+0

# Discrete variable

0
95
3
+622

Please explain the weird layout,  i have no idea how to approach

Oct 28, 2018

#1
+3593
+1

It's pretty straightforward.

P(X) must sum to 1.  You're given the first 4 values.  Determine the 5th.  The rest are 0.

$$E[X] = \sum~x p(x) = \sum \limits_{x=1}^4~x\cdot \dfrac{1}{x+2}+5k$$

Use the value of $$k=P(5)$$ and evaluate that sum.

Oct 28, 2018
#2
+622
0

can u explain how k= 1/20 plz im still very confused, thanks

YEEEEEET  Oct 28, 2018
#3
+3593
+1

the probabilities must sum up to 1

$$1 = P[1]+P[2]+P[3]+P[4]+P[5] = \dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6} + k\\ \\ 1 = \dfrac{20+15+12+10}{60}+k\\ \\ 1 = \dfrac{57}{60}+k\\ \\ k=\dfrac{3}{60} = \dfrac{1}{20}$$

Rom  Oct 28, 2018