Question

The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.954 grams and a standard deviation of 0.292 grams. Find the probability of randomly selecting a cigarette with 0.37 grams of nicotine or less. Round your answer to four decima

Answer 1

Let X be the random variable representing the amount (in grams) of nicotine contained in a randomly chosen cigarette.

P(X ≤ 0.37) = P((X - 0.954)/0.292 ≤ (0.37 - 0.954)/0.292) = P(Z ≤ -2)

where Z follows the standard normal distribution with mean 0 and standard deviation 1. (We just transform X to Z using the rule Z = (X - mean(X))/sd(X).)

Given the required precision for this probability, you should consult a calculator or appropriate z-score table. You would find that

P(Z ≤ -2) ≈ 0.0228

You can also estimate this probabilty using the empirical or 68-95-99.7 rule, which says that approximately 95% of any normal distribution lies within 2 standard deviations of the mean. This is to say,

P(-2 ≤ Z ≤ 2) ≈ 0.95

which means

P(Z ≤ -2 or Z ≥ 2) ≈ 1 - 0.95 = 0.05

The normal distribution is symmetric, so this means

P(Z ≤ -2) ≈ 1/2 × 0.05 = 0.025

which is indeed pretty close to what we found earlier.