One piece of PVC pipe is to be inserted inside another piece. The length of the

Question

One piece of PVC pipe is to be inserted inside another piece. The length of the first piece is normally distributed with mean value 22 in. and standard deviation 0.8 in. The length of the second piece is a normal random variable with mean and standard deviation 16 in. and 0.5 in., respectively. The amount of overlap is normally distributed with mean value 1 in. and standard deviation 0.2 in. Assuming that the lengths and amount of overlap are independent of one another, what is the probability that the total length after insertion is between 32.65 in. and 35.35 in.?

I know E(z)=22+16-1=37 v(z)=0.93

how to do next?

Explanation / Answer

Your E(X) and Var(X) values are correct.

Now by using these two values we have to calculate P(32.65 < X < 35.35).

E(X) = 37

Var(X) = 0.93

sd(X) = sqrt(Var(X)) = sqrt(0.93) = 0.9644

First convert X = 32.65 and X = 35.35 into z-score.

z = (X - E(X)) / sd(X)

z = (32.65 - 37) / 0.9644 = -4.51

z = (35.35 - 37) / 0.9644 = -1.71

Now we have to find P(-4.51 < Z < -1.71).

P(-4.51 < Z < -1.71) = P(Z <=-1.71) - P(Z <=-4.51)

These probabilities we can find by using EXCEL.

syntax is ,

=NORMSDIST(z)

where z is test statistic value.

P(Z <=-1.71) = 0.043543

P(Z <=-4.51) = 0.000003

P(-4.51 < Z < -1.71) = 0.04354

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