Question

The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.974 g and a standard deviation of 0.325 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 48 cigarettes with a mean nicotine amount of 0.918 g. Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly seleting 32 cigarettes with a mean of 0.917 g or less.

P(M < 0.917 g) = __________

Answer 1

P(M < 0.917 g) = 0.1611.

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 z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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.

Mean of 0.974 g and a standard deviation of 0.325 g.

This means that
`\mu = 0.974, \sigma = 0.325`

Sample of 32:

This means that
`n = 32, s = (0.325)/(√(32))`

Fnd the probability of randomly selecting 32 cigarettes with a mean of 0.917 g or less.

This is the p-value of Z when X = 0.917. So

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

By the Central Limit Theorem

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

`Z = (0.917 - 0.974)/((0.325)/(√(32)))`

`Z = -0.99`

`Z = -0.99` has a p-value of 0.1611.

So

P(M < 0.917 g) = 0.1611.