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

Answer:

3.67% probability of randomly seleting 37 cigarettes with a mean of 0.809 g 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 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:

$Z = \frac{X - \mu}{\sigma}$

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 $s = \frac{\sigma}{\sqrt{n}}$.

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:

$\mu = 0.895, \sigma = 0.292, n = 37, s = \frac{0.292}{\sqrt{37}} = 0.048$

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.

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.809 - 0.895}{0.048}$

$Z = -1.79$

$Z = -1.79$ has a pvalue of 0.0367

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