Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $103$?

share|cite|improve this question
Use \overline{Y} to get $\overline{Y}$. – Brian M. Scott Jan 27 '13 at 21:38
what did you try? – Seyhmus Güngören Jan 27 '13 at 22:32
I believe I found the probability that any arbitrary $Y_i$ will exceed 103 is $0.4286$. I am having trouble finding what $\overline{Y}$. I am also very uncertain how to find the probability of exactly $3$. – user59633 Jan 27 '13 at 23:03

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 $$\binom{9}{3}p^3(1-p)^6.$$

share|cite|improve this answer
Thank you so much! This makes much more sense now! – user59633 Jan 28 '13 at 2:26
Good! It takes a little while to see how these things work, but after you have it figured out, it stays with you. – André Nicolas Jan 28 '13 at 2:35

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.