The Mean Square Terrors team of the Statistics Department at School A has been c

Question

The Mean Square Terrors team of the Statistics Department at School A has been competing in a soccer league for the past few years. The team is known to be a very defensive team, with goal scoring occurring to a Poisson process with a rate of 2 goals per hour.

a. What is the probability that the team scores the first goal in less than 15 minutes into a soccer game?

b. As an incentive to score more goals in their next match, the Statistics Department has promised to reward the team if they can score 3 or more goals within 90 minutes of a soccer game. What is the probability that the team wins the reward promised by the department?

c. The Mean Square Terrors team plays a visiting team from another university next week. Suppose that this visiting team has goal scoring that occurs to a Poisson process with rate 1.5 goals per hour. Find the probability that in this 90-minute game, the third goal scored in the game takes place within the first 45 minutes. State any assumption that you need in your calculation.

Explanation / Answer

Solution

Let X = Number of goals scored by the MST Team in one hour. Then, the given information implies that X ~ Poisson (?), where ? is known to be 2 per hour……….(A)

Back-up Theory

If a random variable X ~ Poisson(?), i.e., X has Poisson Distribution with mean ? then

probability mass function (pmf) of X is given by P(X = x) = e – ?.?x/(x!) …………..(1)

where x = 0, 1, 2, ……. , ?

Values of p(x) for various values of ? and x can be obtained by using Excel Function, POISSON(x,Mean,Cumulative)

If X = number of times an event occurs during period t, Y = number of times the same event occurs during period kt, and X ~ Poisson(?), then Y ~ Poisson (k?) …………….. (2)

Mean = ?

If X1 ~ Poisson (?1), X2 ~ Poisson (?2) and X1, X2 are independent, then

(X1 + X2) ~ Poisson (?1 + ?2)………………………………………………………..(3)

Part (a)

Let Y = Number of goals scored by the MST Team in 15 minutes. Then, vide (A) and (2),

Y ~ Poisson (0.5)

So, probability that the team scores the first goal in less than 15 minutes into a soccer game

= P(Y ? 1)

= 1 – P(Y = 0)

= 1 – 0.778801

= 0.221199 ANSWER

Part (b)

Let Z = Number of goals scored by the MST Team in 90 minutes. Then, vide (A) and (2),

Z ~ Poisson (3)

So, probability that the team wins the reward promised by the department

= P(Z ? 3)

= 1 – P(Z ? 2)

= 1 – 0.42319

= 0.57681 ANSWER

Part (c)

Let W = Number of goals scored by the Visiting Team in 45 minutes. Then, given ‘visiting team has goal scoring that occurs to a Poisson process with rate 1.5 goals per hour’ and vide (2), W ~ Poisson (1.125).

Also, if Z1 = Number of goals scored by the MST Team in 45 minutes. Then, vide (A) and (2), Z1 ~ Poisson (1.5).

Assuming the scoring of the two teams are independent, S = (Z1 + W) ~ Poisson (2.625)

So, probability that in this 90-minute game, the third goal scored in the game takes place within the first 45 minutes

= P(S ? 3)

= 1 – P(S ? 2)

= 1 – 0.512172

= 0.487828 ANSWER

As already stated, we assume, the scoring of the two teams are independent.

DONE

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