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The weights of certain machine components are normally distributed with a mean of 8.04 g and a standard deviation of 0.08 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)

Sagot :

lublana

Answer:

The bottom 3 is separated by weight 7.8896 g and the top 3 is separated by weight 8.1904 g.

Step-by-step explanation:

We are given that

Mean, [tex]\mu=8.04 g[/tex]

Standard deviation, [tex]\sigma=0.08g[/tex]

We have to find the two weights that separate the top 3% and the bottom 3%.

Let x be the weight of  machine components

[tex]P(X<x_1)=0.03, P(X>x_2)=0.03[/tex]

[tex]P(X<x_1)=P(Z<\frac{x_1-8.04}{0.08})[/tex]

=0.03

From z- table we get

[tex]P(Z<-1.88)=0.03, P(Z>1.88)=0.03[/tex]

Therefore, we get

[tex]\frac{x_1-8.04}{0.08}=-1.88[/tex]

[tex]x_1-8.04=-1.88\times 0.08[/tex]

[tex]x_1=-1.88\times 0.08+8.04[/tex]

[tex]x_1=7.8896[/tex]

[tex]\frac{x_2-8.04}{0.08}=1.88[/tex]

[tex]x_2=1.88\times 0.08+8.04[/tex]

[tex]x_2=8.1904[/tex]

Hence, the bottom 3 is separated by weight 7.8896 g and the top 3 is separated by weight 8.1904 g.

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