Need help on stats (probability)?

Question

This is the problem:
I am the car salesperson.
My probability of selling a car is 1/10
n is the number of time I try to sell the car.
i.e. If I have to talk to 3 persons to sell a car n=3 (fail the first 2 attempts and succeed on the 3rd attempt)
p(3) should be (9/10)^2(1/10)
Find E(n) and Var(n).

Thank you.

Also I calculated E(n^3) to be 5950 but the right answer is 5410.
Could somebody help me on this also.
Thank you so much.

Answer 1

Your talking about a geometric distribution. n is the number of trials it takes to get to a success where the trials are independent.

If X has a geometric distribution with probability of success p, then E(X) = 1/p and Var(X) = (1-p)/p^2. Since p = 1/10 in this case, then

E(n) = 1/(1/10) = 10.
Var(n) = (1-1/10)/(1/10)^2
= (9/10)/(1/100)
= 90.

edit: The skewness of a geometric is

(2-p)/sqrt(1-p)

So applying that to this,

skewness = 1.9/sqrt(0.9)

skewness is calculated as E(n-mu)^3/Var(n)^(3/2)

so

E(n-mu)^3 = E(n^3 - 3*mu*n^2 + 3*n*mu^2 - mu^3)
= E(n^3) - 3*mu*E(n^2) + 3*mu^2*E(n) - mu^3
= E(n^3) - 3*10*(90+10^2) + 3*10^3 - 10^3
= E(n^3) - 5700 + 3000 - 1000
= E(n^3) - 3700

Var(n)^(3/2) = 90^(3/2)
= sqrt(729000)
= 270*sqrt(10)

So

1.9/sqrt(0.9) = (E(n^3) - 3700)/(270sqrt(10))

E(n^3) - 3700 = 1710

E(n^3) = 5410

Whew! Lucky I checked back, huh?

Answer 2

you are right about p(3)

This is a geometric distribution where there is a probability to sell the car on every attempt, but no limitations on attempts until a success

by definition E(n) = 1/p (in this case p = 1/10) = 10
and Var(n) = (1-p)/p^2 = (1-.1)*100 = 90

Answer 3

Are you choosing with replace, or choosing from 5, then 4, then 3? If 40% of the 5 automobiles are defective, then 2 are defective and 3 at the instant are not defective. with out changing the automobiles, your hazard is 0. With replace, that's 0.40^3 = 0.064