Question

The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.939 g and a

standard deviation of 0.307 g. Find the probability of randomly selecting a cigarette with 0.417 g of nicotine
or less.
P(X < 0.417 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

Using the formula to find the Z-score of the distribu,

`\begin{gathered} Z=(x-\mu)/(\sigma) \\ \text{where x = 0.417g} \\ \mu=0.939g \\ \sigma=0.307g \\ \text{Substituting all these given data in the formula, we have;} \\ Z=(0.417-0.939)/(0.307) \\ Z=(-0.522)/(0.307) \\ Z=-1.7003 \\ Z=-1.700 \end{gathered}`

The probability of selecting 0.417g of nicotine or less will therefore be;

`\begin{gathered} P(x\leq Z)=P(x\leq-1.7) \\ \text{From Z-score table, P(x}\leq-1.7)\text{ = 0.044565} \\ P(x\leq Z)=0.0446 \end{gathered}`

Therefore, the probability of selecting 0.417g of nicotine or less to 4 decimal places is 0.0446