Answer
A. Suppose the standardized μ is 62 and a standard deviation of σ =14,
a. X = 74, M = 82, σ = 8
z =$\frac{X-μ}{σ}$ =$\frac{74-82}{8}$ =$\frac{-8}{8}$ = -1.0
Z =-1.0 indicates that it is located below the mean by exactly 1 standard deviation, and this location corresponds to X = 62-14 = 48 (below the mean by 14 points).
X = 40,μ = 50, σ = 20
z =$\frac{X-μ}{σ}$ =$\frac{40-50}{20}$ =$\frac{-10}{20}$ = -0.5
Z =-0.5 indicates that it is located below the mean by 0.5 standard deviation.and this location corresponds to X = 62 - 7 = 55 ( below the mean by 7 points).
This means that a score of X = 40 on an exam with μ = 50 and σ = 20 will lead to better grades.
B. Suppose the standardized μ is 57.5 and a standard deviation of σ =11,
X = 51, μ = 45, σ = 2
z =$\frac{X-μ}{σ}$ =$\frac{51-45}{2}$ =$\frac{6}{2}$ = +3
Z =3 indicates that it is located above the mean by exactly 3 standard deviation, and this location corresponds to X = 57.5+33 = 90.5 (above the mean by 33 points).
X = 90, μ = 70, σ =20
z =$\frac{X-μ}{σ}$ =$\frac{90-70}{20}$ =$\frac{20}{20}$ =+1
Z =1 indicates that it is located above the mean by exactly 1 standard deviation, and this location corresponds to X = 57.5+11 = 68.5 (above the mean by 11 points).
This means that a score of X = 51 on an exam with μ = 45 and σ = 2 will lead to better grades.
C. Suppose the standardized μ is 35 and a standard deviation of σ =5,
X = 62, μ = 50, σ =8
z =$\frac{X-μ}{σ}$ =$\frac{62-50}{8}$ =$\frac{12}{8}$ =1.5
Z =1.5 indicates that it is located above the mean by exactly 1.5 standard deviation, and this location corresponds to X = 35+12 =47 (above the mean by 12 points).
X = 23, μ = 20, σ = 2
z =$\frac{X-μ}{σ}$ =$\frac{23-20}{2}$ =$\frac{3}{2}$ =1.5
Z =1.5 indicates that it is located above the mean by exactly 1.5 standard deviation, and this location corresponds to X = 35+12 = 47 (above the mean by 12 points).
This means that both scores will obtain the same positions in terms of grades.
