EDS115

’m31 years old. Whatproportionoftheotherguys in classare less than my age?

Z = (31 – 26) / 3 = 1.67

Look up 1.67 in table (Appendix a in Healey). ”Area betweenmean and Z, 0.4525”

0.5+0.4525 = 0.9525   I’m older than 95 % oftheother men in theclass Finding proportion of area above a score If score is below the mean = Find area between the mean and Z  = Add 0.5 for the top half of the distribution   = If want percentage of scores above your score, multiply by 100 Z = (25 – 26) / 5 = (-)0.2 Look up 0.2 in table (Appendix a in Healey). ”Area beyond, 0.0793”, and add 0.5. 58 % ofthegirls in yourclassare older thanyou. Find proportion between two scores If one score is above the mean and one below the mean = Find area above Z for the first score (tail)  = Find area below Z for the second score (tail)   = Add these two proportions together and subtract from 1.00   = If want percentage, multiply by 100 If both scores are above or below the mean = Find area between the mean and Z for one score  = Find area between the mean and Z for the second score   = Subtract the smaller area from the larger   = If want percentage, multiply by 100 It is very easy and accurate method to get the mean value. I like this method because it is also very easy.