Question

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.

Answer 1

The z-score formula is

`Z=(x-\mu)/(\sigma)`

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

Thus, in our case, setting x=0.9

`Z=(0.9-0.975)/(0.294)=-0.2551\ldots`

Using a z-score, table,

`\Rightarrow P(M<0.9g)=0.4013=40.13\%`

The probability is 0.4013

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

On the other hand,

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

Then,

`\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}`

In that case, the probability is 0.0359