# How to find the value for average rate

I'm doing some textbook problems on my own and there are no steps given to the solutions to some of the problems. If anyone could help me solve the following problem, much would be appreciated.

"A liquid culture medium contains on the average m bacteria per ml. A large number of samples is taken, each of 1 ml, and bacteria are found to be present in 90% of the samples. Estimate m.

The answer given: m = 2.3026

The knowledge we're supposed to use (What we have so far learned in this chapter are: Binomial, Poisson, Exponential and Normal Distributions) is a bit limited, so if it is possible to provide an explanation within those constraints, that would be ideal.

• So you are looking for a distribution which has a nonzero values 90 % of the time (and a zero value 10 % of the time == PDF of 0.1 for 0). Jan 14, 2019 at 10:12
• Okay, problem solved! Thanks for your help. I used the Poisson distribution: Step1: $Pr[X=0]=0.1$ Step2: $[(e^−λ)(λ^x)]/x!=0.1$ Since $x=0$, that means the equation simplified to $e^−λ=0.1$ and then solved for $\lambda$ using log rules. Jan 14, 2019 at 10:54

Since you figured it out, I will post the answer. Poisson is the easiest option, as you figured out, since it has the least number of parameters. So using the Poisson formula

$$P(k \leq 0)=P(k=0)=0.1$$

Which means

$$e^{-\lambda}\frac{\lambda^k}{k!}=0.1$$

and since k=0

$$e^{-\lambda}=0.1$$ $$\lambda=-\ln(0.1)$$

• Could it have been done another way? You said, "Poisson is the easiest option" and it sounds like there is an alternative. I'd appreciate if you explain/post the alternative solution as well. Jan 14, 2019 at 12:13
• @SooKyungAhn No, Poisson is the only available option in your case. That was poor wording on my part. The binomial distribution would also be an option, but you would additionally require the sample size to perform the same process. Jan 14, 2019 at 12:24