Question

Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of 1.8. (Round your answers to four decimal places.)
(a) If the distribution is normal, what is the probability that the sample mean hardness for a random sample of 17 pins is at least 51?
answer is 0.011

(b) What is the (approximate) probability that the sample mean hardness for a random sample of 45 pins is at least 51?

Answer 1

Answer:

a) 0.011 = 1.1% probability that the sample mean hardness for a random sample of 17 pins is at least 51

b) 0.0001 = 0.1% probability that the sample mean hardness for a random sample of 45 pins is at least 51

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 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.

In this problem, we have that:

$\mu = 50, \sigma = 1.8$

(a) If the distribution is normal, what is the probability that the sample mean hardness for a random sample of 17 pins is at least 51?

Here $n = 17, s = \frac{1.8}{\sqrt{17}} = 0.4366$

This probability is 1 subtracted by the pvalue of Z when X = 51. So

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

By the Central Limit Theorem

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

$Z = \frac{51 - 50}{0.4366}$

$Z = 2.29$

$Z = 2.29$ has a pvalue of 0.9890

1 - 0.989 = 0.011

0.011 = 1.1% probability that the sample mean hardness for a random sample of 17 pins is at least 51

(b) What is the (approximate) probability that the sample mean hardness for a random sample of 45 pins is at least 51?

Here $n = 17, s = \frac{1.8}{\sqrt{45}} = 0.2683$

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

$Z = \frac{51 - 50}{0.0.2683}$

$Z = 3.73$

$Z = 3.73$ has a pvalue of 0.9999

1 - 0.9999 = 0.0001

0.0001 = 0.1% probability that the sample mean hardness for a random sample of 45 pins is at least 51