Question

Given a normal distribution with a MEAN of 20 and SD of 80, N=80. Find out
(1) what percent of the cases will lie between the scores of 12 and 26?
(2) what percent of the cases is expected to have score more then 30?

Answer 1

1)
X1 = 12      X2 = 26           μ = 20      σ = 80 or 8 ?       N = 80

Z1 = normal distribution variable = (X1 - μ) / σ = -0.1
Z2 = (X2 - μ)/σ = +0.075

Probability of a given case being X1 < X < X2 =  ZTable (-0.1 < Z < 0.075)
           = ZTable (0 < Z < 0.1) + ZTable (0 < Z < 0.075)
           = 0.03980 + 0.02989 = 0.06969

Percentage of cases = 6.969 %  or   nearly 7%
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2)
       X > 30 
      Z = (X - μ)/σ = 0.125
   ZTable (Z > 0.125) = 0.5 - ZTable (0<Z < 0.125) = 0.5 - 0.04974 = 0.45026

Number of persons with score X > 30 = 0.45026 * N = 36.0208 
Percentage of cases = 36.0208 * 100 / 80 = 45.026 %, same as probability * 100.