Thompson and Thompson is a steel bolts manufacturing company.  Their current steel bolts have a mean diameter of 147147 millimeters, and a variance of 2525.

If a random sample of 4444 steel bolts is selected, what is the probability that the sample mean would differ from the population mean by more than 1.11.1 millimeters? Round your answer to four decimal places.

We are dealing with a problem involving the sampling distribution of the sample mean. The key here is to use the **Central Limit Theorem (CLT)**, which states that for a large enough sample size, the sampling distribution of the sample mean will be approximately normal, even if the population distribution is not normal.

### Given Data:
- Population mean (\(\mu\)): 147 millimeters
- Population variance (\(\sigma^2\)): 25
- Sample size (\(n\)): 44
- We are interested in the probability that the sample mean differs from the population mean by more than 1 millimeter. That is, \( P(|\bar{X} - \mu| > 1) \).

### Step 1: Standard Error of the Mean
The standard error of the mean (SEM) is calculated as:

\[
\text{SEM} = \frac{\sigma}{\sqrt{n}}
\]
where:
- \(\sigma\) is the standard deviation of the population, and
- \(n\) is the sample size.

The population standard deviation \(\sigma\) is the square root of the variance:

\[
\sigma = \sqrt{25} = 5
\]

So, the standard error of the mean is:

\[
\text{SEM} = \frac{5}{\sqrt{44}} \approx 0.7548
\]

### Step 2: Convert the problem to a standard normal distribution
We are interested in the probability that the sample mean differs from the population mean by more than 1 millimeter. That means we want to find \( P(|\bar{X} - 147| > 1) \).

Using the **Z-score** formula to standardize this:

\[
Z = \frac{\bar{X} - \mu}{\text{SEM}}
\]

So, for \( \bar{X} - \mu = 1 \):

\[
Z = \frac{1}{0.7548} \approx 1.324
\]

Now we want to find the probability that the Z-score is greater than 1.324 (since we are looking for the sample mean to differ by more than 1 millimeter, we are interested in the two tails of the normal distribution):

\[
P(Z > 1.324) \quad \text{and} \quad P(Z < -1.324)
\]

From standard normal tables, the probability corresponding to a Z-score of 1.324 is approximately 0.0934 for one tail. Since we want the total probability in both tails:

\[
P(|Z| > 1.324) = 2 \times 0.0934 = 0.1868
\]

### Final Answer:
The probability that the sample mean would differ from the population mean by more than 1 millimeter is approximately **0.1868** or **18.68%**.

---

Would you like more details on how to interpret this result or how to use Z-tables for different problems? 

Here are some related questions for further exploration:

1. What is the Central Limit Theorem and when does it apply?
2. How do you calculate the standard error of the mean for different sample sizes?
3. What is the significance of Z-scores in statistics?
4. How do you use a Z-table to find probabilities for normal distributions?
5. What are confidence intervals and how do they relate to the standard error?

**Tip**: Whenever you are working with a large sample size, the sampling distribution of the sample mean will tend to be normally distributed, regardless of the population's distribution!

Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 147 millimeters, and a variance of 25. If a random sample of 44 steel bolts is selected, what is the probability that the sample mean would differ from the population mean by more than 1 millimeter? Round your answer to four decimal places.

Calculate the probability that the sample mean diameter of steel bolts differs from the population mean by more than 1 millimeter using the Central Limit Theorem. This problem includes calculating the standard error of the mean, Z-scores, and applying normal distribution concepts. Suitable for college statistics students or advanced high school learners.

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