Chapter 11, Problem 11.1AYK

(a)

To determine

To explain:

Thewidest distribution also be taller or the shortest of the given data.

Explanation of Solution

Given:

  $\begin{array}{l}m=550ml\\ \sigma =100ml\end{array}$

Concept used:

Formula

The normal distribution:-

  $z=\frac{x-\mu }{\sigma }$

Calculation:

  $95\%$ Of the men's bladder volumes falls between $350$ and $750$ .

Mean $=550$

S.D $=100$

According to empirical rule $95\%$ of the data points falls between $2$ standard deviations.

  $\begin{array}{l}{z}_1=-2.0\\ z_2=2.0\end{array}$

By applying normal distribution:-

  $z=\frac{x-\mu }{\sigma }$

  $\begin{array}{l}{x}_1=350\\ x_2=750\end{array}$

(b)

To determine

To find:

Thevalue of each distribution’s standard deviation of the given data.

Answer to Problem 11.1AYK

  $x_1=175,x_2=625$

Explanation of Solution

Given:

  $\begin{array}{l}m=550ml\\ \sigma =100ml\end{array}$

Concept used:

Formula

The normal distribution:-

  $z=\frac{x-\mu }{\sigma }$

Calculation:

  $99.7\%$ Of the women's bladder volumes falls between $175$ and $625$ .

Mean $=400$

S.D $=75$

According to empirical rule $99.7\%$ of the data points falls between $3$ standard deviations.

  $\begin{array}{l}{z}_1=-3.0\\ z_2=3.0\end{array}$

By applying normal distribution:-

  $z=\frac{x-\mu }{\sigma }$

  $\begin{array}{l}{x}_1=175\\ x_2=625\end{array}$