Homework SoursePlease explain 5 The scores on a college aptitude test foll
Please explain :)
5. The scores on a college aptitude test follow N(490, 70). Use the “NORM.DIST” function in Excel and
properties of areas to answer the following questions. Give your answers as percentages rounded
to one decimal place.
a) What percentage of test takers will have scores below 400?
b) What percentage of scores will be above 550?
c) What proportion of test takers will score between 473 and 571?
d) Suppose 7,500 college applicants took this aptitude test. Approximately how many of these
would you expect to have scores between 473 and 571?
Solution
A)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 400
u = mean = 490
s = standard deviation = 70
Thus,
z = (x - u) / s = -1.285714286
Thus, using a table/technology, the left tailed area of this is
P(z < -1.285714286 ) = 0.099271397 [ANSWER]
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b)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 550
u = mean = 490
s = standard deviation = 70
Thus,
z = (x - u) / s = 0.857142857
Thus, using a table/technology, the right tailed area of this is
P(z > 0.857142857 ) = 0.195682969 [ANSWER]
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c)
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 473
x2 = upper bound = 571
u = mean = 490
s = standard deviation = 70
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -0.242857143
z2 = upper z score = (x2 - u) / s = 1.157142857
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.404058033
P(z < z2) = 0.876393
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.472334967 [ANSWER]
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d)
There are approximately 0.472334967*7500 = 3542.512253 students. [ANSWER, 3542.512253]
