Chapter 2 Solutions - 35 Chapter 2 2.1 Eleanors standardized score z = z= 27 18 = 1.5 6 680 500 = 1.8 is higher than Geralds standardized score 100 2.2

Chapter 2 Solutions - 35 Chapter 2 2.1 Eleanors...

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Chapter 2 2.3 (a) Judy’s bone density score is about one and a half standard deviations below the average score for all women her age. The fact that your standardized score is negative indicates that your bone density is below the average for your peer group. The magnitude of the standardized score tells us how many standard deviations you are below the average (about 1.5). (b) If we let denote the standard deviation of the bone density in Judy’s reference population, then we can solve for σin the equation 9489561.45σ=. Thus, 5.52σ2.4 σ ± . 35