Question

The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.963 g and a standard deviation of 0.301 g. Find the probability of randomly selecting a cigarette with 0.572 g of nicotine or less.
P(X < 0.572 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

To find the probability of randomly selecting a cigarette with 0.572 g of nicotine or less, we can use the Z-score formula. The Z-score is calculated using the following formula:

\[
Z = \frac{(X - \mu)}{\sigma}
\]

where:
- \( X \) is the value we are interested in (0.572 g),
- \( \mu \) is the mean (0.963 g), and
- \( \sigma \) is the standard deviation (0.301 g).

Substituting the given values:

\[
Z = \frac{(0.572 - 0.963)}{0.301} = \frac{-0.391}{0.301} \approx -1.296
\]

Next, we need to find the probability corresponding to the Z-score of -1.296. We can look this value up in the standard normal distribution table or use a calculator or software that provides cumulative probabilities for the normal distribution.

Using standard normal distribution tables or a calculator:

- The cumulative probability \( P(Z < -1.296) \) is approximately 0.0974.

Thus, the probability of randomly selecting a cigarette with 0.572 g of nicotine or less is:

\[
P(X < 0.572) \approx 0.0974
\]

Finally, rounding to four decimal places, we get:

\[
P(X < 0.572 g) = 0.0974
\]