I am an undergraduate student of Economics. Today I was trying to solve 1 exercise related to Poisson process that I found confusing and I would be very grateful for your help, as my Mathematics knowledge is quite basic.
So the exercise is:
The average number of calls (which follow the Poisson process) per hour is 4,25. 24% of the calls are unpleasant.
1) What is a probability that in 1 hour will be received exactly 6 normal and 1 unpleasant call?
2) If call duration is Normally distributed, what is 5th and 95th centile? (In Normal distribution 95th centile is z=1.96)
MY SOLUTION: 1) to find a $\theta^1$ for unpleasant calls and $\theta^2$ for normal calls:
$\theta^1 = 0.24 * 4.25 = 1.02$
$\theta^2= 4.25 - 1.02 = 3.23$
X - average number of Normal calls per hour
Y - average number of Unpleasant calls per hour
$P(X=6, Y=1 | X+Y=7) = \frac{\frac{e^{-3.23} * 3.23^6}{6!} * e^{-1.02} * 1.02}{ \frac{e^{-4.25} * 4.25^7}{7!}} $
Please tell me what do you think because I really feel I am mistaken somewhere and not sure if the Poisson parameter $\theta$ can be splitted in 2 parts.
Also if somebody could give any idea about 2) question I would be very grateful.