Answered

Get the information you need with the help of IDNLearn.com's extensive Q&A platform. Get timely and accurate answers to your questions from our dedicated community of experts who are here to help you.

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 :

Lublanaidn

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.

Thank you for joining our conversation. Don't hesitate to return anytime to find answers to your questions. Let's continue sharing knowledge and experiences! Your search for solutions ends here at IDNLearn.com. Thank you for visiting, and come back soon for more helpful information.