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The nicotine content in a single cigarette of a particular brand has a distribution with mean \(0.8 \mathrm{mg}\) and standard deviation \(0.1 \mathrm{mg}\). If 100 of these cigarettes are analyzed, what is the probability that the resulting sample mean nicotine content will be less than \(0.79 ?\) less than \(0.77\) ?

Short Answer

Expert verified
The probability that the resulting sample mean nicotine content will be less than 0.79 mg is approximately 0.1587. The probability that it will be less than 0.77 mg is approximately 0.0013 or 0.13%.

Step by step solution

01

Identify Parameters of Distribution

It's given that the mean \(\mu\) nicotine content in a single cigarette is 0.8 mg with a standard deviation \(\sigma = 0.1 mg\). The sample size \(n = 100\).
02

Calculate Standard Error of Mean

The standard error of the mean can be calculated using the formula \(SE = \frac{\sigma}{\sqrt{n}}\). Substituting the known values in the formula, we get \(SE = \frac{0.1}{\sqrt{100}} = 0.01 mg.\)
03

Calculate Z-Scores

Now we'll calculate Z-Score for the given nicotine values using the formula \(Z = \frac{X - \mu}{SE}\). For \(X = 0.79 mg\), the Z-Score is \(Z = \frac{0.79 - 0.8}{0.01} = -1\). Similarly, for \(X = 0.77 mg\), the Z-Score is \(Z = \frac{0.77 - 0.8}{0.01} = -3\).
04

Find Probability from Z-Scores

Looking up these Z-Scores in the Z-Table, we can find the probabilities associated with these Z-Scores. The probability \(P(X < 0.79 mg)\) is equivalent to \(P(Z < -1)\), and from the table this is approximately 0.1587. The probability \(P(X < 0.77 mg)\) is equivalent to \(P(Z < -3)\), and from the table this is approximately 0.0013 or 0.13%

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