Chapter 5.1, Problem 1.1CYU

(a)

To determine

To explain:the probability equal to 0.61.

Answer to Problem 1.1CYU

A large sample of U.S. adult about $61\%$ eat breakfast

Explanation of Solution

Given:

A randomely selected U.S. adult usually eats breakfast is $0.61$

Concept used:

Definition of odds:-odds is the ratio of two positive integer which have no comman factor

Let A be the event of a sample space then the odd favorable to A

Calculation:

Let A be the event of a sample space then the odd favorable to A

  $=\frac{P\left(A\right)}{P\left(A^c\right)}$

That is $P\left(A\right):P\left(A^c\right)$

The books of odds probability $0.61$ means in this setting

That is $\begin{array}{l}=\frac{0.61}{100}\times 100\times 100\\ =61\%\end{array}$

This means that a large sample of U.S. adult about $61\%$ eat breakfast

(b)

To determine

To explaint the reason due to which the given probability 0.61 does means that exactly 61 out of 100 eats breakfast.

Explanation of Solution

Given:

A randomely selected U.S. adult usually eats breakfast is $0.61$

Concept used:

Definition of odds:-odds is the ratio of two positive integer which have no comman factor

Let A be the event of a sample space then the odd favorable to A

Calculation:

Let A be the event of a sample space then the odd favorable to A

  $=\frac{P\left(A\right)}{P\left(A^c\right)}$

That is $P\left(A\right):P\left(A^c\right)$

The books of odds probability $0.61$ means in this setting

That is $\begin{array}{l}=\frac{0.61}{100}\times 100\times 100\\ =61\%\end{array}$

Although this expect around $61$ adults will eat breakfast the exact number will differ from sample to sample