advertisement

Normal Family of Distributions Now suppose we also know the population standard deviations of the scores on the two tests: ABC: = 100, = 10 XYZ: = 150, = 15 Problems: 1. Use tables in textbook to solve the similar problem for the XYZ test. Answer: solve 0.85=(X-150)/15 for X, getting X=162.75. So the Catalog will say you need a score of 163. 2. Pilots for a certain type of aircraft must be under 5'7". If a normal population of men has = 5'9" and = 3.5", what percent of this population is excluded from piloting such aircraft. Answer: Let X be the height of a man. P[X is at least 5x12+7=67 inches tall] is excluded. P[ Z X 69 67 69 0.57] 1 .2843 .7157. 3.5 3.5 Problems: Draw sketches for all. Use normal tables to work. 3. Find c such that P{–c < Z c} = .95 for standard normal Z. Answer: P[Z< -c]=.05/2=.025 gives –c= -1.96. So c=1.96. 4. If X ~ NORM(200, 25), find P{190 < X < 230}. Answer: 190 200 X 200 230 200 P[ ] P[.4 Z 1.2] P{Z 1.2] P[ Z .4] .8849 .3446 .5403. 25 25 25 5. Find d such that P{–d < Z d} = .99 for Z ~ NORM(0, 1). Answer: P[Z<-d]=.005 gives –d=-2.60. So d=2.60. 6. If = 100 and P{95 < Y 105} = .6826, then find . Answer: P[-5<Y-100<5]=P[|Z|<5/Hence P[Z<-5/ Table 1 gives -5/and so is 5. 7. If = 0.1, then find such that P{W > 16} = .90. Answer: P[Z>(16-)/0.1]=0.90 and Table 1 gives Z 1.28. Hence solving (16-)/0.1=-1.28 for gives 16.128. 0.90 8. If X ~ NORM(65, 3), then find P{X = 66} and P{65.5 < X 66.5}. Answer: P[X=66]=0 and P[(65.5 65) / 3 Z (66.5 65) / 3] P[.167 Z .5] P{Z .5] P[ Z .167] .6915 .5675 .124.