Question

The population mean and standard deviation are given below. Find the indicated probability and determine whether a sample mean in the given range below would be considered unusual. If​ convenient, use technology to find the probability.
For a sample of n=40​, find the probability of a sample mean being less than 12,752 or greater than 12,755 when μ=12, 752 and σ=1.2.

For the given sample, the probability of a sample mean being less than 12, 752 or greater than 12,755 is:________

Would the given sample mean be considered unusual?

a. The sample mean would not be considered unusual because there is a probability less than 0.05 of the sample mean being within this range.
b. The sample mean would be considered unusual because there is a probability less than 0.05 of the sample mean being within this range.
c. The sample mean would not be considered unusual because there is a probability greater than 0.05 of the sample mean being within this range.
d. The sample mean would be considered unusual because there is a probability greater than 0.05 of the sample mean being within this range.

Answer 1

Answer:

For the given sample, the probability of a sample mean being less than 12, 752 or greater than 12,755 is: 0.5

c. The sample mean would not be considered unusual because there is a probability greater than 0.05 of the sample mean being within this range.

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:

$\mu = 12752, \sigma = 1.2, n = 40, s = \frac{1.2}{\sqrt{40}} = 0.1897[tex]For the given sample, the probability of a sample mean being less than 12, 752 or greater than 12,755 is:Less than 12752 is the pvalue of Z when X = 12752. So[tex]Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

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

$Z = \frac{12752 - 12752}{0.1897}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5.

Greater than 12755

1 subtracted by the pvalue of Z when X = 12755. So

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

$Z = \frac{12755 - 12752}{0.1897}$

$Z = 15.8$

$Z = 15.8$ has a pvalue of 1

1-1 = 0

For the given sample, the probability of a sample mean being less than 12, 752 or greater than 12,755 is:

0.5 + 0 = 0.5

Would the given sample mean be considered unusual?

A sample mean is unusual if there is a 0.05 or less probability of being in the interval. In this problem, this probability is greater, so the correct answer is.

c. The sample mean would not be considered unusual because there is a probability greater than 0.05 of the sample mean being within this range.