One piece of pvc pipe is to be inserted inside another piece. the length of the first piece is 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 one another, what is the probability that the total length after insertion is between 34.5 in. and 35 in.?

0.159 is the probability that the total length after insertion is between 34.5 inch and 35 inch.  

Step-by-step explanation:

We are given the following information in the question:

The length of the first piece is normally distributed with mean value 20 inch and standard deviation 0.5 inch.

The length of the second piece is a normal with mean and standard deviation 15 inch  and 0.4 inch respectively.

The amount of overlap is normally distributed with mean value 1 inch and standard deviation 0.1 inch.

The lengths and amount of overlap are independent of one another.

Let the total length of PVC pipe be represented by Y.

The sum of two independent normally distributed random variables is normal, with its mean being the sum of the two means, and its variance being the sum of the two variances

Then, Y follows a normal distribution with

Y = Length of first pipe + Length of second Pipe - Overlap

P(total length after insertion is between 34.5 inch and 35 inch)

0.159 is the probability that the total length after insertion is between 34.5 inch and 35 inch.  

your answer is a

step-by-step explanation:

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step-by-step explanation:

the multiplicative inverse of a fraction a/b is b/a. for the multiplicative inverse of a real number, divide 1 by the number. for example, the reciprocal of 5 is one fifth 1/5 or 0.2, and the reciprocal of 0.25 is 1 divided by 0.25, or 4.