Question

The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.904 g and a standard deviation of 0.299 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 50 cigarettes with a mean nicotine amount of 0.815 g.Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly selecting 50 cigarettes with a mean of 0.815 g or less. P(M < 0.815 g) =

Answer 1

The Solution:

The formula for Z score is

$Z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt[]{n}}}$

In this case,

$\begin{gathered} \bar{x}=sample\text{ mean=0.815}g \\ \mu=population\text{ mean=0.904g} \\ \sigma=\text{ standard deviation=0.299g} \\ n=\text{ number of cigarettes=50} \end{gathered}$

Substituting these values in the formula, we get the Z-score.

$Z=\frac{\bar{0.815}-0.904}{\frac{0.299}{\sqrt[]{50}}}=\frac{\bar{0.815}-0.904}{\frac{0.299}{\sqrt[]{50}}}=\frac{-0.089}{0.0422849855}=-2.104766$

From the Z-distribution tables, we have

$P(Z<-2.104766)=0.017656$

Thus, the probability of randomly selecting 50 cigarettes with a mean of 0.815g or less is 0.017656.