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.

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

Step-by-step 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 and standard deviation , the zscore of a measure X is given by:

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 and standard deviation , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard deviation .

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

Sample of 32 filtered cigarettes

This means that

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

By the Central Limit Theorem

has a pvalue of 0.0526

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

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