Chapter 7.3, Problem 57E

(a)

To determine

To find: the probability that a randomly selected bottle contains less than 295 ml.

Answer to Problem 57E

15.87%

Explanation of Solution

Given:

  $\begin{array}{c}\mu =298\\ \sigma =3\\ x=295\end{array}$

Formula used:

  $z=\frac{x-\mu }{\sigma }$

Calculation:

The z-score is

  $\begin{array}{c}z=\frac{x-\mu }{\sigma }\\ =\frac{295-298}{3}\\ =-1.00\end{array}$

The associating probability using the normal probability $P\left(Z<-1.00\right)$ is given in the row starting with $-1.0$ and in the column starting with $.00$ of the standard normal probability

  $P\left(x<295\right)=P\left(Z<-1.00\right)=0.1587=15.87\%$

(b)

To determine

To Calculate: the probability that the mean contents of six randomly selected bottles is less than 295 ml.

Answer to Problem 57E

0.71%

Explanation of Solution

Given:

  $\begin{array}{c}\mu =298\\ \sigma =3\\ n=6\\ x=295\end{array}$

Formula used:

  $z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}$

Calculation:

Since the population distribution is normal, the sampling distribution of the sample mean $\overline{x}$ is also normal.

The z-score is

  $\begin{array}{c}z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}\\ =\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}\\ =\frac{295-298}{3/\sqrt{6}}\\ =-2.45\end{array}$

The associating probability using the normal probability $P\left(Z<-2.45\right)$ is given in the row starting with $-2.4$ and in the column starting with $.05$ of the standard normal probability $P\left(\overline{X}<295\right)=P\left(Z<-2.45\right)=0.0071=0.71\%$