These are the solutions to the standard normal distribution exercises. You are strongly advised to work out your own solutions before you look at these.
Suppose the heights of adult females is normally distributed with a mean of 66 inches and a standard deviation of 1.75 inches. Suppose that the heights of adult males is normally distributed with a mean of 70 inches and a standard deviation of 2.2 inches. Jill is 70 inches tall and Jack is 75 inches tall. Which person is taller relative to those of their same gender?
Jill:
Jack:
Relatively speaking, Jill is taller than Jack because her standard score is greater.
Suppose the average height (in inches) of females is 66 with a standard deviation of 1.75.
Find the standard scores corresponding to the following female heights:
A. x = 69 inches
z = (x - mean) / standard deviation = (69 - 66) / 1.75 = 1.71
B. x = 63 inches
z = (x - mean) / standard deviation = (63 - 66) / 1.75 = -1.71
C. x = 5 feet
z = (x - mean) / standard deviation = (60 - 66) / 1.75 = -3.43
Find the following probabilities:
D. P(x <= 66)
z = (x - mean) / standard deviation = (66 - 66) / 1.75 = 0
P(x <= 66) = P(z <=0) = 0.5000E. P(x >= 70)
z = (x - mean) / standard deviation = (70 - 66) / 1.75 = 2.29
P(x >= 70) = P(z >= 2.29) = 0.0110F. P(65 < x < 67)
z1 = (x - mean) / standard deviation = (65 - 66) / 1.75 = -0.57
z2 = (x - mean) / standard deviation = (67 - 66) / 1.75 = 0.57
P(65 < x < 67) = P(-0.57 < z < 0.57) = 0.4314
Suppose the average height of males is 70 inches with a standard deviation of 2.2 inches.
Find the standard scores corresponding to the following male heights:
A. x = 76
z = (x - mean) / standard deviation = (76 - 70) / 2.2 = 2.73
B. x = 69
z = (x - mean) / standard deviation = (69 - 70) / 2.2 = -0.45
C. x = 6 feet
z = (x - mean) / standard deviation = (72 - 70) / 2.2 = 0.91
Find the following probabilities:
D. P(x < 65)
z = (x - mean) / standard deviation = (65 - 70) / 2.2 = -2.27
P(x < 65) = P(z < -2.27) = 0.0116E. P(67 < x < 73)
z = (x - mean) / standard deviation = (67 - 70) / 2.2 = -1.36
z = (x - mean) / standard deviation = (73 - 70) / 2.2 = 1.36
P(67 < x < 73) = P(-1.36 < z < 1.36) = 0.8262F. P(x >= 71)
z = (x - mean) / standard deviation = (71 - 70) / 2.2 = 0.45
P(x >= 71) = P(z >= 0.45) = 0.3264
Suppose the average IQ score is 110 with a standard deviation of 11.
Find the standard scores corresponding to the following IQ scores:
A. x = 93
z = (x - mean) / standard deviation = (93 - 110) / 11 = -1.55
B. x = 105
z = (x - mean) / standard deviation = (105 - 110) / 11 = -0.45
C. x = 110
z = (x - mean) / standard deviation = (110 - 110) / 11 = 0.00
D. x = 129
z = (x - mean) / standard deviation = (129 - 110) / 11 = 1.73
Find the following probabilities:
E. P(x < 90)
z = (x - mean) / standard deviation = (90 - 110) / 11 = -1.82
P(x < 90) = P(z < -1.82) = 0.0344F. P(x >= 120)
z = (x - mean) / standard deviation = (120 - 110) / 11 = 0.91
P(x >= 120) = P(z >= 0.91) = 0.1814G. P(115 <= x <= 140)
z = (x - mean) / standard deviation = (115 - 110) / 11 = 0.45
z = (x - mean) / standard deviation = (140 - 110) / 11 = 2.73
P(115 <= x <= 140) = P(0.45 <= z <= 2.73) = 0.3232
