A racing car consumes a mean of 86
 gallons of gas per race with a standard deviation of 7
 gallons.  

If 41
 racing cars are randomly selected, what is the probability that the sample mean would differ from the population mean by more than 3.1
 gallons? Round your answer to four decimal places.


To solve this problem, we will apply the **Central Limit Theorem** (CLT). The CLT states that the distribution of the sample mean for a large enough sample size will be approximately normally distributed, even if the population distribution is not normal.

### Given:
- Population mean (\(\mu\)) = 86 gallons
- Population standard deviation (\(\sigma\)) = 7 gallons
- Sample size (\(n\)) = 41
- The sample mean differs from the population mean by more than 3.1 gallons.

We are looking for the probability that the sample mean will differ from the population mean by more than 3.1 gallons.

### Step 1: Find the standard error of the mean (SE)
The standard error of the mean is calculated as:

\[
SE = \frac{\sigma}{\sqrt{n}}
\]

Substituting the values:

\[
SE = \frac{7}{\sqrt{41}} \approx 1.093
\]

### Step 2: Calculate the z-scores
We need to calculate the z-scores corresponding to the sample mean differing from the population mean by 3.1 gallons, which means we need to calculate for both \( \mu + 3.1 \) and \( \mu - 3.1 \).

For \( \mu + 3.1 \):

\[
z = \frac{(86 + 3.1) - 86}{SE} = \frac{3.1}{1.093} \approx 2.836
\]

For \( \mu - 3.1 \):

\[
z = \frac{(86 - 3.1) - 86}{SE} = \frac{-3.1}{1.093} \approx -2.836
\]

### Step 3: Find the probability
The z-scores we found represent how many standard errors away the sample mean is from the population mean. Using a standard normal distribution table or a calculator:

- The probability of a z-score being less than -2.836 is approximately 0.0023.
- The probability of a z-score being greater than 2.836 is also approximately 0.0023.

Therefore, the total probability that the sample mean will differ from the population mean by more than 3.1 gallons is the sum of these two probabilities:

\[
P(\text{sample mean differs by more than 3.1 gallons}) = 0.0023 + 0.0023 = 0.0046
\]

### Final Answer:
The probability that the sample mean would differ from the population mean by more than 3.1 gallons is approximately **0.0046** (or 0.46%).

---

Would you like further details or have any questions?

Here are 5 related questions to expand your understanding:
1. What is the Central Limit Theorem and why is it important in statistics?
2. How does increasing the sample size affect the standard error of the mean?
3. What is a z-score and how is it used in probability calculations?
4. How do you interpret the results of a z-score in the context of a normal distribution?
5. What is the significance of the population standard deviation in this problem?

**Tip:** The larger the sample size, the smaller the standard error, which makes the sample mean more likely to be closer to the population mean.

A racing car consumes a mean of 86 gallons of gas per race with a standard deviation of 7 gallons. If 41 racing cars are randomly selected, what is the probability that the sample mean would differ from the population mean by more than 3.1 gallons? Round your answer to four decimal places.

Learn how to calculate the probability that the sample mean for fuel consumption of racing cars differs from the population mean by more than 3.1 gallons. This problem involves concepts like the Central Limit Theorem, Z-Scores, and Standard Error of the Mean. Suitable for advanced high school and introductory college students.

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