Question

The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.889 g and a standard deviation of 0.307 g. The company that produces these cigarettes claims that it has now reduced the amount of nicotine. In what range would you expect to find the middle 70% of amounts of nicotine in these cigarettes (assuming the mean has not changed)?

Between and . (Enter your answers in ascending order...smaller on left, larger on right. Also, enter your answers accurate to four decimal places.) If you were to draw samples of size 55 from this population, in what range would you expect to find the middle 70% of most average amounts of nicotine in the cigarettes in the sample?

Between and . (Enter your answers in ascending order...smaller on left, larger on right. Also, enter your answers accurate to four decimal places.)

Answer 1

then

From z table

P(-1.04 < z < 1.04) = 0.70

z= -1.04 , X= -1.04* 0.307 + 0.889 = 0.5697

For z = 1.04, X = 1.04* 0.307 + 0.889 = 1.2083

Middle 70 % will lie between 0.6597 and 1.2083

The distribution of sample mean follow Normal

with mean = 0.889 and standard error

Then

From z table

P(-1.04 < z < 1.04) = 0.70

z= -1.04,

when z = 1.04 ,

Middle 70 % will lie between 0.8459 and 0.9321