Algebra 2- Binomial distribution and z-score.

According to a national survey, 92% of U.S. dog owners consider their pets to be members of the family. In a random survey of 150 dog owners, what is the probability that at most 132 of them consider their pets to be members of the family?

Question

Algebra 2- Binomial distribution and z-score.

According to a national survey, 92% of U.S. dog owners consider their pets to be members of the family. In a random survey of 150 dog owners, what is the probability that at most 132 of them consider their pets to be members of the family?

anonymous · Answer

Approximating with a normal distribution with mean $np=150\times0.92=138$ and standard deviation $\sqrt{npq}=\sqrt{150\times0.92\times0.08}\approx 3.32$, you have
$$P(X\le132)=P\left(\frac{X-138}{3.32}\le\frac{132-138}{3.32}\right)\approx P\left(Z\le-1.81\right)$$