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Expectations

Theory and Formula


Question



Q1. There are six slips in a box and numbers 1,1,2,2,3,3 are written on these slips . Two slips are taken at random from the box . The expected values of the sum of numbers on the two slips is :

(a) 5

(b) 3

(c) 4

(d) 6

Solution.
Total number of combination 6C2 = 15

(1,1)(1,2)(1,2)(1,3)(1,3)

(2,1)(2,1)(2,2)(2,3)(2,3)

(3,1)(3,1)(3,2)(3,2)(3,3)
x P(x) x.P(x)
2 1/15 2/15
3 4/15 12/15
4 5/15 20/15
5 4/15 20/15
6 1/15 6/15
Expected values of the sum of numbers on the two slips = E(x) = ∑xp(x) =
60/15
= 4
Solution

Q2. Mathematical expectation is

(a) Independent of change of origin and scale

(b) Dependent on change of origin and scale

(c) Independent of origin but not of scale

(d) None of these .

Solution.b
Dependent on change of origin and scale
Solution

Q3. x and y are two independent random variables with mean 6 and 3 respectively . Find the expected value of (6x + 3y)

(a) 20

(b) 30

(c) 45

(d) 33

Solution. c
E(6x + 3y) = 6E(x) + 3E(y)

= 6x6+3x3 = 36 + 9 = 45
Solution

Q4. The mean and variance of a random variable x are 3 and 4 respectively . find the S.D of (2x - 7).

(a) 4

(b) 3

(c) 2

(d) 1

Solution.a
= v(2x - 7)

= v(2x) - v(7)

= 4v(x) ∴ v(7) = 0

= 4 x 4

= 16

So S.D is 4
Solution

Q5. Vaibhav plays a game of tossing a dice . If the number less than 3 appears , he is getting Rs. a , otherwise he has to pay Rs, 10 . If the game is fair , find a :

(a) 28

(b) 22

(c) 20

(d) 15

Solution.c
Game is fair if E(x) = 0
x P(x)
a 2/6
-10 4/6
E(x) = ∑ xP(x) =
2a/6
-
40/6


2a/6
-
40/6
= 0 ⇒ 2a = 40

a = 20
Solution