I would help if It wasn’t so confusing.
C it would be c because that has more and the others have less
Answer: 78.89%
Explanation:
Given : Sample size : n= 1200
Sample mean : ![\overline{x}=2.45](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%3D2.45%20)
Standard deviation : ![\sigma=0.07](https://tex.z-dn.net/?f=%5Csigma%3D0.07)
We assume that it follows Gaussian distribution (Normal distribution).
Let x be a random variable that represents the shaft diameter.
Using formula,
, the z-value corresponds to 2.39 will be :-
![z=\dfrac{2.39-2.45}{0.07}\approx-0.86](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7B2.39-2.45%7D%7B0.07%7D%5Capprox-0.86)
z-value corresponds to 2.60 will be :-
![z=\dfrac{2.60-2.45}{0.07}\approx2.14](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7B2.60-2.45%7D%7B0.07%7D%5Capprox2.14)
Using the standard normal table for z, we have
P-value = ![P(-0.86](https://tex.z-dn.net/?f=P%28-0.86%3Cz%3C2.14%29%3DP%28z%3C2.14%29-P%28z%3C-0.86%29)
![=P(z](https://tex.z-dn.net/?f=%3DP%28z%3C2.14%29-%281-P%28z%3C0.86%29%29%3DP%28z%3C2.14%29-1%2BP%28z%3C0.86%29%5C%5C%5C%5C%3D0.9838226-1%2B0.8051054%5C%5C%5C%5C%3D0.788928%5Capprox0.7889%3D78.89%5C%25)
Hence, the percentage of the diameter of the total shipment of shafts will fall between 2.39 inch and 2.60 inch = 78.89%