Chapter 18, Problem 48E

(a)

To determine

To find out what fraction of all bags sold are underweight.

Answer to Problem 48E

  $4.78\%$ .

Explanation of Solution

It is given in the question that the weight of potato chips in a medium size bag is stated to be $10$ ounces. The amount that the packaging machine puts in these bags is believed to have a normal model with mean $10.2$ ounces and standard deviation $0.12$ ounces. Thus, let us first find the z -score for the following as:

  $\begin{array}{l}{z}_{10}=\frac{10-10.2}{0.12}\\ =-1.667\end{array}$

Thus, the fraction of all bags sold are underweight can be calculated as follows:

  $\begin{array}{l}P(z<-1.667)=normalcdf(-E99,-1.667,0,1)\\ =0.0478\end{array}$

Thus, only about $4.78\%$ of the bags sold are underweight.

(b)

To determine

To find out what is the probability that none of the three is underweight.

Answer to Problem 48E

  $0.863$ .

Explanation of Solution

It is given in the question that the weight of potato chips in a medium size bag is stated to be $10$ ounces. The amount that the packaging machine puts in these bags is believed to have a normal model with mean $10.2$ ounces and standard deviation $0.12$ ounces. Thus, from part (a) we have that only about $4.78\%$ of the bags sold are underweight. Thus, the probability that none of the three is underweight can be calculated as:

  $\begin{array}{l}$ P(\text{none of the three bags are underweight})=1-{(0.0478)}^3$=0.863\end{array}$

Thus, the probability that none of the three bags is underweight is $0.863$ .

(c)

To determine

To find out what is the probability that the mean weight of the three bags is below the stated amount.

Answer to Problem 48E

  $0.0019$ .

Explanation of Solution

It is given in the question that the weight of potato chips in a medium size bag is stated to be $10$ ounces. The amount that the packaging machine puts in these bags is believed to have a normal model with mean $10.2$ ounces and standard deviation $0.12$ ounces. Thus, from part (a) we have that only about $4.78\%$ of the bags sold are underweight. Let us first find the z -score as:

  $\begin{array}{l}z=\frac{10-10.2}{\frac{0.12}{\sqrt{3}}}\\ =-2.887\end{array}$

Thus, the probability that the mean weight of the three bags is below the stated amount is as:

  $\begin{array}{l}P(z<-2.887)=normalcdf(-E99,-2.887,0,1)\\ =0.0019\end{array}$

Thus, the probability that the mean weight of the three bags is below the stated amount is about $0.0019$ .

(d)

To determine

To find out what is the probability that the mean weight of a $24$ -bag case of potato chips is below $10$ ounces.

Answer to Problem 48E

The probability that the mean weight of a $24$ -bag case of potato chips is below $10$ ounces is zero.

Explanation of Solution

It is given in the question that the weight of potato chips in a medium size bag is stated to be $10$ ounces. The amount that the packaging machine puts in these bags is believed to have a normal model with mean $10.2$ ounces and standard deviation $0.12$ ounces. Thus, from part (a) we have that only about $4.78\%$ of the bags sold are underweight. Now, for the mean weight we have to first find the following as:

  $\begin{array}{l}{\mu }_{\overline{y}}=10.2\\ {\sigma }_{\overline{y}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{0.12}{\sqrt{24}}\\ =0.024\end{array}$

Thus, the z -score will be then as:

  $\begin{array}{l}z=\frac{10-10.2}{0.024}\\ =-8.333\end{array}$

And thus the probability that the mean weight of a $24$ -bag case of potato chips is below $10$ ounces is calculated as:

  $\begin{array}{l}P(z<-8.333)=normalcdf(-E99,-8.333,0,1)\\ =0\end{array}$

Thus, the probability that the mean weight of a $24$ -bag case of potato chips is below $10$ ounces is zero.