Statistics

jam

2011-10-19 11:31 pm

There are 5 coins in a pocket, which are $1, $2, $2, $5 and $10. David will randomly select two coins from the pocket and the other three coins will be given to Tom.
(a) Prepare the probability distribution function of the total amount of money David will get.
(b) Find the expected value and varience of the amount of money David will get.
(c) Find the expected value and varience of the amount of money Tom will get.

回答 (1)

用戶10012

2011-10-21 3:12 am

✔ 最佳答案

Let X = Amount of money David will get from 2 coins.
Possible values of X, x = $3, $4, $6, $7, $11, $12 and $15.
(a) P(X = 3) = P(1st coin $1 and 2nd coin $2) or P(1st coin $2 and 2nd coin $1)
= (1/5)(2/4) + (2/5)(1/4) = 2/10.
Similarly,
P(X = 4) = 2(1/5)(1/4) = 1/10
P(X = 6) = 2(1/5)(1/4) - 1/10
P(X = 7) = 2/10
P(X = 11) = 1/10
P(X = 12) = 2/10
P(X = 15) = 1/10.
(b) Expected value = E(X) = 3(2/10) + 4(1/10) + 6(1/10) + 7(2/10) + 11(1/10) + 12(2/10) + 15(1/10) = $8.
E(X^2)) = 3^2(1/10) + 4^2(1/10) + 6^2(1/10) + 7^2(2/10) + 11^2(1/10) + 12^2(2/10) + 15^2(1/10) = 80.2
So Var(X) = 80.2 - (8^2) = 16.2
(c) Let Y = Amount of money Tom will get.
So X + Y = $20
Y = 20 - X
Expected value of Y = E(Y) = E(20 - X) = 20 - E(X) = 20 - 8 = $12.
Variance of Y = Var(Y) = Var(20 - X) = (-1)^2 Var(X) = Var(X) = 16.2

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