MATH 1040 - Introduction to Statistics

Practice Question 12.1.3

Here are three distributions. One of them is invalid. Determine which is the invalid distribution and what can be done to fix it.

Distribution 1 A B C D E F
Probability 0.24 0.28 0.24 0.19 -0.11 0.16
Distribution 2 A B C D E F
Probability 0.04 0.09 0.13 0.16 0.26 0.32
Distribution 3 A B C D E F
Probability 0.13 0.17 0.24 0.21 0.16 0.09

Answer to Practice Question 9.2.1

Let’s start by finding the sums of each distribution.

Here is the sum of Distribution 1: \(0.24+0.28+0.24+0.19+(-0.11)+0.16=1.00\)

Here is the sum of Distribution 2: \(0.04+0.09+0.13+0.16+0.26+0.32 = 1.00\)

Here is the sum of Distribution 3: \(0.13+0.17+0.24+0.21+0.16+0.09 = 1.00\)

The probabilities in all three distributions add up to 1.0. So, they all pass that test.

However, \(P(E) = -0.11\) in Distribution 1. All probabilities need to be between 0 and 1, so distribution 1 is invalid.