Question

the amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.904 grams and a standard deviation of 0.299 grams. find the probability of randomly selecting a cigarette with 0.515 grams of nicotine or less. round your answer to four decimals.

$p(x < 0.515) =$

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Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 0.515\), \(\mu=0.904\), and \(\sigma = 0.299\).

$$ z=\frac{0.515 - 0.904}{0.299}=\frac{-0.389}{0.299}\approx - 1.30 $$

Step2: Find the probability using the standard normal distribution table

We want to find \(P(X\lt0.515)\), which is equivalent to \(P(Z\lt - 1.30)\) using the standard normal distribution (\(Z\) - distribution).
Looking up the value of \(P(Z\lt - 1.30)\) in the standard normal table, we get \(P(Z\lt - 1.30)=0.0968\)

Answer:

\(0.0968\)