In the answers below, the phrase "area of [a,b]" is short for "area under the normal curve for the bin [a,b]."
a. (-∞, -2.00] Ans: 0.0288 b. (-∞, -3.00] Ans: 0.0013 c. (-∞, 0.00] Ans: 0.5000
d. (-∞, 1.00] Ans: 0.8413 e. (-∞, 3.00] Ans: 0.9987 f. (-∞, 1.64] Ans: 0.9495
g. (-∞, -0.72] Ans: 0.2358 h. (-∞, ∞) Ans: 1.000
i. [2.87, ∞) = 1 - area of (-∞, 2.87] = 1 - 0.9972 = 0.0027
j. [0.78, ∞) = 1 - area of (-∞, -0.78] = 1 - 0.2177 = 0.7823
k. [-1.12, ∞) = 1 - area of (-∞, -1.12] = 1 - 0.1214 = 0.8786
l. [-0.04, ∞) = 1 - area of (-∞, -0.04] = 1 - 0.4880 = 0.5120
m. [-1.00, 1.00] = area of (-∞, 1.00] - area of (-∞, -1.00] = 0.8413 - 0.1587 = 0.6826
n. [-2.00, 2.00] = area of (-∞, 2.00] - area of (-∞, -2.00] = 0.9772 - 0.0228 = 0.9544
o. [-3.00, 3.00] = area of (-∞, 3.00] - area of (-∞, -3.00] = 0.9987 - 0.0013 = 0.9974
p. [-3.49, 3.49] = area of (-∞, 3.49] - area of (-∞, -3.49] = 0.9998 - 0.0002 = 0.9996
q. [-1.54, 2.45] = area of (-∞, 2.45] - area of (-∞, -1.54] = 0.9946 - 0.0618 = 0.9828
r. [0.23, 2.09] = area of (-∞, 2.09] - area of (-∞, 0.23] = 0.9817 - 0.5910 = 0.3907
s. [-0.95, 1.37] = area of (-∞, 1.37] - area of (-∞, -0.95] = 0.9147 - 0.1711 = 0.7436
t. [-3.32, -3.09] = area of (-∞, -3.09] - area of (-∞, -3.32] = 0.0010 - 0.0005 = 0.0005
a. 0.138 Ans: (-∞, -1.09] b. 0.376 Ans: (-∞, -0.31] c. 0.500 Ans: (-∞, 0.00]
d. 0.652 Ans: (-∞, 0.39] e. 0.998 Ans: (-∞, 2.88]
a. 0.007 Ans: area of [-∞, 2.46] = 1 - 0.007 = 0.993, so area of [2.46, ∞) = 1 - 0.993 = 0.007.
b. 0.287 Ans: area of [-∞, 0.56] = 1 - 0.287 = 0.713, so area of [0.56, ∞) = 1 - 0.713 = 0.287.
c. 0.701 Ans: area of [-∞, -1.97] = 1 - 0.701 = 0.309, so area of [-1.97, ∞) = 1 - 0.309 = 0.701.
d. 0.998 Ans: area of [-∞, -3.49] = 1 - 0.998 = 0.002, so area of [-3.49, ∞) = 1 - 0.002 = 0.998.
a. 0.137
Ans: If area of [-z, z] = 0.137, then area of (-∞, -z] and [z, -∞) together is 1 - 0.0137 = 0.9863. Also area of (-∞, -z] = area of [z, ∞), so area of (-∞, -z] = 0.9863 / 2 = 0.4931. Therefore z = 0.02.
b. 0.379
Ans: By the same reasoning as Part a, if the area of [-z, z] is 0.379, the area of (-∞, -z] = (1 - 0.379) / 2 = 0.3105, so z = 0.52.
c. 0.652
Ans: By the same reasoning as Part a, if the area of [-z, z] is 0.652, the area of (-∞, -z] = (1 - 0.652) / 2 = 0.174, so z = 0.92.
d. 0.992
Ans: By the same reasoning as Part a, if the area of [-z, z] is 0.992, the area of (-∞, -z] = (1 - 0.992) / 2 = 0.004, so z = 3.38
Ans: We want the area of [-z, z] to be 0.5000, so the area of (-∞, -z] = (1 - 0.5000) / 2 = 0.25. This means that z = 0.67; Q1 = -0.67, Q3 = 0.67. IQR = 0.67 - (-0.67) = 1.34
Ans: The inner fence above Q3 is at Q3 + 1.5 * IQR = 0.37 + 2.01 = 2.38. The inner fence below Q1 is at -2.38.
Ans: The area of (-∞, -2.38] is 0.0087 = area of [2.38, ∞), so the proportion of mild outliers is 2 * 0.0087 = 0.0174.
b. 85 Ans: area of (-∞, z] = 0.85, so z = 1.04. This means that the IQ score is 1.04 SDs above average: the IQ score is 100 + 1.04 * 15 = 115.6
c. 95 Ans: area of (-∞, z] = 0.95, so z = 1.64. This means that the IQ score is 1.64 SDs above average: the IQ score is 100 + 1.64 * 15 = 126.6
d. 99.9 Ans: area of (-∞, z] = 0.999, so z = 3.08. This means that the IQ score is 3.08 SDs above average: the IQ score is 100 + 3.08 * 15 = 146.2