Stanford–Binet IQ Test scores are normally distributed with a mean score of 100
ID: 3022598 • Letter: S
Question
Stanford–Binet IQ Test scores are normally distributed with a mean score of 100 and a standard deviation of 16.
Write the equation that gives the z score corresponding to a Stanford–Binet IQ test score.
Find the probability that a randomly selected person has an IQ test score. (Round answers to 4 decimal places.)
Suppose you take the Stanford–Binet IQ Test and receive a score of 136. What percentage of people would receive a score higher than yours? (Answer as a percentage, rounded to two decimal places, without the percentage (%) symbol.)
(b)Write the equation that gives the z score corresponding to a Stanford–Binet IQ test score.
Explanation / Answer
b)
z = (x - u)/o [ANSWER]
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c)
1.
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 140
u = mean = 100
s = standard deviation = 16
Thus,
z = (x - u) / s = 2.5
Thus, using a table/technology, the right tailed area of this is
P(z > 2.5 ) = 0.006209665 [ANSWER]
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2.
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 88
u = mean = 100
s = standard deviation = 16
Thus,
z = (x - u) / s = -0.75
Thus, using a table/technology, the left tailed area of this is
P(z < -0.75 ) = 0.226627352 [ANSWER]
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3.
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 72
x2 = upper bound = 128
u = mean = 100
s = standard deviation = 16
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -1.75
z2 = upper z score = (x2 - u) / s = 1.75
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.040059157
P(z < z2) = 0.959940843
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.919881686 [ANSWER]
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4.
z1 = lower z score = -1.5
z2 = upper z score = 1.5
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.066807201
P(z < z2) = 0.933192799
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.866385597 [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 = 136
u = mean = 100
s = standard deviation = 16
Thus,
z = (x - u) / s = 2.25
Thus, using a table/technology, the right tailed area of this is
P(z > 2.25 ) = 0.012224473 = 1.22% [ANSWER]
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