Probability Distribution of a Random Variable

Problem 5.12:   Suppose that the probability distribution of a random variable x can be described by the formula p(x) = x/15; for each of the values x = 1, 2, 3, 4, and 5.   For example, then, P(x = 2) = p(2) = 2/15.
a. Write out the probability distribution of x.   1/15, 2/15, 3/15, 4/15, 5/15
b. Show that the probability distribution of x satisfies the properties of a discrete probability distribution.   1. p(x) is greater than or equal to zero for each value of x; and
2. 1/15 + 2/15 + 3/15 + 4/15 + 5/15 = 15/15 = 1.
c. Calculate the mean of x.   1(1/15) + 2(2/15) + 3(3/15) + 4(4/15) + 5(5/15) = 3.67
d. Calculate the variance, 2x, and the standard deviation, x.
Data -Mean Deviation from Mean Deviations Squared
0.066667 3.66667 -3.600003 12.9600216
0.133333 3.66667 -3.533337 12.48447036
0.2 3.66667 -3.46667                 12.01780089
0.266667 3.66667 -3.400003 11.5600204
0.333333 3.66667 -3.333337 11.11113556
                60.1334488 Total of squared deviations
                / 4 Divided by number of data samples less 1
                15.03336 Equals the variance

              15.03336
                = 3.877 Calculate the positive square root of the variance to find the standard deviation

Bowerman, B. L., O’Connell, R. T., Orris, J. B., Murphree, E. S. (2010). Essentials of business statistics. McGraw-Hill Primis Custom Publishing, 38503.