Assuming a normal distribution:
Hello,
I am solving the following problem: “A Salesman has scheduled two appointments to sell encyclopedias. His first appointment will lead to a sale with probability .3, and his second will lead independently to a sale with probability .6. Any sale made is equally likely to be either for the deluxe model, which costs $1000, or the standard model which costs $500. Determine the probability mass function of X, the total dollar value of all sales and determine the mean and variance of X.”
I solved:
1. P(X=0) = 0.28
2. P(X=500) = 0.27
3. P( X=1000) = 0.315
4. P( X=1500)= 0.09
5. P(X=2000)= 0.045
E[X] = $675, since 0(0.28) + 0.27(500) + (0.315)(1000) + 0.09(1500) + 0.045(2000) = $675
Var[X] = E[X2] – (E[X])^2
E[X2] = 0.28^2(0) + 0.27^2(500) + 0.315^2(1000) + 0.09^2(1500) + 0.045^2(2000) = 151.875
So, Var[X] = 151.875 – (675^2) = A ridiculously low negative number.
Obviously this is incorrect, or at least I think it is.
Could someone please tell me where my error lies?
Hi, I didn't check the probabilities but this looks wrong since you are never going to be squaring probabilities but rather the various events of X:
E[X2] = 0.28^2(0) + 0.27^2(500) + 0.315^2(1000) + 0.09^2(1500) + 0.045^2(2000) = 151.875