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 0.882 g and a standard deviation of 0.299 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 45 cigarettes with a mean nicotine amount of 0.793 g. Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly selecting 45 cigarettes with a mean of 0.793 g or less. P( ¯¯¯ X X¯ < 0.793 g) = ---------------------- Round to 4 decimal places. NOTE: Answers obtained using exact z-scores or z-scores rounded to 2 decimal places are accepted.

Explanation / Answer

here std error of mean =std deviation/(n)1/2 =0.0446

hence P(Xbar<0.793)=P(Z<(0.793-0.882)/0.0446)=P(Z<-1.9968)=0.0229

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