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.