Question

When a production process is operating correctly, thenumber of units produced per hour has a normal distri- bution with a mean of 92.0 and a standard deviation of3.6. A random sample of 4 different hours was taken.a. Find the mean of the sampling distribution of thesample means. b. Find the variance of the sampling distribution ofthe sample mean.c. Find the standard error of the sampling distribu-tion of the sample mean.d. What is the probability that the sample mean ex-ceeds 93.0 units?

Answer 1

Answer:

a) The mean of the sampling distribution of the sample means is 92.

b) The variance of the sampling distribution of the sample mean is 3.24.

c) The standard error of the sampling distribution of the sample mean is 1.8.

d) 28.77% probability that the sample mean ex-ceeds 93.0 units.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$, a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$

Normal probability distribution

Problems of normally distributed samples can be 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.

In this problem, we have that:

$\mu = 92, \sigma = 3.6$.

a. Find the mean of the sampling distribution of thesample means.

The mean is the same as the mean of the population. So the mean of the sampling distribution of the sample means is 92.

b. Find the variance of the sampling distribution ofthe sample mean.

The standard deviation is $s = \frac{\sigma}{\sqrt{n}} = \frac{3.6}{\sqrt{4}} = 1.8$

The variance is $s^{2} = 1.8^{2} = 3.24$

c. Find the standard error of the sampling distribution of the sample mean.

This is the same as the standard deviation of the sample. So the standard error of the sampling distribution of the sample mean is 1.8.

d. What is the probability that the sample mean ex-ceeds 93.0 units?

This is 1 subtracted by the pvalue of Z when X = 93. So

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

$Z = \frac{93 - 92}{1.8}$

$Z = 0.56$

$Z = 0.56$ has a pvalue of 0.7123.

So there is a 1-0.7123 = 0.2877 = 28.77% probability that the sample mean ex-ceeds 93.0 units.