# Probability Problem (Poisson Distribution)

It has been observed that statistics majors in a certain university produce $60$ hypothesis test results, on average, daily. What is the probability that a particular statistics major from that university $a$) will produce at least one hypothesis test results tomorrow $b$) produces $70$ hypothesis test results in two days My solution at $a$ is ($e^{-60}$$(60)^0 ) / 0! and at b is (e^{-120}$$(120)^{70}$) / $70!$ . Are my answers correct or I misinterpret the said problem?

• Are you told the process is Poisson? Maybe a student produces between $55$ and $65$ uniformly. Or possibly s/he produces $420$ on Monday but $0$ on every other day. Or $84$ on each weekday but takes the weekend off. And so on. If you are told it is Poisson, that's one thing....otherwise it's a big leap. – lulu Feb 11 '17 at 13:34
• Also, the numbers seem odd. If for some reason we guess that each day has that same distribution as each other, then surely the answer in both cases is $1-\epsilon$. I think there must be some missing information. – lulu Feb 11 '17 at 13:42
• Well I just posted what my prof gave to me and though I just assumed that it is Poisson because of the time and space given – Codex Feb 11 '17 at 13:51
• To be Poisson requires, among other things, that the probability of occurrence in any fixed time interval depends only on the length of that interval. That seems unrealistic in this case. Of course, in an abstract problem one can declare whatever distribution is desired, but if you are meant to model the physical situation I would say there isn't nearly enough information. But, besides that, the numbers really seem wrong. Like I say, with any likely distribution that is at least roughly the same on each day, both answers are clearly a hair less than $1$. – lulu Feb 11 '17 at 13:58
• I agree on that. I think this is created just to correlate that it is a Poisson type of distribution. Just a quick question, if it's not a Poisson distribution, what other type of distribution it would be? – Codex Feb 11 '17 at 14:45

Putting details of this somewhat fanciful 'homework overload' situation aside, let's look at your attempt to apply the Poisson distribution:

In (a), suppose $X \sim \mathsf{Pois}(60),$ and that you seek $P(X \ge 1) = 1 - P(X = 0) = 1 - e^{-60}.$ Your result is correct. The numerical answer is very nearly 1, because $P(X=0) \approx 0.$

In (b), suppose $Y \sim \mathsf{Pois}(120),$ and that you seek $P(Y = 70) = e^{-120}(120)^{70}/70!.$ Your result is correct. Specifically, you correctly doubled the Poisson rate to match the 2-day time period. (Note: If you try to compute $70!$ on a typical calculator, it will overflow the capacity of the internal software because $70! > 10^{100}.$)

If the problem intended $P(Y \ge 70),$ then the answer is much larger. You could use software for an exact answer or a normal approximation (with $\mu = 70$ and $\sigma = \sqrt{70}).$

Using R statistical software, one gets the following numerical results.

1 - dpois(0, 60)                 # (a)
## 1
dpois(70, 120)                   # (b) P(Y=70)
## 2.233286e-07
1 - ppois(69, 120)               # (b) P(Y >= 70) exact
## 0.9999997
1 - pnorm(69.5, 120, sqrt(120))  # (b) P(Y >= 70) norm aprx w/ continuity correction
## 0.999998
factorial(70)                    # 70! is HUGE
## 1.197857e+100