Ladies Home Journal magazine reported that $66\%$of all dog owners greet their dogs 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 X is a binomial random variable.

(b) Only$4$ of the owners in the random sample greeted their dogs first. Does this give convincing evidence against the Ladies Home Journal claim? Calculate an appropriate probability to support your answer.

Let $X=$ the number of owners who greet their dogs first.

Binary (success/failure), independent trials, a fixed number of trials, and the equal probability of success for each trial are the four conditions for a binomial setting.

Success = greeting the dog first; failure = not greeting the dog first.

Because the dog owners were chosen at random, they are independent.

Fixed number of trials: $12$.

Probability of success: $66\%$or $0.66$.

Thus all $4$conditions have been met, which means that $X$is a binomial random variables.

Let $X=$ the number of owners greet their dogs first.

$n=12$

$p=66\%=0.66$

Definition binomial probability:

localid="1650079286605" $$\begin{gathered} P(X\le 4)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4) \\ =0.0213 \\ \approx 2.13\% \end{gathered}$$

Command Ti83-Ti84 calculator: binomcdf $(12,0.66,4)$.

Since the probability is less than $5\%$, there is convincing evidence against the Ladies Home Journal claim.