Chapter 9

Central Limit Theorem

Exercise 9.2 #3 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 trials process with 100 experiments each having expected value of 10 pounds and standard deviation of 1. Find the approximate probability that the rope will support a weight

  1. of 1000 pounds
  2. of 970 pounds

Solution

Based off the text we know the following: \(\mu = 10\) \(\sigma = 1\)

We are running 100 independent trials therefore: \(\mu = 10 * 100 = 1000\) \(\sigma = \sqrt100 = 10\)

In regards to the rope being able to support 1000 pounds, therefore will take the weight and subtract it from the number of trials and divide it by the square root of the number of trials which is then multiplied by the standard deviation to the second power.

(a = (1000 - 100 * 10)/sqrt(100 * 1 ^2))
## [1] 0

Lastly using pnorm we calculate the probability.

pnorm(a, lower.tail = FALSE)
## [1] 0.5

In regards to the rope being able to support 970 pounds we will follow the same methods from example A, this time substituting 1000 with 970.

(b = (970 - 100 * 10)/sqrt(100 * 1 ^2))
## [1] -3
pnorm(b, lower.tail = FALSE)
## [1] 0.9986501