ACT, Inc., the producer of the ACT test of readiness for college work, also prod

Question

ACT, Inc., the producer of the ACT test of readiness for college work, also produces tests for 8th and 9^th grade students designed to help them to plan for the future. Two of these tests measure reading and mathematics achievement. Each has scores that range from 1 to 25. For students tested in the fall of their 8^th grade year, the reading test has mean 13.9 and standard deviation 3.63. the mean score on the math test is 14.4 and the standard deviation is 3.46. Scores are scaled to follow roughly a normal distribution. If a student's reading score X and math score Y were independent, what would be the distribution of the total X + Y? Using the distribution found in (a), what percent of 8^th graders have a total score of 30 or higher? in fact, the scores X and Y are not independent - they are strongly correlated. in this case, does the mean of X + Y still have the value you found in (a)? Does the standard deviation still have the value you found in (a)? Would you expect the true percentage of students with a total score of 30 or higher to be larger or smaller than what you found in (b)? What additional information would be required for us to be able to compute the actual probability? Clearly explain each of your answers.

Explanation / Answer

a) reading X - 13.9 , 3.63

Math Y - 14.4 ,3.46

X +Y - N( 13.9+14.4 , 3.63^2 + 3.46^2)

N( 28.3,25.1485)

mean = 28.3 ,variance = 25.1485 ,sd =5.01482801

b) P(X+Y > 30)

P(Z > (30 -28.3)/5.01482801)

=P(Z> 0.33899)

= 0.3673

c) mean will remain same .

sd will change as cov (X,Y) is not zero.

Var(X +Y ) = Var(X)+Var(Y ) + 2 Cov(X, Y ).

since data are highly co-related , cov(X,Y) > 0

hence var (X+Y) is greater in this case

hence z-score will decrease , so P(Z > z*) will increase

we require cov(X,Y) in this case to compute actual probability.

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