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Let's go step by step to solve the problem you provided in the image.

### Problem Context:
- The amounts of nicotine in a brand of cigarettes are normally distributed with a **mean** \( \mu = 0.921 \) g and a **standard deviation** \( \sigma = 0.307 \) g.
- A sample of 30 cigarettes is taken, and the **mean nicotine** in the sample is found to be \( M = 0.803 \) g.
- You are asked to find the **probability** that a sample of 30 cigarettes has a **mean of 0.803 g or less** given that the population parameters have not changed.

### Solution:

This is a **sampling distribution of the sample mean** problem. Since the population mean and standard deviation are known, and the sample size is large (30), we can use the **Central Limit Theorem** to approximate the sampling distribution of the sample mean as normal.

#### Step 1: Calculate the Standard Error of the Mean (SE)
The standard error (SE) is the standard deviation of the sample mean distribution and is given by:
\[
SE = \frac{\sigma}{\sqrt{n}}
\]
where:
- \( \sigma = 0.307 \) g (population standard deviation),
- \( n = 30 \) (sample size).

So,
\[
SE = \frac{0.307}{\sqrt{30}} \approx 0.0560 \, g
\]

#### Step 2: Find the Z-score
The **Z-score** formula for a sample mean is:
\[
Z = \frac{M - \mu}{SE}
\]
where:
- \( M = 0.803 \) g (sample mean),
- \( \mu = 0.921 \) g (population mean),
- \( SE \approx 0.0560 \) g.

Substitute the values:
\[
Z = \frac{0.803 - 0.921}{0.0560} \approx -2.107
\]

#### Step 3: Find the Probability
Now, we need to find the probability corresponding to this Z-score using the **standard normal distribution** table (or a calculator). The Z-score of \( Z = -2.107 \) corresponds to a cumulative probability (area to the left) of approximately **0.0175**.

Thus, the probability \( P(M < 0.803) \approx 0.0175 \).

### Final Answer:
\[
P(M < 0.803) = 0.0175
\]
This means the probability of selecting a sample of 30 cigarettes with a mean nicotine content of 0.803 g or less is approximately 1.75%.

### Conclusion (Next Part of the Question):
Since the probability is **below 5%**, it suggests that this outcome is quite **unlikely to have occurred by chance**. Therefore, it would be reasonable to conclude that the new product likely has **lower nicotine** content.

**Answer: Yes. The probability of this data is unlikely to have occurred by chance alone.**

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Would you like any further details or clarifications?

### Related Questions:
1. How does increasing the sample size affect the standard error of the mean?
2. What is the relationship between Z-scores and percentiles in the standard normal distribution?
3. How would the result change if the standard deviation of the nicotine content were larger?
4. What would happen if the sample mean were closer to the population mean?
5. Why is the Central Limit Theorem important in this type of problem?

**Tip:** When working with probabilities, always consider how extreme or unusual the result is based on the significance level (usually 5%).

The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.921 g and a standard deviation of 0.307 g. The company claims that it has now reduced the amount of nicotine. A sample of 30 cigarettes has a mean nicotine amount of 0.803 g. Assuming that the given mean and standard deviation have not changed, find the probability of randomly selecting 30 cigarettes with a mean of 0.803 g or less.

Learn how to calculate the probability of obtaining a sample mean of 0.803 g of nicotine or less in a normally distributed sample of 30 cigarettes. Understand key concepts like standard error, Z-scores, and the Central Limit Theorem. Suitable for college-level statistics.

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