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.