Chapter 11.1, Problem 29E

(a)

To determine

To calculate:

The probability value for all four boxes of cereal not more than $19.4$ ounces

Answer to Problem 29E

The probability value for all four boxes of cereal not more than $19.4$ ounces is $0.82\%$ .

Explanation of Solution

Given information:

  $P\left(x\le 19.4\right)$

Formula used:

  $Z-$ Score is given by the following expression:

  $Z=\frac{x-\mu }{\sigma }$

Where,

Standard deviation: $\sigma$

Mean: $\mu$

Calculation:

The weight of $19.4ounces$ has a corresponding following $Z-$ Score formula to calculate the probability value for all four boxes of cereal not more than $19.4$ ounces

  $\begin{array}{l}P\left(x\le 19.4\right)=P\left(z\le \frac{19.4-\mu }{\sigma }\right)\\ P\left(x\le 19.4\right)=P\left(z\le -2.4\right)\end{array}$

Use the standard normal table:

  $P\left(x\le 19.4\right)=0.0082=0.82\%$

(b)

To determine

To explain:

The machine is working properly or not

Answer to Problem 29E

Yes, the machine is functioning properly.

Explanation of Solution

Know that:

  $\begin{array}{l}\mu =20ounces\\ \sigma =0.25ounces\end{array}$

Yes, the machine is functioning properly, because the probability that a box contains an quantity of cereal considerably less than the mean is very small.