Question

The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.919 g and a standard deviation of 0.327 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 50% 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 59 from this population, in what range would you expect to find the middle 50% 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

z value at 50% = +/- 0.67

z = (x - mean)/sigma
-0.67 = (x - 0.919)/0.327
x = 0.327 * -0.67 + 0.919
x = 0.6999

z = (x - mean)/sigma
0.67 = (x - 0.919)/0.327
x = 0.327 * 0.67 + 0.919
x = 1.1381

between 0.6999 and 1.1381

for n = 59

z value at 50% = +/- 0.67

z = (x - mean)/sigma
-0.67 = (x - 0.919)/0.0426
x = 0.0426 * -0.67 + 0.919
x = 0.8905

z = (x - mean)/sigma
0.67 = (x - 0.919)/0.0426
x = 0.0426 * 0.67 + 0.919
x = 0.9475

between 0.8905 and 0.9475