Chapter 8, Problem 32CR

The nicotine content in a single cigarette of a particular brand has a distribution with mean 0.8 mg and standard deviation 0.1 mg. If 100 randomly selected cigarettes of this brand are analyzed, what is the probability that the resulting sample mean nicotine content will be less than 0. 79? less than 0.77?

To determine

Find the probability that the resulting sample mean nicotine content will be less than 0.79.

Find the probability that the resulting sample mean nicotine content will be less than 0.77.

Answer to Problem 32CR

The probability that the resulting sample mean nicotine content will be less than 0.79 is 0.1587.

The probability that the resulting sample mean nicotine content will be less than 0.77 is 0.0013.

Explanation of Solution

Calculation:

It is given that, a random sample of 100 cigarettes is selected from a population of mean and standard deviation of $\mu =0.8\text{mg}$ and $\sigma =16.5\text{mg}$, respectively.

Assume that a random variable X is drawn from a population with mean $\mu$ and standard deviation $\sigma$. Now, the sample mean $\overline{x}$ for sample of size n has mean ${\mu }_{\overline{x}}\left(=\mu \right)$ and standard deviation ${\sigma }_{\overline{x}}\left(=\frac{\sigma }{\sqrt{n}}\right)$.

Thus, for $n=100$ the mean of $\overline{x}$ is,

$\begin{array}{c}{\mu }_{\overline{x}}=\mu \\ =0.8\end{array}$

Thus, for $n=100$ the standard deviation of $\overline{x}$ is,

$\begin{array}{c}{\sigma }_{\overline{x}}=\frac{0.1}{\sqrt{100}}\\ =\frac{0.1}{10}\\ =0.01\end{array}$

The z-score of sampling mean $\overline{x}$ is defined as $z=\frac{\overline{x}-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}$.

It is known that, for a sample size equal to or more than 30, the sampling distribution would follows normal distribution approximately.

Thus, as the sample size is 100$\left(>30\right)$, then the distribution of $\overline{x}$ will follow approximately normal distribution.

The z-score of sampling mean $\overline{x}$ is defined as $z=\frac{\overline{x}-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}$.

Now, the probability that the resulting sample mean nicotine content will be less than 0.79, implies that, $P\left(\overline{x}<0.79\right)$.

Thus,

$\begin{array}{c}P\left(\overline{x}<0.79\right)=P\left(\frac{\overline{x}-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}<\frac{0.79-0.8}{0.01}\right)\\ =P\left(z<\frac{-0.01}{0.01}\right)\\ =P\left(z<-1\right)\end{array}$

The step by step procedures to obtain $P\left(z<-1\right)$ from the table “Standard Normal Probabilities (Cumulative z Curve Areas)” are given below:

• Consider the 1^st column of .00 from the table.
• Consider the row of –1.0 from the table.
• Choose the intersection value of 1^st column and row of –1.0.

Hence, $P\left(z<-1\right)=0.1587$.

Thus, $P\left(\overline{x}<0.79\right)=0.1587$

Therefore, the probability that the resulting sample mean nicotine content will be less than 0.79 is 0.1587.

Now, the probability that the resulting sample mean nicotine content will be less than 0.77, implies that, $P\left(\overline{x}<0.77\right)$.

Thus,

$\begin{array}{c}P\left(\overline{x}<0.77\right)=P\left(\frac{\overline{x}-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}<\frac{0.77-0.8}{0.01}\right)\\ =P\left(z<\frac{-0.03}{0.01}\right)\\ =P\left(z<-3\right)\end{array}$

The step by step procedures to obtain $P\left(z<-3\right)$ from the table “Standard Normal Probabilities (Cumulative z Curve Areas)” are given below:

• Consider the 1^st column of .00 from the table.
• Consider the row of –3.0 from the table.
• Choose the intersection value of 1^st column and row of –3.0.

Hence, $P\left(z<-3\right)=0.0013$.

Thus,

$P\left(\overline{x}<0.77\right)=0.0013$

Therefore, the probability that the resulting sample mean nicotine content will be less than 0.77 is 0.0013.