a) 16% b) 97.5% c) 80%
Problem 2.
a) 10% b) 60% c) 1%
Problem 3.
a) -0.67 b) 0.67 c) -0.13 d) 2.05
Problem 4.
a) 2.05 b) 0.00 c) -0.67 d) 1.28
Problem 5.
a) 84 - 16 = 68% b) 96 - 10 = 86% c) 84 - 50 = 34%
Problem 8.
a) The standardized score is -0.41, which is the 34th percentile.
b) The standardized score is -0.5, which is the 31st percentile.
c) The standardized score is 1.2, which is approximately th 88.5th percentile.
Problem 9.
a) By Table 8.1 you need a standardized score of at least 2.05, which corresponds to an IQ of 100 + 2.05*16 = 132.8.
b) Again, you need a standardized score of 2.05 or a GRE score of at least 497 + 2.05*497 = 732.75
Problem 12.
a) -0.67 to + 0.67
b) The range from -0.67 to + 0.67 (one IQR) covers 1.34 standard deviations.
c) The IQR covers 1.34 standard deviations, so 2*IQR is 2.68 standard deviations. The standardized score of the lower whisker, -2.68, corresponds to about the 0.4 percentile and standardized score of the upper whisker, 2.68, corresponds to the 99.6th percentile.
Problem 14.
They should accept students with a standardized score of 0.52 or higher, corresponding to a GRE score of 497 + (0.52)*(115) = 556.8 or higher.
Problem 17.
Notice that 50% is two standard deviations above the mean, or a standardized score of 2.00, which is about the 97.75th percentile. So about 2.25% or between 2% and 2.5% of such surveys, should show a majority favoring the candidate.