# Discrete Probability Distribution II

• Oct 13th 2010, 03:47 AM
cloud5
Discrete Probability Distribution II
Could anyone teach me how to solve these?(Crying)

Attachment 19313

Attachment 19314

Q9: (i)0.1108, (ii)0.6227, 2, 0.2681
Q10: (i)0.0803, (ii)0.0498, (iii)0.0032
• Oct 13th 2010, 04:07 AM
Quote:

Originally Posted by cloud5
continued from the previous thread... Could anyone teach me how to solve these?(Crying)

Attachment 19313Attachment 19314

Q9: (i)0.1108, (ii)0.6227, 2, 0.2681
Q10: (i)0.0803, (ii)0.0498, (iii)0.0032

# 9

You are asked to determine the most likely value of X or the mode of this poisson distribution.

$\displaystyle \frac{P(X=x+1)}{P(X=x)}=\frac{\lambda}{x+1}$ in which lambda is 2.2

Consider for two cases, if P(X=x+1)>P(X=x), then 2.2>x+1, x<1.2 so x=1

P(X=2)>P(X=1)>P(X=0)

If P(X=x+1)<P(X=x), then 2.2<x+1 , x>1.2 so x=2,3,4,...

P(x=2)>P(X=3)>P(X=4)>...

Do you see that P(X=2) has the highest probability? That implies that the value of X most likely to occur is 2. Calculate its probability.

(10) Adjust the mean to get 1/12.
• Oct 13th 2010, 06:24 PM
cloud5
Quote:

# 9
$\displaystyle \frac{P(X=x+1)}{P(X=x)}=\frac{\lambda}{x+1}$ in which lambda is 2.2

$\displaystyle \frac{P(X=x+1)}{P(X=x)}=\frac{\lambda}{x+1}$

How did you find this solution?
• Oct 14th 2010, 03:27 AM
$\displaystyle \frac{P(X=x+1)}{P(X=x)}=\frac{\lambda}{x+1}$
$\displaystyle P(X=x+1)=\frac{e^{{-\lambda}}{(\lambda)^{x+1}}}{(x+1)!}$ -- 1
$\displaystyle P(X=x)=\frac{{e^{-\lambda}}{\lambda^x}}{x!}$ -- 2