The average number of moves a person makes in his or her lifetime is 12 and the standard deviation is 3.0. Assume that the sample is taken from a large population and the correction factor can be ignored. Round the final answers to 4 decimal places and intermediate $z$ value calculations to 2 decimal places.

AFind the probability that the mean of a sample of 25 people is less than 10.

$P(\bar{X}<10)=$

BFind the probability that the mean of a sample of 25 people is greater than 10.

$P(\bar{X}>10)=$

CFind the probability that the mean of a sample of 25 people is between 11 and 12.

$P(11<\bar{X}<12)=$

To analyze this problem, we look at the distribution of the sample mean $\bar{X}$. According to the Central Limit Theorem, the sampling distribution of the mean for a large enough sample size ($n=25$) is approximately normal, even if the underlying population distribution is not.
Given:

Population mean $\mu = 12$
Population standard deviation $\sigma = 3.0$
Sample size $n = 25$

First, we calculate the standard error of the mean ($\sigma_{\bar{X}}$):
\[\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{3.0}{\sqrt{25}} = \frac{3.0}{5} = 0.6\]

Calculate P($\bar{X} < 10$)

For part (a), we want to find $P(\bar{X} < 10)$.
First, calculate the z-score for $\bar{X} = 10$:
\[z = \frac{\bar{X} - \mu}{\sigma_{\bar{X}}} = \frac{10 - 12}{0.6} = \frac{-2}{0.6} \approx -3.33\]
Using the standard normal distribution table for $z = -3.33$:
\[P(Z < -3.33) = 0.0004\]
So, the probability that the sample mean is less than 10 is approximately $0.0004$.

Calculate P($\bar{X} > 10$)

For part (b), we need $P(\bar{X} > 10)$.
Since the total area under the normal curve is 1, we use the complement rule:
\[P(\bar{X} > 10) = 1 - P(\bar{X} < 10)\]
Using the result from part (a):
\[P(\bar{X} > 10) = 1 - 0.000429 \approx 0.9996\]
Thus, the probability that the sample mean is greater than 10 is $0.9996$.

Calculate P(11 < $\bar{X}$ < 12)

For part (c), we need $P(11 < \bar{X} < 12)$.
Calculate the z-scores for 11 and 12:

For $\bar{X} = 11$: $z_1 = \frac{11 - 12}{0.6} = \frac{-1}{0.6} \approx -1.67$
For $\bar{X} = 12$: $z_2 = \frac{12 - 12}{0.6} = 0.00$

Now find the probability using the z-table:
\[P(11 < \bar{X} < 12) = P(-1.67 < Z < 0) = P(Z < 0) - P(Z < -1.67)\]
From the table, $P(Z < 0) = 0.5000$ and $P(Z < -1.67) = 0.0478$.
\[P(11 < \bar{X} < 12) = 0.5000 - 0.0478 = 0.4522\]
So, the probability that the sample mean is between 11 and 12 is $0.4522$.

Examples

This type of statistical analysis is fundamental in quality control and social science research. For instance, if an e-commerce company tracks the average number of clicks a user makes before buying a product, they can use these probabilities to determine if a specific group of 25 users is behaving atypically, helping them identify if a recent website update significantly changed user engagement. Understanding these distributions allows businesses to make data-driven decisions confidently.