Question

The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.932 g and a standard deviation of 0.329 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 30 cigarettes with a mean nicotine amount of 0.878 g. Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly seleting 30 cigarettes with a mean of 0.878 g or less.

Answer 1

The probability of randomly selecting 30 cigarettes with a mean of 0.878 g or less is 0.1177 or 11.77%.

Step-by-step explanation:

We are given that the mean of the nicotine amounts in this certain brand of cigarette is 0.932 g and the standard deviation is 0.329 g. The company claims that the amount of nicotine is reduced and provides a sample of 30 cigarettes with a mean nicotine amount of 0.878 g. We need to find the probability of randomly selecting 30 cigarettes with a mean of 0.878 g or less.

We can use the Central Limit Theorem for this problem. Since we are dealing with a sample size of 30, the distribution of sample means will be approximately normal even if the population distribution is not normal. The mean of the sample means will be equal to the population mean, and the standard deviation of the sample means will be equal to the population standard deviation divided by the square root of the sample size.

First, we find the z-score for the mean nicotine amount of 0.878 g using the formula: z = (x - mu) / (sigma / sqrt(n)). Substituting the given values, we have z = (0.878 - 0.932) / (0.329 / sqrt(30)) = -1.198.

To find the probability of randomly selecting 30 cigarettes with a mean of 0.878 g or less, we need to find the area to the left of the z-score -1.198 in the standard normal distribution. By looking up this z-score in the z-table, we find that the probability is 0.1177 or 11.77%.