9. Populations and Probabilities

Question 1

(a) This is a valid probability model since all probabilities are nonnegative and sum to 1.

(b) This is not a valid probability model since P(C) is negative, even though the total is 1.

(c) This is not a valid probability model since the sum of the probabilities is 1.6, not 1.

(d) This is not quite a valid probability model since the sum of the probabilities is 0.99, not 1. To accurately assign equal probabilities fractions (1/3) should be used.

 

Question 2

The surprisal is -log₂(p) = 2.5. Rearranging gives p = 2^-2.5 = 0.1768.

 

Question 3

From Section 9.5 we know the height of the uniform density is 1/40. The probability to the right of a particular value x is

P(X ≥ x) = base × height = (190 - x) × (1/40).

We want this to be p. Solving gives 40p = 190 - x, so x = 190 - 40p.

For example, how tall would you be if you were in the top 10% of heights in this distribution? Substituting p = 0.1 into the formula gives x = 190 - 40(0.1) = 186, so the top 10% of heights consists of females taller than 186 cm.

 

Question 4

This question is much harder than the previous one. From Section 9.5 we know the height of the triangular density is 1/20. If x is 170 or more then the height of the density curve at x is (1/20)×((190 - x)/20). You can check that this gives height 0 for x=190 and height 1/20 for x=170. The area to the right of x is then

P(X ≥ x) = (1/2) × base × height = (1/2) × (190 - x) × (1/20)×((190 - x)/20) = (1/800)×(190 - x)².

We want this to be p. Solving gives 800p = (190 - x)², so x = 190 - √(800p).

For example, how tall would you be if you were in the top 10% of heights in this distribution? Substituting p = 0.1 into the formula gives x = 190 - √(80) = 181.1, so the top 10% of heights consists of females taller than 181.1 cm.

This formula only works for x ≥ 170 (p ≤ 0.5). For x ≤ 170 (p ≥ 0.5) the area to the right is

P(X ≥ x) = 1 - (1/800)×(x - 150)².

Solving this to get x in terms of p gives x = 150 + √(800(1 - p)).