Act scores for 1,171,460 members of the 2004 high school graduating class who took the test closely followed the normal distribution with mean 20.9 and standard deviation 4.8. Choose two students independently and at random from this group. A) what is the expected difference in their scores b) what is the standard deviation in their scores c) find the probability that the difference in the two students scores is greater than 6. Thank you so much for your help!!
Question
Answer:
a) Let x = first score and y = second score E[x] = 20.9 and E[y] = 20.9
E[x-y] = E[x] – E[y] = 20.9 – 20.9 = 0
b) Standard deviation
= Var[x-y] = Var[x] + Var [y]
= 4.8^2 + 4.8^2 = 46.08
SD[x-y] = sqrt(Var[x-y])
= sqrt(46.08)
= 6.8
c) Z = +/- (mean-x)/SD = +- (0-6)/6.8 = +/- 0.88
From Z table: P(Z<-0.88 or Z>0.88)
= 2*P(Z>0.88)
= 2*0.1894
= 0.3788
solved
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