The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.904 g and a standard deviation of 0.295 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)

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the amounts of nicotine in a certain brand of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(0.904,0.295)[/tex]

Where [tex]\mu=0.904[/tex] and [tex]\sigma=0.295[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

We want to find a two values that accumulates the middel 50% of the data. So we need to find a quantile who accumulates (1-0.5)/2 = 0.25 of the area on each tail and we got:

[tex] z = \pm 0.674[/tex]

Since P(Z<-0.674) = 0.25 and P(Z>0.674) =0.25. And using this value we can find the possible values of X

[tex] -0.674 = \frac{X -0.904}{0.295}[/tex]

[tex] X= 0.904 -0.674*0.295 = 0.705[/tex]

[tex] 0.674 = \frac{X -0.904}{0.295}[/tex]

[tex] X= 0.904 +0.674*0.295 = 1.103[/tex]

So we expect about 50% of the middle values within 0.705 and 1.103 g