3.5 - The Central Limit Theorem Score: $25 / 338 / 9$ answered Question 9 The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of $10 \mathrm{mg}$ and a standard deviation of $1.43 \mathrm{mg}$. The company that produces these cigarettes claims that it has now reduced the amount of nicotine. The supporting evidence consists of a sample of 44 cigarettes with a mean nicotine amount of $9.634 \mathrm{mg}$. Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly seleting 44 cigarettes with a mean of $9.634 \mathrm{mg}$ or less. \[ P(M<9.634 \mathrm{mg})= \] Based on the result above, is it valid to claim that the amount of nicotine is lower? Yes. The probability of this data is unlikely to have occurred by chance alone. No. The probability of obtaining this data is high enough to have been a chance occurrence. Question Help: $\square$ Message instructor $D$ Post to forum Submit Question

Step 1 ：Given the population mean (μ) = 10 mg, population standard deviation (σ) = 1.43 mg, sample size (n) = 44, and sample mean (x̄) = 9.634 mg.

Step 2 ：We first calculate the standard error (SE), which is the standard deviation of the distribution of sample means. The formula for standard error is \(SE = \frac{\sigma}{\sqrt{n}}\).

Step 3 ：Substituting the given values, we get \(SE = \frac{1.43}{\sqrt{44}} \approx 0.215\) mg.

Step 4 ：Next, we calculate the z-score, which measures how many standard errors away our sample mean is from the population mean. The formula for the z-score is \(z = \frac{(x̄ - μ)}{SE}\).

Step 5 ：Substituting the given values, we get \(z = \frac{(9.634 - 10)}{0.215} \approx -1.70\).

Step 6 ：The z-score tells us that our sample mean is approximately 1.70 standard errors below the population mean.

Step 7 ：Finally, we use the standard normal distribution (z-distribution) to find the probability that a z-score is less than -1.70. Looking this up in a z-table or using a calculator, we find \(P(M<9.634 \text{ mg}) = P(Z<-1.70) \approx 0.045\).

Step 8 ：This means there is approximately a 4.5% chance of randomly selecting 44 cigarettes with a mean of 9.634 mg or less, assuming the population mean and standard deviation have not changed.

Step 9 ：Based on this result, it is valid to claim that the amount of nicotine is lower. The probability of obtaining this data is low enough (less than 5%) to suggest that it is unlikely to have occurred by chance alone. Therefore, it is reasonable to conclude that the company has indeed reduced the amount of nicotine in their cigarettes. The final answer is \(\boxed{0.045}\).