Answer: 78.89%
Explanation:
Given : Sample size : n= 1200
Sample mean : [tex]\overline{x}=2.45 [/tex]
Standard deviation : [tex]\sigma=0.07[/tex]
We assume that it follows Gaussian distribution (Normal distribution).
Let x be a random variable that represents the shaft diameter.
Using formula, [tex]z=\dfrac{x-\mu}{\sigma}[/tex], the z-value corresponds to 2.39 will be :-
[tex]z=\dfrac{2.39-2.45}{0.07}\approx-0.86[/tex]
z-value corresponds to 2.60 will be :-
[tex]z=\dfrac{2.60-2.45}{0.07}\approx2.14[/tex]
Using the standard normal table for z, we have
P-value = [tex]P(-0.86<z<2.14)=P(z<2.14)-P(z<-0.86)[/tex]
[tex]=P(z<2.14)-(1-P(z<0.86))=P(z<2.14)-1+P(z<0.86)\\\\=0.9838226-1+0.8051054\\\\=0.788928\approx0.7889=78.89\%[/tex]
Hence, the percentage of the diameter of the total shipment of shafts will fall between 2.39 inch and 2.60 inch = 78.89%
