Chapter 6, Problem T6.12SPT

(a)

To determine

To explain: X can be modelled by a binomial distribution.

Explanation of Solution

Given information:

X is the number of owners who greet their dogs first.

Sample size $=\text{ 12}$

Percentage of dog owners who greet their dogs first $=\text{ 66}\%$

Concept used:

  $\text{1}0\%\text{ condition }$

  $\text{n }<0.\text{1}0\text{N}$

The rule says that if the sample is less than $\text{1}0\%$ of the population, then it is reasonable to assume that the trials are independent and can be modelled by binomial distribution irrespective of the without replacement sample.

Here, the sample size ( $=\text{ 12}$ ) is far less than $\text{1}0\%$ of all dog owners.

Moreover, here the trial of each adult is independent. Whether one dog greets their dogs first does not give any idea about the other dog owner.

So, $\text{X }~\text{ Bin }\left(\text{12},0.\text{66}\right)$

Hence, X can be modelled by a binomial distribution.

(b)

To determine

To calculate: Probability that exactly 6 owners greet their owners first.

Answer to Problem T6.12SPT

Probability that exactly 6 owners greet their owners first is 0.117980.

Explanation of Solution

Calculation:

  $\text{P }\left(\text{X }=\text{ 6}\right)=12{{\displaystyle C}}_6\times {{\displaystyle 0.66}}^6\times {{\displaystyle \left(1-0.66\right)}}^{12-6}$

  $\text{P }\left(\text{X }=\text{ 6}\right)=0.\text{11798}0$

(c)

To determine

To check: Whether there is any evidence against the Ladies Home Journal claim or not.

Answer to Problem T6.12SPT

There seems to be strong evidence against the Ladies Home Journal claim.

Explanation of Solution

Given information:

4 owners in reality greeted their digs first.

Calculation:

Finding the probability of the above fact according to Ladies Home Journal claim

  $P\left(X\le 4\right)=\text{ P }\left(\text{X }=0\right)+\text{ P }\left(\text{X }=\text{ 1}\right)+\text{ P }\left(\text{X }=\text{ 2}\right)+\text{ P}\left(\text{X }=\text{ 3}\right)+\text{ P }\left(\text{X }=\text{ 4}\right)$

So,

  $\text{P }(\text{X } \le 4)=$

  $\begin{array}{l}{{\displaystyle C}}_0\times {{\displaystyle 0.66}}^0\times {{\displaystyle \left(1-0.66\right)}}^{12-0}+{{\displaystyle C}}_1\times {{\displaystyle 0.66}}^1\times {{\displaystyle \left(1-0.66\right)}}^{12-1}+{{\displaystyle C}}_2\times {{\displaystyle 0.66}}^2\times {{\displaystyle \left(1-0.66\right)}}^{12-2}+\\ {{\displaystyle C}}_3\times {{\displaystyle 0.66}}^3\times {{\displaystyle \left(1-0.66\right)}}^{12-3}+{{\displaystyle C}}_4\times {{\displaystyle 0.66}}^4\times {{\displaystyle \left(1-0.66\right)}}^{12-4}\end{array}$

  $\text{P }(\text{X } \le \text{ 4) = 0}\text{.021265}$

It can be seen that the probability according to Ladies Home Journal is very low giving evidence to strongly reject the claim.