Question

The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.894 g and a standard deviation of 0.302 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 60% 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 54 from this population, in what range would you expect to find the middle 60% 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

Solution :

Given that ,

mean = $\mu$ = 0.894

standard deviation = $\sigma$ = 0.302

n = 84

$\mu$_{$\bar x$} = 0.894

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n = 0.302/ $\sqrt$ 84 = 0.0330

middle 60 of score is

P(-z < Z < z) = 0.60

P(Z < z) - P(Z < -z) = 0.60

2 P(Z < z) - 1 = 0.60

2 P(Z < z) = 1 + 0.60 = 1.60

P(Z < z) = 1.60 / 2 = 0.8

P(Z <0.84 ) = 0.8

z  ±0.84

Using z-score formula  

x= z * $\sigma$ + $\mu$

x= - 0.84 *0.0330+0.894

x= 0.8663

z = 0.84

Using z-score formula  

x= z * $\sigma$ + $\mu$

x= 0.84*0.0330+0.894

x= 0.9217

smaller value =0.8663 larger value = 0.9217