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建立人际资源圈Probability_Distribution_of_a_Random_Variable
2013-11-13 来源: 类别: 更多范文
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.
