Question

The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.933g and a standard deviation of 0.301g. The company claims it has reduced the amount of nicotine. Given a sample of 43 cigarettes with a mean nicotine amount of 0.869g, find the probability of randomly selecting 43 cigarettes with a mean of 0.869g or less, P(M<0.869g).

(a) 0.192
(b) 0.308
(c) 0.692
(d) 0.808

Answer 1

To find the probability of randomly selecting 43 cigarettes with a mean nicotine amount of 0.869g or less (P(M<0.869g)), we need to calculate the standard error of the mean (SEM) and use it to calculate the z-score. With the z-score, we can then find the probability using a z-table or a calculator.

Step-by-step explanation:

To find the probability of randomly selecting 43 cigarettes with a mean nicotine amount of 0.869g or less (P(M<0.869g)), we will use the concept of sampling distribution. The sampling distribution of means follows a normal distribution.

1. First, we need to calculate the standard deviation of the sampling distribution, known as the standard error of the mean (SEM). The formula for SEM is SEM = standard deviation / √sample size.
2. Next, we calculate the z-score, which is the number of standard deviations a particular observation is from the mean. The formula for z-score is z = (sample mean - population mean) / SEM.
3. Finally, we use the z-score to find the probability using a z-table or a calculator.

Let's calculate the probability:

1. SEM = 0.301g / √43 ≈ 0.046g
2. z = (0.869g - 0.933g) / 0.046g ≈ -1.391
3. Using a z-table or a calculator, we find that the probability of randomly selecting 43 cigarettes with a mean of 0.869g or less is approximately 0.165 or 16.5%