Question

The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.917 g and a standard deviation of 0.303 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 0.872 g. Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly seleting 37 cigarettes with a mean of 0.872 g or less.

Answer 1

`z = (0.872-0.917)/((0.303)/(√(37)))= -0.903`

And we can find this probability to find the answer:

`P(z<-0.903)`

And using the normal standar table or excel we got:

`P(z<-0.903)=0.1833`

Explanation:

Let X the random variable that represent the amounts of nocatine of a population, and for this case we know the distribution for X is given by:

`X \sim N(0.917,0.303)`

Where
`\mu=0.917` and
`\sigma=0.303`

We have the following info from a sample of n =37:

`\bar X= 0.872` the sample mean

And we want to find the following probability:

`P(\bar X \leq 0.872)`

And we can use the z score formula given by;

`z=(x-\mu)/((\sigma)/(√(n)))`

And if we find the z score for the value of 0.872 we got:

`z = (0.872-0.917)/((0.303)/(√(37)))= -0.903`

And we can find this probability to find the answer:

`P(z<-0.903)`

And using the normal standar table or excel we got:

`P(z<-0.903)=0.1833`