The amounts of nicotine in a certain brand of cigarette are normally distributed

Question

The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.961 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 43 cigarettes with a mean nicotine amount of 0.902 g.

Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly seleting 43 cigarettes with a mean of 0.902 g or less. P(M < 0.902 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.

Explanation / Answer

Mean is 0.961 and s is 0.321. N is 43 and therefore standard error SE is s/sqrt(N)=0.321/sqrt(43)=0.049

P(x<0.902)=P(z<(0.902-0.961)/0.049)=P(z<-1.20) or 1-P(z<1.2). from normal distribution table we get 1-0.8849= 0.1151

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