# Combining two normal distrubtions, 1 of which is for a project which may not happen

#### mr e guest

##### New Member
Hello,

I have to 2 projects:
Project A: Median: $450k, Variance: 625 Project B: Median:$750k, Variance: 1250
Combined Freqency Distribution: Median: \$1200k, Variance: 1875

Please could you tell me how to find the Combined Frequency Distribution (for both projects combined) if Project B only has a 60% likelihood of happening?

Mr E Guest

#### mr e guest

##### New Member
Re: Combining two normal distrubtions, 1 of which is for a project which may not happ

sorry I'm guessing from the no replies that the question doesn't really make sense? or is this a very hard problem to solve? Should I approve the problem in a different way?

Thanks,

Mr E

#### BGM

##### TS Contributor
Re: Combining two normal distrubtions, 1 of which is for a project which may not happ

Your question does make sense, and I hope you understand that there are quite a number of new posts in TS everyday so we may miss your thread as it sink beyond the most updated 10.

So I assume that in the original setting, you are having two random variables $$A, B$$ (representing the two projects) and you have some results about the sum of them

$$C = A + B$$

(which you called combined of them). The result itself is always valid if they are independent and you are talking about mean instead of median.

The situation you mentioned I guess means the following mixture:

$$C = \left\{\begin{matrix} A + B, & \text{with probability 0.6} \\ A, & \text{with probability 0.4} \end{matrix} \right.$$

The distribution of this mixture can always be found (btw finding a distribution in probability context means find its cumulative distribution function), you just need to provide the distribution of each of them. If you are only interested in the mean and variance, you may apply the law of total variance to calculate it, by rewriting

$$C = (A + B)I + A(1 - I) = A + BI$$

where

$$I \sim \text{Bernoulli}(0.6)$$ which is independent of $$A, B$$.

So in this case

$$E[C] = E[A + BI] = E[A] + EE = E[A] + 0.6E$$

$$Var[C] = Var[A + BI]$$

$$= Var[A] + Var[BI]$$

$$= Var[A] + Var[E[BI|I]] + E[Var[BI|I]]$$

$$= Var[A] + Var[EI] + E[VarI^2]$$

$$= Var[A] + E^2Var + VarE[I^2]$$

$$= Var[A] + E^2 \times 0.6 \times 0.4 + Var \times 0.6$$

So you can just plug in the values given to obtain the result.