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Smalltown Elevator produces elevator rails. To meet specifications, an elevator rail must be between 0.995 inches and 1.005 inches in diameter. Suppose that the diameter of an elevator rail follows a normal random variable with mean of 1 inch and standard deviation of 0.003 inches. Rounded to the nearest one tenth of one percent, what fraction of all elevator rails will meet specifications?

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JeanaShupp

Answer: 90.5%

Step-by-step explanation:

Given: Mean : [tex]\mu = 1\text{ inch}[/tex]

Standard deviation : [tex]\sigma = 0.003\text{ inch}[/tex]

The formula to calculate z is given by :-

[tex]z=\dfrac{x-\mu}{\sigma}[/tex]

For x= 0.995

[tex]z=\dfrac{0.995-1}{0.003}=-1.66666666667\approx-1.67[/tex]

The P Value =[tex]P(z<-1.67)=0.0474597[/tex]

For x= 1.005

[tex]z=\dfrac{1.005-1}{0.003}=1.66666666667\approx1.67[/tex]

The P Value =[tex]P(z<1.67)= 0.9525403[/tex]

[tex]\text{Now, }P(0.995<X<1.005)=P(X<1.005)-P(X<0.995)\\\\=P(z<1.67)-P(z<-1.67)\\\\=0.9525403-0.0474597=0.9050806[/tex]

In percent ,

[tex]P(0.995<X<1.005)=0.9050806\times100=90.50806\%\approx90.5\%[/tex]

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