Question

The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.918 g and a standard deviation of 0.285 g. Find the probability of randomly selecting a cigarette with 0.662 g of nicotine or less.
P(X < 0.662 g) =  
Enter your answer as a number accurate to 4 decimal places. NOTE: Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer 1

Let X denotes the amounts of nicotine in a certain brand of cigarette.

$X \sim N(0.918,{0.285^2})$  

Since $\mu = 0.918g{\text{ and }}\sigma = 0.285g$

The probability of randomly selecting a cigarette with 0.662 g of nicotine or less is:

$P(X < 0.662) = P\left( {\frac{{X - 0.918}}{{0.285}} < \frac{{0.662 - 0.918}}{{0.285}}} \right)$

  $= P(Z < - 0.898)$

$= 1 - P(Z < 0.898)$

$= 1 - 0.8154{\text{ (Using z - tables)}}$

  $= 0.1846$.

Hence, the required probability is 0.1846.