Question

The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.956 g and a standard deviation of 0.321 g. The company that produces these cigarettes claims that it has now reduced the amount of nicotine.

In what range would you expect to find the middle 95% of amounts of nicotine in these cigarettes (assuming the mean has not changed)?
Between _____ and ____.

If you were to draw samples of size 56 from this population, in what range would you expect to find the middle 95% of most average amounts of nicotine in the cigarettes in the sample?
Between ____ and____ .

Enter your answers as numbers. Your answers should be accurate to 4 decimal places.

Answer 1

To find the middle 95% of amounts of nicotine in the cigarettes, use the Empirical Rule for a normal distribution. The range would be between 0.314 g and 1.598 g. To find the middle 95% of most average amounts of nicotine in a sample of size 56, use the Central Limit Theorem. The range would be between 0.872 g and 1.040 g.

Step-by-step explanation:

To find the range in which we would expect to find the middle 95% of amounts of nicotine in the cigarettes, we can use the Empirical Rule for a normal distribution. According to the Empirical Rule, approximately 95% of the data falls within two standard deviations of the mean. In this case, the mean is 0.956 g and the standard deviation is 0.321 g. So, we can calculate the range as follows:

Lower Range = Mean - (2 * Standard Deviation) = 0.956 - (2 * 0.321) = 0.314 g

Upper Range = Mean + (2 * Standard Deviation) = 0.956 + (2 * 0.321) = 1.598 g

Therefore, we would expect to find the middle 95% of amounts of nicotine in these cigarettes between 0.314 g and 1.598 g.

To find the range in which we would expect to find the middle 95% of most average amounts of nicotine in a sample of size 56 drawn from this population, we can use the Central Limit Theorem. According to the Central Limit Theorem, when the sample size is large enough (typically greater than 30), the distribution of sample means is approximately normal, regardless of the shape of the population distribution.

In this case, the sample mean is still 0.956 g and the standard deviation becomes the standard deviation of the population divided by the square root of the sample size (0.321 / √56 = 0.042 g).

Range of sample means = Mean ± (2 * Standard Deviation)

Lower Range = 0.956 - (2 * 0.042) = 0.872 g

Upper Range = 0.956 + (2 * 0.042) = 1.040 g

Therefore, we would expect to find the middle 95% of most average amounts of nicotine in the cigarettes in a sample of size 56 between 0.872 g and 1.040 g.