How to calculate probability of this random variable if all is known is the mean and variance?

Question

Suppose a light bulb turns on with a certain probability. The number of light bulbs that turn on per hour is a random variable with mean 1000 and variance 1000. What is the probability that more than 1200 light bulbs will turn on in an hour?

Kind of stuck with this question as all is known is the mean and variance. Nothing about the distribution is known, but $E(X) = \operatorname{Var}(X)$ suggests that $X\sim \operatorname{Poisson}(1000)$. Am I allowed to jump to this assumption though?

Answer 1

Poisson distribution? I agree with @user1, that it's reasonable to assume the number of lightbulbs turning on in an hour is Poisson. It is typical to talk of Poisson events as taking place with a certain average number within a particular period of time.

Also, Poisson distributions are among those that have mean and variance equal. It would add to the strength of this supposition if you have recently covered Poisson distributions. So I'd say to go ahead and "jump."

Desired probability. So consider $X \sim \mathsf{Pois}(\lambda = 1000).$ Then you seek $P(X > 1200) = 1 - P(X \le 1200) \approx 0.$

Normal approximation: With such a large mean as $\lambda = 1000,$ one can use the normal approximation to Poisson distributions to find this probability in terms of a normal distribution with $\mu = 1000, \sigma = \sqrt{1000}= 31.62278.$

So you can get the answer by standardizing and using printed normal tables. Because $(1200 - 1000)/31,62278 \approx 6.32$ standard deviations above the mean, you can guess that the probability is essentially $0.$

Software: You can also use statistical computer software or a statistical calculator to find the exact Poisson probability. In R, where ppois is a Poisson CDF, the computation is as shown below:

1 - ppois(1200, 1000)
[1] 3.884939e-10

In terms of the normal approximation, R gives nearly $0$ again:

1 - pnorm(1200, 1000, sqrt(1000))
[1] 1.269814e-10

If you do this by standardizing and using normal tables, you will see that z-scores above about 3.5 are off the table, so you would have to understand that a z-score above $6$ corresponds to an answer near $0$ (which might be one point of this exercise).

Below is a plot of the Poisson probabilities (with values between 860 and 1200) along with the density function of he approximating normal distribution.

x = 860:1200; pdf = dpois(x, 1000)
plot(x, pdf, type="h", col="blue")
 abline(h=0, col="green2")
 curve(dnorm(x,1000,sqrt(1000)), add=T, col="brown", lwd=2, lty="dashed")
 abline(v = 1200)