Homework W8 Score: $8 / 10$ $4 / 5$ answered Question 3 The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of $0.952 \mathrm{~g}$ and a standard deviation of $0.324 \mathrm{~g}$. Find the probability of randomly selecting a cigarette with $0.369 \mathrm{~g}$ of nicotine or less. \[ \mathrm{P}(X<0.369 \mathrm{~g})= \] 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 accested. Question Help: $\square$ Video Submit Question

Step 1 ：Given the mean \(\mu = 0.952\) g, the standard deviation \(\sigma = 0.324\) g, and the value from the dataset \(X = 0.369\) g.

Step 2 ：We calculate the z-score using the formula: \(Z = \frac{X - \mu}{\sigma}\).

Step 3 ：Substitute the given values into the formula: \(Z = \frac{0.369 - 0.952}{0.324} = -1.798\).

Step 4 ：The z-score of -1.798 represents how many standard deviations the value 0.369 g is from the mean.

Step 5 ：We need to find the probability that the z-score is less than -1.798. This is equivalent to finding the area to the left of -1.798 under the standard normal curve.

Step 6 ：We can look up this value in a standard normal (Z) table, or use a calculator or software that can calculate it.

Step 7 ：The probability that a z-score is less than -1.798 is approximately 0.0362.

Step 8 ：So, the probability of randomly selecting a cigarette with 0.369 g of nicotine or less is 0.0362, or 3.62% when expressed as a percentage.

Step 9 ：\(\boxed{0.0362}\) or \(\boxed{3.62\%}\) is the final answer.