Question

The fracture strength of tempered glass averages 14 (measured in thousands ofpounds per square inch) and has standard deviation 2.
a. What is the probability that theaverage fracture strength of 100 randomly selected pieces of this glass exceeds 14.5?
b. Find an interval that includes, with probability 0.95, the average fracture strength of 100 randomly selected pieces of this glass.

Answer 1

Answer:

a) 0.62% probability that theaverage fracture strength of 100 randomly selected pieces of this glass exceeds 14.5

b) The interval that includes, with probability 0.95, the average fracture strength of 100 randomly selected pieces of this glass is between 13.61 and 14.39 thousands of pounds per square inch

Step-by-step explanation:

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

To solve question b, we find a normal confidence interval.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes 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.

a. What is the probability that theaverage fracture strength of 100 randomly selected pieces of this glass exceeds 14.5?

We have that $\mu = 14, \sigma = 2, n = 100, s = \frac{2}{\sqrt{100}} = 0.2$

This probability is 1 subtracted by the pvalue of Z when X = 14.5. So

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

By the Central Limit Theorem

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

$Z = \frac{14.5 - 14}{0.2}$

$Z = 2.5$

$Z = 2.5$ has a pvalue of 0.9938

1 - 0.9938 = 0.0062

0.62% probability that theaverage fracture strength of 100 randomly selected pieces of this glass exceeds 14.5

b. Find an interval that includes, with probability 0.95, the average fracture strength of 100 randomly selected pieces of this glass.

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$.

So it is z with a pvalue of $1-0.025 = 0.975$, so $z = 1.96$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96*\frac{2}{\sqrt{100}} = 0.39$

The lower end of the interval is the sample mean subtracted by M. So it is 14 - 0.39 = 13.61

The upper end of the interval is the sample mean added to M. So it is 14 + 0.39 = 14.39

The interval that includes, with probability 0.95, the average fracture strength of 100 randomly selected pieces of this glass is between 13.61 and 14.39 thousands of pounds per square inch