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.

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dfbustos dfbustos · Answer 1 · 2020-06-07T02:58:55+02:00

Answer:

[tex] z = \frac{0.872-0.917}{\frac{0.303}{\sqrt{37}}}= -0.903[/tex]

And we can find this probability to find the answer:

[tex] P(z<-0.903)[/tex]

And using the normal standar table or excel we got:

[tex] P(z<-0.903)=0.1833 [/tex]

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:

[tex]X \sim N(0.917,0.303)[/tex]

Where [tex]\mu=0.917[/tex] and [tex]\sigma=0.303[/tex]

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

[tex] \bar X= 0.872[/tex] the sample mean

And we want to find the following probability:

[tex] P(\bar X \leq 0.872)[/tex]

And we can use the z score formula given by;

[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

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

[tex] z = \frac{0.872-0.917}{\frac{0.303}{\sqrt{37}}}= -0.903[/tex]

And we can find this probability to find the answer:

[tex] P(z<-0.903)[/tex]

And using the normal standar table or excel we got:

[tex] P(z<-0.903)=0.1833 [/tex]