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Discrete variable

0

745

3

+845

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

YEEEEEET Oct 28, 2018

3⁺⁰ Answers

+6252

+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.

Rom Oct 28, 2018

+845

0

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

YEEEEEET Oct 28, 2018

+6252

+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

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