# 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?

1
by purn

2014-10-18T05:33:47+05:30

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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.