The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.961 g and a standard deviation of 0.296 g . The company that produces these cigarettes claims that it has now reduced the amount of nicotine. The supporting evidence consists of a sample of 49 cigarettes with a mean nicotine amount of 0.881 g . Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly seleting 49 cigarettes with a mean of 0.881 g or less. \( \mathrm{P}(M<0.881 \mathrm{~g})=\square \) Enter your answer as a number accurate to 4 decimal places. NOTE: Answers obtained using exact \( Z \)-scores or \( z \)-scores rounded to 3 decimal places are accepted. Assume that any probability below \( 5 \% \) is sufficient to conclude that the new product really have lower nicotine. 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 solve this problem, we can use the concept of the sampling distribution of the sample mean. We know that the population mean (μ) is 0.961 g and the standard deviation (σ) is 0.296 g. For a sample size (n) of 49, the standard error (SE) is calculated using the formula: \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{0.296}{\sqrt{49}} = \frac{0.296}{7} = 0.0423\text{ g } \] Now, we want to find the probability \( P(M < 0.881) \). First, we need to calculate the Z-score: \[ Z = \frac{M - \mu}{SE} = \frac{0.881 - 0.961}{0.0423} = \frac{-0.080}{0.0423} \approx -1.89 \] Next, we can find the probability corresponding to a Z-score of -1.89 using the standard normal distribution table or a calculator: \[ P(Z < -1.89) \approx 0.0294 \] Thus, \( P(M < 0.881 \text{ g}) \approx 0.0294 \). Given that this probability is less than 0.05, we can conclude that the data suggests the nicotine amount is likely lower than claimed. So the final answer is: \( \mathrm{P}(M<0.881 \mathrm{~g})=0.0294 \) Now, when considering the claims regarding the nicotine content, since the probability of obtaining this sample mean by chance is low (below 5%), it is valid to claim that the amount of nicotine is lower.