For a distribution in which X 100 SX 16 and N 500 a What

For a distribution in which X = 100, SX = 16, and N = 500: (a) What is the relative frequency of scores between 76 and the mean? (b) How many participants are expected to score between 76 and the mean? (c) What is the percentile of someone scoring 76? (d) How many subjects are expected to score above 76?

Solution

Sol)

Given mean=100 sd=16 and n=500

a) z = (X - Mean)/SD
z = (76 - 100)/16 = - 1.5
z value corresponding to mean = 0
P(76 < X < 100) = P(- 1.5 < z < 0)
= P(z < 0) - P(z < - 1.5)
= 0.5000 - 0.0668
= 0.4332
Relative frequency desired = 0.4332 in decimal form
or 0.4332 * 100 = 43.32% in percentage form

b) Expected number of participants desired = N*p
= 500 * 0.4332
= 216.6 or 217 on rounding

c) P(Z < z) indicates percentile
Therefore, P(Z < - 1.5) = 0.0668
0.0668*100 = 6.68
Percentile desired = P6.68 or P7 rounding

d) P(X > 76)
= P(z > - 1.5)
= 1 - P(z < - 1.5)
= 1 - 0.0668
= 0.9332
Expected number of participants desired = 500 * 0.9332 = 466.6 or 467 on rounding