For adults, intelligence scores are distributed approximately normally with μ = 100 and σ = 15.†
In each part of this question, carry out any calculations using
two places after the decimal point for z scores
and four places after the decimal point for proportions. For each
part, be sure to include an appropriate sketch
a) What proportion of intelligence scores is lower than 120?
b) What proportion of intelligence scores is higher than 128?
c) What proportion of intelligence scores is between 73 and 82?
d) What proportion of intelligence scores is between 116.5 and 117.5?
e) What proportion of intelligence scores is exactly equal to 117?
f) How high must an intelligence score be to be in the highest 2%?
g) What intelligence scores bracket the center 98%?
Solution: Given that μ = 100 and σ = 15
z scores = (x-μ)/σ
a) P(X < 120) = P(Z < (120-100)/15)
= P(Z < 1.33)
= 0.9082
b) P(X > 120) = P(Z > (128-100)/15)
= P(Z > 1.87)
= 0.0307
c) P(73 < X < 82) = P((73-100)/15 < Z <
(82-100)/15))
= P(-1.8 < Z < -1.2)
= 0.0792
d) P(116.5 < X < 117.5) = P((116.5-100)/15 < Z <
(117.5-100)/15))
= P(1.1 < Z < 1.16)
= 0.0147
e) P(X = 117) = P(116.5 < X < 117.5) = 0.0147
f) for Z = 2.05
X = μ + Z*σ = 100 + 2.05*15 = 130.75
g) for Z = +/- 2.33
x = 100 +/- 2.33*15 = (65.05,134.95)
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