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?

  • 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

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 $0.0066\%$

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