Question

The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.972 g and a standard deviation of 0.289 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 36 cigarettes with a mean nicotine amount of 0.929 g. Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly selecting 36 cigarettes with a mean of 0.929 g or less. P( ¯ x < 0.929 g) =

Answer 1

P( ¯ x < 0.929 g) = 0.1867

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 random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sample means with size n of at least 30 can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`

In this problem, we have that:

`\mu = 0.972, \sigma = 0.289, n = 36, s = (0.289)/(√(6)) = 0.0482`

Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly selecting 36 cigarettes with a mean of 0.929 g or less. P( ¯ x < 0.929 g) =

This is the pvalue of Z when X = 0.929. So

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

By the Central Limit Theorem

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

`Z = (0.929 - 0.972)/(0.0482)`

`Z = -0.89`

`Z = -0.89` has a pvalue of 0.1867. So

P( ¯ x < 0.929 g) = 0.1867