Answer:
The probability that the the total length after insertion is between 34.5 and 35 inches is 0.1589.
Step-by-step explanation:
Let the random variable <em>X</em> represent the length of the first piece, <em>Y</em> represent the length of the second piece and <em>Z</em> represents the overlap.
It is provided that:
![X\sim N(20,\ 0.50^{2})\\Y\sim N(15,\ 0.40^{2})\\Z\sim N(1,\ 0.10^{2})](https://tex.z-dn.net/?f=X%5Csim%20N%2820%2C%5C%200.50%5E%7B2%7D%29%5C%5CY%5Csim%20N%2815%2C%5C%200.40%5E%7B2%7D%29%5C%5CZ%5Csim%20N%281%2C%5C%200.10%5E%7B2%7D%29)
It is provided that the lengths and amount of overlap are independent of each other.
Compute the mean and standard deviation of total length as follows:
![E(T)=E(X+Y-Z)\\=E(X)+E(Y)-E(Z)\\=20+15-1\\=34](https://tex.z-dn.net/?f=E%28T%29%3DE%28X%2BY-Z%29%5C%5C%3DE%28X%29%2BE%28Y%29-E%28Z%29%5C%5C%3D20%2B15-1%5C%5C%3D34)
![SD(T)=\sqrt{V(X+Y-Z)}\\=\sqrt{V(X)+V(Y)+V(Z)}\\=\sqrt{0.50^{2}+0.40^{2}+0.10^{2}}\\=0.6480741\\\approx 0.65](https://tex.z-dn.net/?f=SD%28T%29%3D%5Csqrt%7BV%28X%2BY-Z%29%7D%5C%5C%3D%5Csqrt%7BV%28X%29%2BV%28Y%29%2BV%28Z%29%7D%5C%5C%3D%5Csqrt%7B0.50%5E%7B2%7D%2B0.40%5E%7B2%7D%2B0.10%5E%7B2%7D%7D%5C%5C%3D0.6480741%5C%5C%5Capprox%200.65)
Since X, Y and Z all follow a Normal distribution, the random variable <em>T</em>, representing the total length will also follow a normal distribution.
![T\sim N(34, 0.65^{2})](https://tex.z-dn.net/?f=T%5Csim%20N%2834%2C%200.65%5E%7B2%7D%29)
Compute the probability that the the total length after insertion is between 34.5 and 35 inches as follows:
![P(34.5](https://tex.z-dn.net/?f=P%2834.5%3CT%3C35%29%3DP%28%5Cfrac%7B34.5-34%7D%7B0.65%7D%3C%5Cfrac%7BT-%5Cmu_%7BT%7D%7D%7B%5Csigma_%7BT%7D%7D%3C%5Cfrac%7B35-34%7D%7B0.65%7D%29%5C%5C%5C%5C%3DP%280.77%3CZ%3C1.54%29%5C%5C%5C%5C%3DP%28Z%3C1.54%29-P%28Z%3C0.77%29%5C%5C%5C%5C%3D0.93822-0.77935%5C%5C%5C%5C%3D0.15887%5C%5C%5C%5C%5Capprox%200.1589)
*Use a <em>z</em>-table.
Thus, the probability that the the total length after insertion is between 34.5 and 35 inches is 0.1589.