Question

The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.895 g and a standard deviation of 0.292 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 37 cigarettes with a mean nicotine amount of 0.809 g. Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly seleting 37 cigarettes with a mean of 0.809 g or less.

Answer 1

3.67% probability of randomly seleting 37 cigarettes with a mean of 0.809 g or less.

Explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

`\mu = 0.895, \sigma = 0.292, n = 37, s = (0.292)/(√(37)) = 0.048`

Find the probability of randomly seleting 37 cigarettes with a mean of 0.809 g or less.

This is the pvalue of Z when X = 0.809.

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (0.809 - 0.895)/(0.048)`

`Z = -1.79`

`Z = -1.79` has a pvalue of 0.0367

3.67% probability of randomly seleting 37 cigarettes with a mean of 0.809 g or less.