Homework SourseA certain placement test has scores centered around a mean o
A certain placement test has scores centered around a mean of 200 points, with a standard deviation of about 50 points. The distribution of scores is bell-shaped. The minimum is zero and maximum is 400. What is the approximate probability a randomly selected test-taker will get under a 300 on the test? b) over a 250? c) between a 50 and a 150?, d) no more than a 100. e) better than 150?, f) between 100 and 250., g) between 150 and 300?, and h) more than 200?
Solution
A)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 300
u = mean = 200
s = standard deviation = 50
Thus,
z = (x - u) / s = 2
Thus, using a table/technology, the left tailed area of this is
P(z < 2 ) = 0.977249868 [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 = 250
u = mean = 200
s = standard deviation = 50
Thus,
z = (x - u) / s = 1
Thus, using a table/technology, the right tailed area of this is
P(z > 1 ) = 0.158655254 [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 = 50
x2 = upper bound = 150
u = mean = 200
s = standard deviation = 50
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -3
z2 = upper z score = (x2 - u) / s = -1
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.001349898
P(z < z2) = 0.158655254
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.157305356 [ANSWER]
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d)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 100
u = mean = 200
s = standard deviation = 50
Thus,
z = (x - u) / s = -2
Thus, using a table/technology, the left tailed area of this is
P(z < -2 ) = 0.022750132 [ANSWER]
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e)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 150
u = mean = 200
s = standard deviation = 50
Thus,
z = (x - u) / s = -1
Thus, using a table/technology, the right tailed area of this is
P(z > -1 ) = 0.841344746 [ANSWER]
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f)
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 100
x2 = upper bound = 250
u = mean = 200
s = standard deviation = 50
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -2
z2 = upper z score = (x2 - u) / s = 1
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.022750132
P(z < z2) = 0.841344746
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.818594614 [ANSWER]
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g)
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 150
x2 = upper bound = 300
u = mean = 200
s = standard deviation = 50
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -1
z2 = upper z score = (x2 - u) / s = 2
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.158655254
P(z < z2) = 0.977249868
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.818594614 [ANSWER]
*****************
h)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 200
u = mean = 200
s = standard deviation = 50
Thus,
z = (x - u) / s = 0
Thus, using a table/technology, the right tailed area of this is
P(z > 0 ) = 0.5 [ANSWER]
