Chapter 7, Problem 92CE

The amount of fill in a half-liter (500 ml) soft drink bottle is normally distributed. The process has a standard deviation of 5 ml. The mean is adjustable. (a) Where should the mean be set to ensure a 95 percent probability that a half-liter bottle will not be underfilled? (b) A 99 percent probability? (c) A 99.9 percent probability? Explain.

To determine

Find the mean, so that the probability that a half-liter bottle will not be underfilled is 95%.

Answer to Problem 92CE

Themean, so that the probability that a half-liter bottle will not be underfilled is 95%is 508.225 ml.

Explanation of Solution

Calculation:

The filling of a half-liter or 500 ml bottle of soft drink has a normal distribution with an adjustable mean and a standard deviation of 5 ml.

Normal distribution:

A continuous random variable X is said to follow normal distribution if the probability density function of X is,

$f\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{-\frac{{\left(x-\mu \right)}^2}{2{\sigma }^2}},-\infty <x<\infty$, with mean $\mu$ and standard deviation $\sigma$.

Standard normal distribution:

A continuous random variable Z is said to follow standard normal distribution if the probability density function of X is,

$f\left(x\right)=\frac{1}{\sqrt{2\pi }}{e}^{-\frac{x^2}{2}},-\infty <x<\infty$, with mean 0 and standard deviation 1.

The z-score:

The z-score for a variable X with mean $\mu$ and standard deviation $\sigma$ can be obtained by standardizing X as follows:

$z=\frac{x-\mu }{\sigma }$.

Denote X as the amount of fill in the half-liter bottle. Here, $\sigma =5$.

There must be a 95% chance that the half-liter bottle will not be underfilled. This implies that $P\left(X\ge 500\right)=0.95$.

Now, $P\left(X\ge 500\right)=0.95$ can be rewritten as follows:

$\begin{array}{c}P\left(X\ge 500\right)=0.95\\ 1-P\left(X<500\right)=0.95\\ P\left(X<500\right)=1-0.95\\ =\mathrm{0.05.}\\ P\left(\frac{X-\mu }{\sigma }<\frac{500-\mu }{\sigma }\right)=0.05\\ P\left(Z<\frac{500-\mu }{5}\right)=0.05\\ P\left(Z<z\right)=\mathrm{0.05.}\left[\text{Let }z=\frac{500-\mu }{5}\right]\end{array}$

Variable value:

Software procedure:

Step-by-step software procedure to obtain variable value using EXCEL is as follows:

• • Open an EXCEL file.
• • In cell G1, enter the formula “=NORM.S.INV(0.05)”.
• Output using EXCEL software is given below:

Thus,

$\begin{array}{c}P\left(Z<-1.645\right)=0.05\\ \frac{500-\mu }{5}=-1.645\\ 500-\mu =-8.225\\ \mu =500+8.225\\ =\mathrm{508.225.}\end{array}$

Hence, themean, so that the probability that a half-liter bottle will not be underfilled is 95% is 508.225 ml.

To determine

Find the mean, so that the probability that a half-liter bottle will not be underfilled is 99%.

Answer to Problem 92CE

Themean, so that the probability that a half-liter bottle will not be underfilled is 99% is 511.63 ml.

Explanation of Solution

Calculation:

There must be a 99% chance that the half-liter bottle will not be underfilled. This implies that $P\left(X\ge 500\right)=0.99$.

Now, $P\left(X\ge 500\right)=0.99$ can be rewritten as follows:

$\begin{array}{c}P\left(X\ge 500\right)=0.99\\ 1-P\left(X<500\right)=0.99\\ P\left(X<500\right)=1-0.99\\ =\mathrm{0.01.}\\ P\left(\frac{X-\mu }{\sigma }<\frac{500-\mu }{\sigma }\right)=0.01\\ P\left(Z<\frac{500-\mu }{5}\right)=0.01\\ P\left(Z<z\right)=\mathrm{0.01.}\left[\text{Let }z=\frac{500-\mu }{5}\right]\end{array}$

Variable value:

Software procedure:

Step-by-step software procedure to obtain variable value using EXCEL is as follows:

• • Open an EXCEL file.
• • In cell G1, enter the formula “=NORM.S.INV(0.01)”.
• Output using EXCEL software is given below:

Thus,

$\begin{array}{c}P\left(Z<-2.326\right)=0.01\\ \frac{500-\mu }{5}=-2.326\\ 500-\mu =-11.63\\ \mu =500+11.63\\ =\mathrm{511.63.}\end{array}$

Thus, the mean, so that the probability that a half-liter bottle will not be underfilled is 99% is 511.63 ml.

To determine

Find the mean, so that the probability that a half-liter bottle will not be underfilled is 99.9%.

Answer to Problem 92CE

Themean, so that the probability that a half-liter bottle will not be underfilled is 99.9% is 515.45 ml.

Explanation of Solution

Calculation:

There must be a 99.9% chance that the half-liter bottle will not be underfilled. This implies that $P\left(X\ge 500\right)=0.999$.

Now, $P\left(X\ge 500\right)=0.999$ can be rewritten as follows:

$\begin{array}{c}P\left(X\ge 500\right)=0.999\\ 1-P\left(X<500\right)=0.999\\ P\left(X<500\right)=1-0.999\\ =\mathrm{0.001.}\\ P\left(\frac{X-\mu }{\sigma }<\frac{500-\mu }{\sigma }\right)=0.001\\ P\left(Z<\frac{500-\mu }{5}\right)=0.001\\ P\left(Z<z\right)=\mathrm{0.001.}\left[\text{Let }z=\frac{500-\mu }{5}\right]\end{array}$

Variable value:

Software procedure:

Step-by-step software procedure to obtain variable value using EXCEL is as follows:

• • Open an EXCEL file.
• • In cell G1, enter the formula “=NORM.S.INV(0.001)”.
• Output using EXCEL software is given below:

Thus,

$\begin{array}{c}P\left(Z<-3.09\right)=0.01\\ \frac{500-\mu }{5}=-3.09\\ 500-\mu =-15.45\\ \mu =500+15.45\\ =\mathrm{515.45.}\end{array}$

Thus, the mean, so that the probability that a half-liter bottle will not be underfilled is 99.9%is 515.45 ml.