At Midvale School for the Gifted, all pupils take an end-of-year test called the

Question

At Midvale School for the Gifted, all pupils take an end-of-year test called the Extraordinary Extension Exercise. The exam consists of ten fiendishly difficult questions. All pupils sit the same exam, but their scores tend to improve as they get older. The head teacher, Professor Partridge, decides to create a model for her pupils' performance on the EEE. There are three classes at Midvale School: class 1, 2, and 3, where class 1 contains the youngest children and class 3 contains the oldest. Let Y_x be a random variable denoting the score out of 10 that a child in class x achieves on the Extraordinary Extension Exercise, where x can be 1, 2, or 3. Professor Partridge uses the following model for Y_x: Y_x ~ Binomial (n = 10, p = 0.3x) for x = 1, 2, 3 Scores from all children in all classes are assumed to be independent. The model is illustrated in the figure overleaf, and the probability functions of Y_1, Y_2, and Y_3 are given in the table. Following fro Question 3, we wish to calculate the probability that a randomly selected child from class 2 gets a higher mark than a randomly selected child from class 3. Decide on a suitable sample space, define appropriate events, select a suitable probability rule, and hence show that the probability needed is: P(Y_2 > Y_3) = sigma^10_y = 0 P(Y_2 > y) P(Y_3 = y). Calculate this probability using the table shown in Question 3. Is it likely that a child in class 2 will do better than a child in class 3?

Explanation / Answer

Yx is a random variable denoting the score out of 10 of a child from class x where x=1,2,3

Yx~Bin(10,0.3x)

so Y1~Bin(10,0.3) Y2~Bin(10,0.6) Y3~Bin(10,0.9) independently

so P[Y1=y]=10Cy0.3y0.710-y   P[Y2=y]=10Cy0.6y0.410-y   P[Y3=y]=10Cy0.9y0.110-y

in the table there are 2 unknown quantities

p4=P[Y2=4]=10C40.640.410-4=0.11 p5=P[Y2=5]=10C50.650.410-5=0.20

so the probability that a randomly selected child from class 2 will get higher marks than a child of class 3 is

P[Y2>Y3]=P[Y3=0]*{P[Y2=1]+P[Y2=2]+.....+P[Y2=10]}+P[Y3=1]*{P[Y2=2]+P[Y2=3]+.....+P[Y2=10]}+P[Y3=2]*{P[Y2=3]+P[Y2=4]+.....+P[Y2=10]}+P[Y3=3]*{P[Y2=4]+P[Y2=5]+.....+P[Y2=10]}+P[Y3=4]*{P[Y2=5]+P[Y2=6]+.....+P[Y2=10]}+P[Y3=5]*{P[Y2=6]+P[Y2=7]+.....+P[Y2=10]}+P[Y3=6]*{P[Y2=7]+P[Y2=8]+.....+P[Y2=10]}+P[Y3=7]*{P[Y2=8]+P[Y2=9]+P[Y2=10]}+P[Y3=8]*{P[Y2=9]+P[Y2=10]}+P[Y3=9]*{P[Y2=10]}

now since P[Y3=0]=P[Y3=1]=...=P[Y3=5]=0

so from the table P[Y2>Y3]=0.01*{0.22+0.12+0.04+0.01}+0.06*{0.12+0.04+0.01}+0.19*{0.04+0.01}+0.39*0.01=0.0275 [answer]

so it is not very likely that a child in class 2 will do better than a child in class 3.

score,y 0 1 2 3 4 5 6 7 8 9 10 P[Y1=y] 0.03 0.12 0.23 0.27 0.20 0.10 0.04 0.01 0.00 0.00 0.00 P[Y2=y] 0.00 0.00 0.01 0.04 0.11 0.20 0.25 0.22 0.12 0.04 0.01 P[Y3=y] 0.00 0.00 0.00 0.00 0.00 0.00 0.01 0.06 0.19 0.39 0.35

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