Probability and net profit

Question

Suppose you are playing a game. The random variable, Y, represents your net profit in dollars. The probability distribution of Y is as follows. (Round your answers to 4 decimal places, if needed.)

y	-3	-2	0	3	5
P(Y = y)	0.2	0.14	?	0.22	0.1

a) What is the mean for net profit?

b) What is the median for net profit?

c) What is the standard deviation for net profit?

Answer 1

First let's find P(Y=0). This will be 1 minus the sum of the other probabilities: 1-0.66 = 0.34

A) Mean = ∑y•P(Y=y) = -.6-.28+.66+.5 = 0.28. This means your average net profit is 28 cents.

B) Median is the value of y where P(Y≤y) = 0.5. This occurs when y = 0.5 because P(Y≤-2) = 0.34 and P(Y≤0) = 0.68

C) Standard Deviation = √∑(y-μ)²•P(Y=y) = √(2.15168+0.727776+0.026656+1.627648+4.901248) = 3.071646 (Just round to the number of decimal points needed)