229k views
5 votes
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) =

User Celso Agra
by
7.4k points

1 Answer

3 votes

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}}}=(-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.

User Anthony Sneed
by
9.0k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories