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

Answer:

$z = \frac{0.872-0.917}{\frac{0.303}{\sqrt{37}}}= -0.903$

And we can find this probability to find the answer:

$P(z$

And using the normal standar table or excel we got:

$P(z$

Step-by-step 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;

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

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

$z = \frac{0.872-0.917}{\frac{0.303}{\sqrt{37}}}= -0.903$

And we can find this probability to find the answer:

$P(z$

And using the normal standar table or excel we got:

$P(z$