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The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.902 g and a standard deviation of 0.316 g. The company that produces these cigarettes claims that it has now reduced the amount of nicotine. The supporting evidence consists of a sample of 41 cigarettes with a mean nicotine amount of 0.823 g.Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly seleting 41 cigarettes with a mean of 0.823 g or less. P(M < 0.823 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.

Sagot :

Let m be a random variable representing the amounts of nicotine in a certain brand of cigarette. Given that x is normally distributed, we would calculate the z score by applying the formula given below

[tex]z\text{ = }\frac{x\text{ - }\mu}{\frac{\sigma}{\sqrt[]{n}}}[/tex]

where

x = sample mean

σ = population mean

n = sample size

μ = population mean

From the information given,

μ = 0.316

x = 0.823

n = 41

σ = 0.902

Bu substituting these values into the formula,

[tex]\begin{gathered} z\text{ = }\frac{0.823\text{ - 0.902}}{\frac{0.316}{\sqrt[]{41}}} \\ z\text{ = - 1.6}01 \end{gathered}[/tex]

We want to find P(m < 0.823)

We would do this by finding the probability value corresponding to a z score of - 1.601 from the standard normal distribution table. From the table,

P(m < 0.823) = 0.054

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