CLT Page 354 #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 1. Find the approximate probability that the rope will support a weight

  1. of 1000 pounds.
  2. of 970 pounds.


Given: The number of strands (\(n = 100\)) Expected value (mean, \(\mu\)) of breaking strength for each strand = 10 pounds Standard deviation (\(\sigma\)) of breaking strength for each strand = 1 pound


Step 1: Calculate the mean and standard deviation of the sum


Step 2: Use the CLT to approximate the distribution}

Using th CLT, the distribution of the sum of the breaking strengths of the strands is approximately normal with mean \(\mu_{sum} = 1000\) pounds and standard deviation \(\sigma_{sum} = 10\) pounds.


Step 3: Calculate the probabilities


Probability that the rope will support 1000 pounds ie \(P(X \leq 1000)\).

Since the mean is 1000 pounds, \(P(X \leq 1000)\) corresponds to the 50th percentile of a normal distribution, which is 0.5 or 50%.


Probability that the rope will support 970 pounds

To find this probability, we first standardize the value using the Z-score formula:

\[ Z = \frac{X - \mu}{\sigma} \]

\[ Z = \frac{970 - 1000}{10} = -3 \]

Now we look up the probability corresponding to \(Z = -3\)

\[ P(X \leq 970) = P(Z \leq -3) \]

The value from the standard normal distribution for \(Z = -3\) is 0.0013 or 0.13%.

# parameters
n <- 100 # Number of strands
mu <- 10 # Mean of breaking strength for each strand (pounds)
sigma <- 1  # Standard deviation of breaking strength for each strand (pounds)

# Calculate the mean and standard deviation of the sum
mu_sum <- n * mu  # Mean of the sum of breaking strengths
sigma_sum <- sqrt(n) * sigma   # Standard deviation of the sum of breaking strengths

# Calculate the probabilities

# (a) Probability that the rope will support 1000 pounds (P(X <= 1000))
probability_1000 <- pnorm(1000, mean = mu_sum, sd = sigma_sum)

# (b) Probability that the rope will support 970 pounds
z_970 <- (970 - mu_sum) / sigma_sum
probability_970 <- pnorm(z_970)

# Print the results
cat("Probability the rope will support 1000 pounds:", probability_1000, "\n")
## Probability the rope will support 1000 pounds: 0.5
cat("Probability the rope will support 970 pounds:", probability_970, "\n")
## Probability the rope will support 970 pounds: 0.001349898