Question

In a study of the health effects of cigarettes, a random sample of 32 filtered cigarettes

was obtained and the tar content was measured. The sample has a mean of 19.2 mg.
Given that the tar content of cigarettes have a mean of 20.1 mg and a standard
deviation of 3.15 mg, what is the probability that 32 filtered cigarettes have a mean of
19.2 mg or less?
Round answer to 4 decimal places.

Answer 1

0.0526 = 5.26% probability that 32 filtered cigarettes have a mean of 19.2 mg 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 normal distributions can be 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.

Given that the tar content of cigarettes have a mean of 20.1 mg and a standard deviation of 3.15 mg

This means that
`\mu = 20.1, \sigma = 3.15`

Sample of 32 filtered cigarettes

This means that
`n = 32, s = (3.15)/(√(32)) = 0.5568`

What is the probability that 32 filtered cigarettes have a mean of 19.2 mg or less?

This is the pvalue of Z when X = 19.2. So

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

By the Central Limit Theorem

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

`Z = (19.2 - 20.1)/(0.5568)`

`Z = -1.62`

`Z = -1.62` has a pvalue of 0.0526

0.0526 = 5.26% probability that 32 filtered cigarettes have a mean of 19.2 mg or less.