Is this a Course 3 problem?

[Gandalf]

I know I've seen things that look similar to this discussed in the Course 3 threads. If you know how to do it and can explain it in Course 1 terms, please help Packet Storm and me by posting your response in this Course 1 thread. Thanks.

Originally Posted by Packet_Storm, in Course 1 forum,

A player gets an incentive contract that pays:

Type of Hit:
P(Single)= 0.14
P(Double)= 0.05
P(Triple)= 0.02
P(Home Run)= 0.03

The player gets paid:
c - single
2c -Double
3c -Triple
4c -Home Run


Num at bats is Poisson w/mean = 200.

E(x) and V(x) of compensation the player will recieve.

Find c so that compensation >= 400,000 is 95%
From that c, what is the expected compensation?


I am a bit lost how to set this problem up.

[Woody]

I hope I did this right. You need to set it up realizing they are looking for Pr(S >= 400,000) = 0.95. So, to normalize it, you must find E(S) and Var(S).

E(X) = (0.14)(c) + (0.05)(2c) + (0.02)(3c) + (0.03)(4c) + (0.76)(0c) = 0.42c
E(X^2) = (0.14)(c^2) + (0.05)[(2c)^2] + (0.02)[(3c)^2] + (0.03)[(4c)^2] = c^2
Var(X) = E(X^2) - E(X) = c^2 - (0.42c)^2 = 0.8236c^2

E(N) = Var(N) = 200 (since it is Poisson)

E(S) = E(X)E(N) = (0.42c)(200) = 84c
Var(S) = Var(N)E(X^2) = 200c^2 (since it is Poisson)

So, Pr(S >= 400,000) = Pr(Z >= [400,000 - E(S)]/[Var(S)^(0.5)]) = Pr(Z >= (400,000 - 84c)/[(200c^2)^(0.5)])

(400,000 - 84c)/[(200c^2)^(0.5)] = 1.645

After doing the algebra, you get c = 3729.12.

Therefore, E(X) = (0.42)(3729.12) = 1566.23 and E(S) = E(X)(200) = 313,246.37.

[Tim><]

You made one mistake: you calculated the c that made the probabity c&lt;= 400,000 = .95.

This can be seen as your mean salary is well below 400k.

Try Mean minus 1.645 standards of distribution - this would mean that 95% of the time, the salary was higher.

Now you get 84C - 1.645 (200^.5)C = 400,000

60.736C = 400,000

C = 6,585.86

E(200X) = 553,212.21
Pr(S(200)&gt;=400,000) = .95

[Woody]

Doh! You're right J-Rab. Thanks for pointing that out.