The IQs of $9$ randomly selected people are recorded. Let $\overline{Y}$ denote their average. Assuming the distribution from which the $Y_i$'s were drawn is normal with a mean of $100$ and a standard deviation of $16$, what is the probability that $\overline{Y}$ will exceed $103$?
What is the probability that any arbitrary $Y_i$ will exceed $103$?
What is the probability that exactly three of the $Y_i$'s will exceed $103$?
Best Answer
Let $Y_1, Y_2, \dots,Y_9$ be the IQ measurements. Then $$\bar{Y}=\frac{Y_1+Y_2+\cdots+Y_9}{9}.$$ It is a standard fact that I imagine you know that under your assumptions, $\bar{Y}$ has normal distribution, with mean the mean of the $Y_i$, and standard deviation $\frac{\sigma}{\sqrt{9}}$, where $\sigma$ is the standard deviation of the $Y_i$.
Thus in our case, the standard deviation of $\bar{Y}$ is $\frac{16}{3}$.
Now we tackle the problem of the probability that $\bar{Y}\gt 103$. Here there is some ambiguity, because published IQ's are usually integers. But we will assume they can be in principle any real number. Then $$\Pr(\bar{Y}\gt 103)=\Pr\left(Z\gt \frac{103-100}{16/3}\right),$$ where $Z$ is standard normal. You can now obtain the probability from a table of the standard normal, or software. I think the answer is about $0.2868$.
For the second problem and third problem, let $p$ be the probability that an individual's IQ exceeds $103$. This is the probability that a standard normal is $\gt \frac{103-100}{16}$, and can be found using a table or software.
I get about $0.4257$. This is close to your answer, so I am sure you did it more or less the right way. I am almost sure that you rounded $\frac{3}{16}$ to $0.18$. It is actually $0.1875$, so three-quarters of the way to $0.19$.
Call exceeding $103$ a success. We want the probability of exactly $3$ successes in $9$ trials. For the answer, we use the Binomial distribution. The required probability is $$\binom{9}{3}p^3(1-p)^6.$$