Question

PROBABILITY

2. Given the following table

x	0	1	2	3
Pr(X=x)	0.25	0.4	0.3	0.05

answer each of the questions.

Find µ_X and σ_X .

Given that

                          Y = -X+2

find µ_Y and σ_Y .

2B. Suppose that in the game of AFL the mean total number of points per match is 185 with a standard deviation of 33. Suppose that a season involves 202 matches. If the total points are normally distributed then answer the following questions.

1. How many games do you expect to have less than 100 points scored in them in a given season?

2. How many games do you expect to have between 200 and 220 points scored in them in a season?

3. How many points should be needed for a match to be in the top 5% of high scoring matches?

4. If a match is one of the bottom 5 lowest scoring matches of the season, what is an upper bound on what you expect the score to be?

2C. Suppose that asteroid impacts on the earth are modelled with a Poi-son distribution. Asteroids of diameter 4m are estimated to enter the Earth’s atmosphere once per year. Asteroids of diameter 1km are estimated to impact the Earth once every 500,000 years.

1. What is the probability of exactly 2 asteroids of 4m diameter entering the atmosphere in a 6 month period?

2. How many years are required for the probability of at least one 1km diameter impact to exceed 1%?

Answer 1

Question 2

(a)  µ_X = 0 * 0.25 + 1 * 0.4 + 2 * 0.3 + 3 * 0.05 = 1.15

and σ²_X = 0.25 * (0 - 1.15)² + 0.4 * (1 - 1.15)² + 0.3 * (2 - 1.15)² + 0.05 * (3 - 1.15)² = 0.7275

So, σ_X = 0.853

if

Y = -X + 2

µ_Y = - µ_X + 2 = -1.15 + 2 = 0.85

Question 2

Mean total number of pointes in matches in a season = 185

standard deviation = 33

season involves number of matches = 202

(a) if x is the number of points scored in a random match

Pr(x < 100 ; 185 ; 33)

Z = (100 - 185)/33 = -2.578

from Z table

Pr(x < 100 ; 185 ; 33) = Pr(Z < -2.578) = 0.005

expected maches to have less than 100 points scored in them in a given season = 0.005 * 202 = 1

(2) Pr(200 < X <220) = Pr(X <220 ; 185 ; 33) - Pr(X < 200 ; 185 ; 33)

Z₂ = (220 - 185)/33 = 1.0606

Z₁ = (200 - 185)/33 = 0.4545

Pr(200 < X <220) = Pr(X <220 ; 185 ; 33) - Pr(X < 200 ; 185 ; 33) = Pr(Z < 1.0606) - Pr(Z < 0.4545)

= 0.8556 - 0.6753 = 0.1803

expected number of games to have between 200 and 220 points scored in them in a season = 0.1803 * 202 = 36.42 or 36

(3) here the top 5% score is if X₀

Pr(x < x₀) = 1 - 0.05 = 0.95

Z score = 1.645

(X₀ - 185)/33 = 1.645

X₀ = 185 + 33 * 1.645 = 239.29

(4) Here the bottom 5 lowest scoring matches would be

Pr(x < x0) = 0.025

Z = -1.96

so upper bound = 185 - 33 * 1.96 = 120.32

2C

(1) expected number of astreoids of 4m diameter to enter the atmosphere in 6 month period = 1/2 = 0.5

POISSON (x = 2) = e^-0.5 0.5²/ 2! = 0.0758

(2) Here letssay the number of years are atleast x₀

so,

writing the question in mathematical terms.

Pr(x >= 1; = x₀/500,000) > 0.01

or

Pr(x = 0; = x₀/500,000) < 0.99

e^-x0/500000 < 0.99

x₀/500,000 >ln (0.99)

x₀   > 5025.17 years