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.

And we can find this probability to find the

And using the normal standar table or excel we got:

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:

Where and

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

the sample mean

And we want to find the following probability:

And we can use the z score formula given by;

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

And we can find this probability to find the

And using the normal standar table or excel we got:

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step-by-step explanation:

0.85 or 85%

step-by-step explanation:

1) divide 23 by 27

2) round the decimal to the nearest 10

3) so your answer would be 0.85 or 85%