Question

Ladies Home Journal magazine reported that $66\%$ of all dog owners greet their dog before greeting their spouse or children when they return home at the end of the workday. Assume that this claim is true. Suppose 12 dog owners are selected at random. Let $X=$ the number of owners who greet their dogs first.

a. Explain why it is reasonable to use the binomial distribution for probability calculations involving $X$.

b. Find the probability that exactly 6 owners in the sample greet their dogs first when returning home from work.

c. In fact, only 4 of the owners in the sample greeted their dogs first. Does this give convincing evidence against the Ladies Home Journal claim? Calculate $P(X\le 4)$ and use the result to support your answer.

Short answer

1. A binomial distribution can be used to model X.
2. The odds of exactly six owners greeting their owners first are$0.117980$.
3. The Ladies Home Journal assertion appears to be backed up by substantial evidence.

Step-by-step solution

Part (a) Step 1: Given information

Given information:

X is the percentage of dog owners that welcome their animals first.

The value of Sample size is $=12$

Dog owners who greet their dogs first (percentage)$=66\%$

Let use the below concept

$10\%$condition

$\mathrm{n}<0.10\mathrm{N}$

Part (a) Step 2:  Calculation

According to the rule, if the sample represents less than $10\%$of the population, it is safe to assume that the trials are independent and may be modelled using the binomial distribution, regardless of the without replacement sample.

The sample size is $(=12)$, which is much less than $10\%$of all dog owners.

Furthermore, each adult's study is conducted independently. The fact that one dog greets another dog first has no bearing on the other dog's owner.

So, $\mathrm{X}~Bin(12,0.66)$

Therefore , a binomial distribution can be used to model $X.$.

Part (b) Step 1: Given information

Given information:

$X$is the percentage of dog owners that welcome their animals first.

The value of Sample size is 12.

Dog owners who greet their dogs first (percentage) $=66\%$

Let use the below Concept

$10\%$ condition

$\mathrm{n}<0.10\mathrm{N}$

Part(b) Step 2: Calculation

When six owners in the sample arrive home from work, there's a good chance they'll meet their dogs first.

$$\begin{gathered} P(X=6)=12C_6\times 0.{66}^6\times (1-0.66{)}^{12-6} \\ P(X=6)=0.117980 \end{gathered}$$

Part (c) Step 1: Given information

Given,

In truth, four owners were the first to meet their new digs.

Part (c) Step 2: Calculation

Determining the likelihood of the foregoing fact, as stated in the Ladies Home Journal

$P(X\le 4)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)$

So,

$$\begin{gathered} \mathrm{P}(\mathrm{X}\le 4)={}^{12}C_0\times 0.{66}^0\times (1-0.66{)}^{12-0}+{}^{12}C_1\times 0.{66}^1\times (1-0.66{)}^{12-1}{}^{12}C_2\times 0.{66}^2\times (1-0.66{)}^{12-2}+{}^{12}C_3\times 0.{66}^3\times (1-0.66{)}^{12-3}+{}^{12}C_4\times 0.{66}^4\times (1-0.66{)}^{12-4} \\ \mathrm{P}(\mathrm{X}\le 4)=0.021265 \end{gathered}$$

The possibility, according to Ladies Home Journal, is extremely low, providing evidence to strongly refute the assertion.