Chapter 7.3, Problem 58E

a)

To determine

To find: The probability that randomly selected box has less than 9.65 ounces of cereal.

Answer to Problem 58E

The required probability is 0.0475.

Explanation of Solution

Given:

Population mean (µ) = 9.70

Population standard deviation (s) = 0.03

Calculation:

Consider, X be the random variable that shows amount of cereal in a box

The probability that randomly selected box has less than 9.65 ounces of cereal can be calculated as:

  $\begin{array}{c}P\left(X<9.65\right)=P\left(\frac{x-\mu }{\sigma }<\frac{9.65-\mu }{\sigma }\right)\\ =P\left(Z<\frac{9.65-9.70}{0.03}\right)\\ =P\left(Z<-1.67\right)\\ =0.0475\end{array}$

Thus, the required probability is 0.0475.

b)

To determine

To find: The probability that mean amount of cereal in randomly selected 5 boxes is less than 9.65 ounces.

Answer to Problem 58E

The required probability is 0.0001

Explanation of Solution

Given:

Sample size ( n ) = 5

Calculation:

The probability that mean amount of cereal in randomly selected 5 boxes is less than 9.65 ounces can be calculated as:

  $\begin{array}{c}P\left(\overline{X}<9.65\right)=P\left(\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}<\frac{9.65-\mu }{\frac{\sigma }{\sqrt{n}}}\right)\\ =P\left(Z<\frac{9.65-9.70}{\frac{0.03}{\sqrt{5}}}\right)\\ =P\left(Z<-3.73\right)\\ =0.0001\end{array}$

Thus, the required probability is 0.0001.