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.

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7

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14

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3

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step-by-step explanation:

1. 3.4/4

2. 0.8/7

3. 0.5 x 3

4. 0.9 x 6

5. 10.8

6. 2.35

7. 0.567

8. 9.35

9. 0.02

10. 0.06

11. 0.16

12. 0.42

13. 0.63

14. 5.6

15. 8.1

16. 9.9

17. 0.28

18. 0.48

19. 0.27

20. 0.49

21. 0.45

22. 0.72

23. 0.9

24. 0.88

25. 0.75

26. 0.16

27. 0.328

28. 0.495

29. 0.72

30. 1.068