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.

Question

P(X < 0.572 g) =
0.0974
Incorrect
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 (P(X < 0.572 g)), we will first calculate the z-score for 0.572 g using the given mean (μ) and standard deviation (σ) of the normal distribution.

1. Calculate the z-score: \[ z = \frac{X - \mu}{\sigma} \] where:

   • \( X = 0.572 , \text{g} \)
   • \( \mu = 0.963 , \text{g} \)
   • \( \sigma = 0.301 , \text{g} \)

   Substituting the values: \[ z = \frac{0.572 - 0.963}{0.301} = \frac{-0.391}{0.301} \approx -1.297 \]

2. Find the probability corresponding to the z-score: We use a standard normal distribution table or a calculator to find the probability \( P(Z < -1.297) \).

   Looking up \( z = -1.297 \) in the z-table or using a standard normal distribution calculator will yield:

   \( P(Z < -1.297) \approx 0.0968 \)

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

\[ P(X < 0.572) \approx 0.0968 \]

Thus, rounding to four decimal places, the answer is:

\[ \boxed{0.0968} \]