The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.975 g and a standard deviation of 0.294 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 50 cigarettes with a mean nicotine amount of 0.9 g.Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly seleting 50 cigarettes with a mean of 0.9 g or less. P(M < 0.9 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.

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IssakC259354go IssakC259354go · Answer 1

The z-score formula is

[tex]Z=\frac{x-\mu}{\sigma}[/tex]

Where μ is the mean and σ is the standard deviation.

Thus, in our case, setting x=0.9

[tex]Z=\frac{0.9-0.975}{0.294}=-0.2551\ldots[/tex]

Using a z-score, table,

[tex]\Rightarrow P(M<0.9g)=0.4013=40.13\%[/tex]

The probability is 0.4013

Since the sample size is greater than 30, we can use the first formula above.

On the other hand,

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

Then,

[tex]\begin{gathered} \Rightarrow Z=\frac{0.9-0.975}{\frac{0.294}{\sqrt[]{50}}}=-1.80384\ldots \\ \Rightarrow P(M<0.9)=0.0359 \end{gathered}[/tex]

In that case, the probability is 0.0359