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 -1 3 5
P(Y = y) 0.2 0.19 ? 0.22 0.15

(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

To find the missing probability in the probability distribution, we know that the sum of all the probabilities must equal 1.

Sum of probabilities = 0.2 + 0.19 + x + 0.22 + 0.15 = 1
0.76 + x = 1
x = 1 - 0.76
x = 0.24

(a) The mean (or expected value) for net profit can be found by multiplying each net profit value by its corresponding probability, and then summing them up.

Mean = (-3 x 0.2) + (-2 x 0.19) + (-1 x 0.24) + (3 x 0.22) + (5 x 0.15)
Mean = -0.6 - 0.38 - 0.24 + 0.66 + 0.75
Mean = 0.19

Therefore, the mean for net profit is $0.19.

(b) To find the median for net profit, we need to arrange the net profit values in order and find the middle value.

-3, -2, -1, 3, 5

The median is the value in the middle, which is -1.

Therefore, the median for net profit is $-1.

(c) The standard deviation for net profit can be calculated using the following formula:

Standard deviation = sqrt(Sum of [(xi - mean)^2 * P(xi)])

Using the mean value of 0.19:

Standard deviation = sqrt((-3 - 0.19)^2 x 0.2 + (-2 - 0.19)^2 x 0.19 + (-1 - 0.19)^2 x 0.24 + (3 - 0.19)^2 x 0.22 + (5 - 0.19)^2 x 0.15)

Standard deviation = sqrt((13.0191 x 0.2) + (4.3169 x 0.19) + (0.0441 x 0.24) + (6.5791 x 0.22) + (20.7361 x 0.15))

Standard deviation = sqrt(2.60382 + 0.819311 + 0.010584 + 1.447802 + 3.110415)

Standard deviation = sqrt(7.991928)

Standard deviation = 2.8289

Therefore, the standard deviation for net profit is $2.8289.