Answer:
(a) The distribution of (Y - X) is <em>N</em> (0.001, 0.0005).
(b) The probability that the pin will not fit inside the collar is 0.023.
Step-by-step explanation:
The random variable <em>X</em> is defined as the diameter of the pin and the random variable <em>Y</em> is defined as the diameter of the collar.
The distribution of <em>X</em> and <em>Y</em> is:
![X\sim N(0.525, 0.0003)\\Y\sim N(0.526, 0.0004)](https://tex.z-dn.net/?f=X%5Csim%20N%280.525%2C%200.0003%29%5C%5CY%5Csim%20N%280.526%2C%200.0004%29)
The random variables <em>X</em> and <em>Y</em> are independent of each other.
(a)
Compute the expected value of (Y - X) as follows:
![E(Y-X)=E(Y)-E(X)=0.526-0.525=0.001](https://tex.z-dn.net/?f=E%28Y-X%29%3DE%28Y%29-E%28X%29%3D0.526-0.525%3D0.001)
The mean of (Y - X) is 0.001.
Compute the variance of (Y - X) as follows:
![V(Y-X)=V(Y)+V(X)-2Cov(X,Y)\\=V(Y)+V(X);\ X\ and\ Y\ are\ independent\\=0.0003^{2}+0.0004^{2}\\=0.00000025](https://tex.z-dn.net/?f=V%28Y-X%29%3DV%28Y%29%2BV%28X%29-2Cov%28X%2CY%29%5C%5C%3DV%28Y%29%2BV%28X%29%3B%5C%20X%5C%20and%5C%20Y%5C%20are%5C%20independent%5C%5C%3D0.0003%5E%7B2%7D%2B0.0004%5E%7B2%7D%5C%5C%3D0.00000025)
![SD(Y-X)=\sqrt{0.00000025}=0.0005](https://tex.z-dn.net/?f=SD%28Y-X%29%3D%5Csqrt%7B0.00000025%7D%3D0.0005)
The standard deviation of (Y - X) is 0.0005.
Thus, the distribution of (Y - X) is <em>N</em> (0.001, 0.0005).
(b)
Compute the probability of [(Y - X) ≤ 0] as follows:
![P(Y-X\leq 0)=P(\frac{(Y-X)-\mu_{Y-X}}{\sigma_{Y-X}}\leq \frac{0-0.001}{0.0005})=P(Z](https://tex.z-dn.net/?f=P%28Y-X%5Cleq%200%29%3DP%28%5Cfrac%7B%28Y-X%29-%5Cmu_%7BY-X%7D%7D%7B%5Csigma_%7BY-X%7D%7D%5Cleq%20%5Cfrac%7B0-0.001%7D%7B0.0005%7D%29%3DP%28Z%3C-2%29%3D0.0228%5Capprox0.023)
*Use a <em>z</em>-table for the probability value.
Thus, the probability that the pin will not fit inside the collar is 0.023.