1
$\begingroup$

I'm not certain if this would classify as a variation of the birthday problem or not but I have an interesting problem.

Say you have a group of 600 people, what is the approximate probability that at least 75 of them were born in September?

The problem further provides these assumptions for part a and b.

a) Months are equally likely to contain birthdays

b) Days are equally likely to be birthdays

The emphasis on approximate is mine. This leads be to believe I could set this up as a binomial variable with $\frac{1}{12}$ probability of birthday being in September for part a and solve using the normal approximation to the binomial.

But for part b, would I just set the probability equal to $\frac{1}{365}$ and multiply the final answer by 30? That somehow doesn't sound right to me.

Any ideas?

$\endgroup$
  • 1
    $\begingroup$ Why wouldn't it be $p=30/365$? Then you could use the binomial distribution. $\endgroup$ – WaveX Feb 17 '18 at 1:48
  • $\begingroup$ Assumptions (a) and (b) are just two different ways to approximate the actual probability in real life that a randomly selected person was born in September. $\endgroup$ – David K Feb 17 '18 at 2:42
3
$\begingroup$

We could use a binomial distribution. Think of this like pulling $600$ people out of thin air, and one by one, you assign them a September birthday, or a non-September birthday, because if months and days are equally likely to contain birthdays, then the probability of "success" is constant.

We have that $n$ is the "trials" (600 people) and $p$ is the probability $\displaystyle \frac{30}{365}=0.0829$ (for September).

However, this method is seriously inconvenient, because this is a discrete (whole number) variable and individual values have to be calculated for $X=75$, $X=76$, etc.

We should use a normal approximation. A normal approximation is good when we have $np>10$ and $n(1-p)>10$.

The standard deviation would be $\sqrt{np(1-p)}=6.72$, and the mean would be $np=49.3$.

The probability that $75$ or more people are born in September corresponds to a z-score of $z=\displaystyle \frac{75-49.3}{6.72}=3.82$.

Z scores above $3$ are pretty rare!

Using a probability table, we have $P(z>3.82)=\boxed{0.0000667}$, or $0.0066\%$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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