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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

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

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 $X.$.

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

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

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

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

Given,

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

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

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

So,

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