Question

The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.977 g and a standard deviation of 0.313 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 . 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 as numbers. Your answers should be accurate to 4 decimal places.

Answer 1

For , middle will range from /2 and 1 - /2 percentiles.

The middle 50% ( = 0.5) will range from 25% percentile to 75% percentile.

z value for 25% percentile (p = 0.25) is -0.6745

Similarly, z value for 75% percentile (p = 0.75) is 0.6745

Amounts of nicotine at 25% percentile = 0.977 - 0.6745 * 0.313 = 0.7659 g

Amounts of nicotine at 75% percentile = 0.977 + 0.6745 * 0.313 = 1.1881 g

So, the middle 50% of amounts of nicotine in these cigarettes is between 0.7659 g and 1.1881 g

Standard deviation of mean = = = 0.0407

Amounts of nicotine at 25% percentile = 0.977 - 0.6745 * 0.0407 = 0.9495 g

Amounts of nicotine at 75% percentile = 0.977 + 0.6745 * 0.0407 = 1.0044 g

So, the middle 50% of most average amounts of nicotine in these cigarettes is between 0.9495 g and 1.0044 g