Question

One normally distributed with mean value 20 in. and standard deviation .5 in. The length of the second piece is a normal rv with mean and standard deviation 15 in. and .4 in., respectively. The amount of overlap is normally distributed with mean value 1 in. and standard deviation .1 in. Assuming that the lengths and amount of overlap are independent of each other, what is the probability that the total length after insertion is between 34.5 and 35 in.

Answer 1

The probability that the the total length after insertion is between 34.5 and 35 inches is 0.1589.

Explanation:

Let the random variable X represent the length of the first piece, Y represent the length of the second piece and Z 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))`

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`

`SD(T)=√(V(X+Y-Z))\\=√(V(X)+V(Y)+V(Z))\\=\sqrt{0.50^(2)+0.40^(2)+0.10^(2)}\\=0.6480741\\\approx 0.65`

Since X, Y and Z all follow a Normal distribution, the random variable T, representing the total length will also follow a normal distribution.

`T\sim N(34, 0.65^(2))`

Compute the probability that the the total length after insertion is between 34.5 and 35 inches as follows:

`P(34.5<T<35)=P((34.5-34)/(0.65)<(T-\mu_(T))/(\sigma_(T))<(35-34)/(0.65))\\\\=P(0.77<Z<1.54)\\\\=P(Z<1.54)-P(Z<0.77)\\\\=0.93822-0.77935\\\\=0.15887\\\\\approx 0.1589`

*Use a z-table.

Thus, the probability that the the total length after insertion is between 34.5 and 35 inches is 0.1589.