The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 10 mg and a standard deviation of 1.45 mg. The company that produces these cigarettes claims that it has now reduced the amount of nicotine. The supporting evidence consists of a sample of 37 cigarettes with a mean nicotine amount of 9.499 mg.
Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly seleting 37 cigarettes with a mean of 9.499 mg or less.

To find the probability of randomly selecting 37 cigarettes with a mean nicotine amount of 9.499 mg or less, we can use the properties of the normal distribution.

1. **Identify the parameters:**
   - Mean (\( \mu \)) = 10 mg
   - Standard deviation (\( \sigma \)) = 1.45 mg
   - Sample size (\( n \)) = 37
   - Sample mean (\( \bar{x} \)) = 9.499 mg

2. **Calculate the standard error (SE):**
   \[
   SE = \frac{\sigma}{\sqrt{n}} = \frac{1.45}{\sqrt{37}} \approx \frac{1.45}{6.08} \approx 0.238
   \]

3. **Calculate the z-score:**
   \[
   z = \frac{\bar{x} - \mu}{SE} = \frac{9.499 - 10}{0.238} \approx \frac{-0.501}{0.238} \approx -2.10
   \]

4. **Find the probability using the z-score:**
   Using standard normal distribution tables or a calculator, we look up the z-score of -2.10.

   The cumulative probability for \( z = -2.10 \) is approximately 0.0179.

Thus, the probability of randomly selecting 37 cigarettes with a mean nicotine amount of 9.499 mg or less is approximately **0.0179**, or **1.79%**.

### Summary
- The probability of selecting 37 cigarettes with a mean of 9.499 mg or less is about 1.79%.

Would you like more details on any part of this calculation, or do you have any questions? Here are some related questions to consider:

1. How does the sample size affect the standard error?
2. What would happen to the probability if the mean nicotine content was lower?
3. How can we interpret the z-score in the context of this problem?
4. What other statistical tests could be performed to validate the company's claim?
5. How would this probability change if the standard deviation were different? 

**Tip:** Always ensure that the assumptions of normality are met when using the normal distribution in practice.

The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 10 mg and a standard deviation of 1.45 mg. The company that produces these cigarettes claims that it has now reduced the amount of nicotine. The supporting evidence consists of a sample of 37 cigarettes with a mean nicotine amount of 9.499 mg. Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly selecting 37 cigarettes with a mean of 9.499 mg or less.

Explore how to calculate the probability of selecting 37 cigarettes with a mean nicotine content of 9.499 mg or less, given a population mean of 10 mg and standard deviation of 1.45 mg. This guide covers essential statistical concepts and the Central Limit Theorem.

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