Question

The nicotine content in a single cigarette of a particular brand has a distribution with mean 0.4 mg and standard deviation 0.1 mg. If 100 randomly selected cigarettes of this brand are analyzed, what is the probability that the resulting sample mean nicotine content will be less than 0.38

Answer 1

Answer:

Step-by-step explanation:

For this case we can define the random variable of interest as: "The nicotine content in a single cigarette " and for this case we know the following parameters:

$\mu = 0.4, \sigma = 0.1$

And for this case we select a sample size of n =100 and we want to find the following probability:

$P(\bar X <0.38)$

And for this case we can use the z score formula given by:

$z =\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$z= \frac{0.38-0.4}{\frac{0.1}{\sqrt{100}}}= -2$

And we can find the required probability with the normal standard table and we got:

$P(z<-2) = 0.0228$