Question

The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.976 g and a standard deviation of 0.321 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 35 cigarettes with a mean nicotine amount of 0.916 g.Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly selecting 35 cigarettes with a mean of 0.916 g or less.P(M < 0.916 g) = Enter your answer as a number accurate to 4 decimal places. NOTE: Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.Based on the result above, is it valid to claim that the amount of nicotine is lower?No. The probability of obtaining this data is high enough to have been a chance occurrence.Yes. The probability of this data is unlikely to have occurred by chance alone.

Answer 1

We get the z score to be;

`Z=(√(n)(x-\mu))/(\sigma)`

Inserting the values, we have;

`\begin{gathered} z-(√(35)(0.916-0.976))/(0.321) \\ \\ z=-1.1058 \end{gathered}`

Using a z score table, we have;

`\begin{gathered} P(\mu<0.916)=P(z<-1.1058) \\ \\ \\ =0.13441 \end{gathered}`

Our judgement for this claim is;

Yes. The probability of this data is unlikely to have occurred by chance alone.