Page 154 - DMTH404_STATISTICS

Statistics

Notes Thus, E(X) = 6000 - 4E(P) = 6000 - 4 ´ 846.25 = 2615 fans.

And E(TR) = 6000E(P) - 4E(P )
2
= 6000 ´ 846.25 - 4 ´ 717031.25 = Rs 22,09,375.00

Example 13: A person applies for equity shares of Rs 10 each to be issued at a premium
of Rs 6 per share; Rs 8 per share being payable along with the application and the balance at the
time of allotment. The issuing company may issue 50 or 100 shares to those who apply for 200
shares, the probability of issuing 50 shares being 0.4 and that of issuing 100 shares is 0.6. In either
case, the probability of an application being selected for allotment of any shares is 0.2 The
allotment usually takes three months and the market price per share is expected to be Rs 25 at the
time of allotment. Find the expected rate of return of the person per month.
Solution.
Let A be the event that the application of the person is considered for allotment, B be the event
1
that he is allotted 50 shares and B be the event that he is allotted 100 shares. Further, let R
2 1
denote the rate of return (per month) when 50 shares are allotted, R be the rate of return when
2
100 shares are allotted and R = R + R be the combined rate of return.
1 2
We are given that P(A) = 0.2, P(B /A) = 0.4 and P(B /A) = 0.6.
1 2
(a) When 50 shares are allotted
The return on investment in 3 months = (25 - 16)50 = 450
450
\ Monthly rate of return = = 150
3
The probability that he is allotted 50 shares
= P A b B g b g. 1 / Ag = 02 04 = 008
= P A P B b
.
.
´
.
1
Thus, the random variable R takes a value 150 with probability 0.08 and it takes a value 0
1
with probability 1 - 0.08 = 0.92
\ E(R ) = 150 × 0.08 + 0 = 12.00
1
(b) When 100 shares are allotted
The return on investment in 3 months = (25 - 16).100 = 900
900
\ Monthly rate of return = = 300
3
The probability that he is allotted 100 shares
= P A b = P A P B b Ag
´
.
.
.
B g b g.
2 2 / = 02 06 = 012
Thus, the random variable R takes a value 300 with probability 0.12 and it takes a value 0
2
with probability 1 - 0.12 = 0.88
\ E(R ) = 300 ´ 0.12 + 0 = 36
2
Hence, E(R) = E(R + R ) = E(R ) + E(R2) = 12 + 36 = 48
1 2 1
Example 14: What is the mathematical expectation of the sum of points on n unbiased
dice?

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