Fill out the table and determine the value of the following quantities, Using the following p.m.f.
f(x)= (x+1)^{2}/91, x=0,1,2,3,4,5
i.Probability Table:
x |
f(x) |
xf(x) |
0 |
||
1 |
||
2 |
||
3 |
||
4 |
||
5 |
ii . What is the E[X]?
iii. What is the E[x^{2}]?
iv. What is Var(x)?
SOLUTION:
From given data,
Fill out the table and determine the value of the following quantities, Using the following p.m.f.
f(x)= (x+1)^{2}/91, x=0,1,2,3,4,5
i.Probability Table:
x | f(x) | x*f(x) | x^{2}*f(x) |
0 | 0.0109 | 0 | 0 |
1 | 0.0439 | 0.0439 | 0.0439 |
2 | 0.0989 | 0.1978 | 0.3956 |
3 | 0.1758 | 0.5274 | 1.5822 |
4 | 0.2747 | 1.0988 | 4.3952 |
5 | 0.3956 | 1.978 | 9.89 |
ii . What is the E[X]?
E[X] = x*f(x) = 0+0.0439+0.1978+0.5274+1.0988+1.978 = 3.8459
E[X] = 3.8459
iii. What is the E[x^{2}]?
E[X^{2}] = x^{2}*f(x) = 0+0.0439+0.3956+1.5822+4.3952+9.89 = 16.3069
E[X^{2}] = 16.3069
iv. What is Var(x)?
Var(x) = E[X^{2}] - (E[X])^{2}
Var(x) = 16.3069 - (3.8459)^{2}
Var(x) = 16.3069 - 14.79094681
Var(x) = 1.5159
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Fill out the table and determine the value of the following quantities, Using the following p.m.f....
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