Question

A piece of rope is made up of 100 strands. Assume that the breaking strength of the rope is the sum of the breaking strengths of the individual strands. Assume further that this sum may be considered to be the sum of an independent trial process with 100 experiments each having an expected value of 10 pounds and standard deviation Find the approximate probability that the rope will support a weight

a. of 1000 pounds.
b. of 970 pounds.

Answer 1

a. of 1000 pounds. P=0.5

b. of 970 pounds. P=0.99865

Explanation:

The question is incomplete: the standard deviation is 1.

The strength of the rope is the sum of the strengh of the 100 strands that make the rope. If the strands have a mean strength of 10 and a standard devition of 1, the strength of the rope will have the following mean and standard deviations:

`E(X)=nE(x)=100*10=1000\\\\V(X)=aV(x)\\\\\sigma_X=√(n)\sigma_x=√(100)*1=10`

We can calculate then, the probability that the rope will support a weight of 1000 pounds:

`z=(X-\mu)/\sigma=(1000-1000)/10=0\\\\P(X>1000)=P(z>0)=0.5`

The probability that the rope will support a weight of 970 pounds is:

`z=(X-\mu)/\sigma=(970-1000)/10=-30/10=-3\\\\P(X>970)=P(z>-3)=0.99865`