Chapter 6.3, Problem 33PS

Solution

a)

To find: The z-score for the provided weights.

The z-scores are -2.4 and 1.6 respectively.

Given information:

  $\begin{array}{l}\text{Mean}\left(\mu \right)=20\\ \text{Standard deviation}\left(\sigma \right)=0.25\end{array}$

Formula used:

The formula of the z-score is:

  $\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ \text{where}\\ x=\text{Row score}\\ \mu \text{=Population mean}\\ \sigma \text{=Population standard deviation}\end{array}$

Calculation:

Consider, X be the random variable that is distributed normally.

The z-scores are calculated as:

  $\begin{array}{c}z=\frac{x-\mu }{\sigma }\\ =\frac{19.4-20}{0.25}\\ =-2.4\\ z=\frac{x-\mu }{\sigma }\\ =\frac{20.4-20}{0.25}\\ =1.6\end{array}$

The required z-scores are -2.4 and 1.6 respectively.

b)

To find: The probability that the weight of randomly selected box is at most 19.4 ounces.

The probability is 0.0082.

Calculation:

Consider, X be the random variable that is distributed normally.

The probability is calculated as:

  $\begin{array}{c}P\left(x\le 19.4\right)=P\left(Z<\frac{19.4-20}{0.25}\right)\\ =P\left(z\le -2.4\right)\\ =0.0082\end{array}$

The required probability is 0.0082

c)

To find: The probability that the weight of randomly selected box is between 19.4 and 20.4 ounces.

The probability is 0.937.

Calculation:

The probability is calculated as:

  $\begin{array}{c}P\left(19.4<x<20.4\right)=P\left(\frac{19.4-\mu }{\sigma }<Z<\frac{20.4-\mu }{\sigma }\right)\\ =P\left(\frac{19.4-20}{0.25}<Z<\frac{20.4-20}{0.25}\right)\\ =P\left(-2.4<Z<1.6\right)\\ =0.937\end{array}$

The required probability is 0.937.