Q1 (a) Given probability distribution function for discrete random variable X as below: 2 4 8 10 P(X = x) 0.08 0.25 0.35 0.2 0.12 (i) Find the mean of X. (ii) Find the variance of X.

MATLAB: An Introduction with Applications
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Author:Amos Gilat
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Q1
(a)
Given probability distribution function for discrete random variable X as below:
2
4
8
10
P(X = x)
0.08
0.25
0.35
0.2
0.12
(i)
Find the mean of X.
(ii)
Find the variance of X.
(b)
Let the continuous random variable X has probability density function as follows,
(k(x + 2)²,
-2 <x < 0
4
f(x) = { 4k,
0<x <;
3
0,
otherwise
(i)
Find the value of constant k.
Find the value P(-1< x< 1)
Transcribed Image Text:Q1 (a) Given probability distribution function for discrete random variable X as below: 2 4 8 10 P(X = x) 0.08 0.25 0.35 0.2 0.12 (i) Find the mean of X. (ii) Find the variance of X. (b) Let the continuous random variable X has probability density function as follows, (k(x + 2)², -2 <x < 0 4 f(x) = { 4k, 0<x <; 3 0, otherwise (i) Find the value of constant k. Find the value P(-1< x< 1)
Formula
Random Variable :
ÉPCE, ) = 1, E(X) -E- Po).
E(X) =Ex· P(x),
E(X²)=Ex² · P(x),
L f(x) dx =1,
E(X)= [" x•P(x) dx,
E(X²)=["x² - P(x) dx ,
Var(X)= E(X²)-[E(X)]².
Special Probability Distributions :
P(x = r)="C, · p" ·q"-", r = 0, 1, ..., n, P(X = r) =
r!
r = 0, 1, ... , o,
Z =
Transcribed Image Text:Formula Random Variable : ÉPCE, ) = 1, E(X) -E- Po). E(X) =Ex· P(x), E(X²)=Ex² · P(x), L f(x) dx =1, E(X)= [" x•P(x) dx, E(X²)=["x² - P(x) dx , Var(X)= E(X²)-[E(X)]². Special Probability Distributions : P(x = r)="C, · p" ·q"-", r = 0, 1, ..., n, P(X = r) = r! r = 0, 1, ... , o, Z =
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