Question

The average breaking strength of a certain brand of steel cable is 2000 pounds, with a standard deviation of 100 pounds. A sample of 20 cables is selected and tested. Find the sample mean that will cut off the upper 95% of all samples of size 20 taken from the population. Assume the variable is normally distributed.

Answer 1

Answer:

The sample mean that will cut off the upper 95% of all samples of size 20 taken from the population is of 1963.2 pounds.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The average breaking strength of a certain brand of steel cable is 2000 pounds, with a standard deviation of 100 pounds.

This means that $\mu = 2000, \sigma = 100$

A sample of 20 cables is selected and tested.

This means that $n = 20, s = \frac{100}{\sqrt{20}} = 22.361$

Find the sample mean that will cut off the upper 95% of all samples of size 20 taken from the population.

This is the 100 - 95 = 5th percentile, which is X when Z has a p-value of 0.05, so X when Z = -1.645. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$-1.645 = \frac{X - 2000}{22.361}$

$X - 2000 = -1.645*22.361$

$X = 1963.2$

The sample mean that will cut off the upper 95% of all samples of size 20 taken from the population is of 1963.2 pounds.