The College Board National Office recently reported that in 2011–2012, the 547,0
ID: 3204889 • Letter: T
Question
The College Board National Office recently reported that in 2011–2012, the 547,038 high school juniors who took the ACT achieved a mean score of 570 with a standard deviation of 110 on the mathematics portion of the test (http://media.collegeboard.com/digitalServices/pdf/research/2013/TotalGroup-2013.pdf). Assume these test scores are normally distributed. What is the probability that a high school junior who takes the test will score at least 600 on the mathematics portion of the test? If required, round your answer to four decimal places. P (x 600) = What is the probability that a high school junior who takes the test will score no higher than 490 on the mathematics portion of the test? If required, round your answer to four decimal places. P (x 490) = What is the probability that a high school junior who takes the test will score between 490 and 560 on the mathematics portion of the test? If required, round your answer to four decimal places. P (490 x 560) = How high does a student have to score to be in the top 10% of high school juniors on the mathematics portion of the test? If required, round your answer to the nearest whole number.
Explanation / Answer
From information given, mu=570, and sigma=110. Substitute the values in Z score formula to compute the z score. Next, look into Z table to find area corresponding to Z score, which gives the required probability.
P(X>=600)=1-P[Z<(600-570)/110] [Z=(X-mu)/sigma, where, X is raw score, mu is population mean, and sigma is population standard deviation]
=1-P(Z<0.27)
=1-0.6064
=0.3936 (ans)
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P(X<=490)=P[Z<=(490-570)/110]
=P(Z<=-0.73)
=0.2327 (ans)
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Compute Z scores for X1=490 and X2=560
Z1=(490-570)/110=-0.73 and Z2=(560-570)/110=-0.09
The two Z scores are of same sign, therefore, find areas between mean and respective z scores and subtract the smaller from the larger.
P(490<=X<=560)=0.2673-0.0359=0.2314 (ans)
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Using given information, one can write, P(x>=x1)=0.10, where, x1 denote the score to be obtained.
Therefore, P(x<x1)=1-0.10=0.90.
Next look into Z table to find Z score corresponding to area 0.90. Substitute the values in Z score formula to compute the raw score x1.
1..28=(x1-570)/110
x1=710.8~711 (ans)
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