The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 8.4 mg and a standard deviation of 1.4 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 8.078 mg.

Assuming that the given mean and standard deviation **have NOT changed**, find the probability of randomly seleting 37 cigarettes with a mean of 8.078 mg **or less**.
P(*M* < 8.078 mg) = 

Based on the result above, is it valid to claim that the amount of nicotine is lower?


No. The probability of obtaining this data is high enough to have been a chance occurrence.

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

To find the probability of randomly selecting 37 cigarettes with a mean nicotine amount of 8.078 mg or less, we can use the properties of the normal distribution. 

1. **Calculate the standard error (SE)** of the sample mean:
   \[
   SE = \frac{\sigma}{\sqrt{n}} = \frac{1.4 \text{ mg}}{\sqrt{37}} \approx \frac{1.4}{6.08} \approx 0.2305 \text{ mg}
   \]

2. **Calculate the z-score** for the sample mean of 8.078 mg:
   \[
   z = \frac{\bar{x} - \mu}{SE} = \frac{8.078 - 8.4}{0.2305} \approx \frac{-0.322}{0.2305} \approx -1.396
   \]

3. **Find the probability** associated with this z-score using the standard normal distribution table:
   - A z-score of -1.396 corresponds to a cumulative probability of approximately 0.0811.

Thus, the probability of randomly selecting 37 cigarettes with a mean nicotine amount of 8.078 mg or less is approximately:
\[
P(M < 8.078 \text{ mg}) \approx 0.0811
\]

### Conclusion
Since the probability of obtaining a sample mean of 8.078 mg or less is about 0.0811 (or 8.11%), which is relatively high, we would conclude:

**No. The probability of obtaining this data is high enough to have been a chance occurrence.**

### Additional Questions
1. How does the sample size affect the standard error of the mean?
2. What is the significance level typically used to determine statistical significance?
3. How would the result change if the mean nicotine level were even lower?
4. What assumptions are made when using the normal distribution for this analysis?
5. How can you conduct a hypothesis test for this situation?

### Tip
Always ensure the conditions for using a normal distribution are met, especially regarding sample size and population distribution.

The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 8.4 mg and a standard deviation of 1.4 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 8.078 mg.

Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly selecting 37 cigarettes with a mean of 8.078 mg or less.
P(M < 8.078 mg) = 

Based on the result above, is it valid to claim that the amount of nicotine is lower?

No. The probability of obtaining this data is high enough to have been a chance occurrence.

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

Calculate the probability of selecting 37 cigarettes with a mean nicotine amount of 8.078 mg or less, assuming the mean and standard deviation haven't changed. This example demonstrates the application of the normal distribution and z-scores in statistics. Suitable for college-level statistics students.

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