,

(x1 - 8.5) / 0.09 = - 1.88 and

(x2 - 8.5) / 0.09 = 1.88

Solving for x1 and x2:

x1=-1.88 (0.09) + 8.5 and

x2=1.88 (0.09) + 8.5

Which yields:

x1 = 8.33 g

x2 = 8.67 g

Answer: The bottom 3 is separated by the weight 8.33 g and the top 3 by the weight 8.67 g.