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(X<x1)=0.03 and P(X>x2)=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(Z<-1.88)=0.03 and P(Z>1.88)=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

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

answer: x=3 and y=4

step-by-step explanation: