Question

7. Hint for c and d: given P(X S x) a percentage, we have P(Z Sz)the percentage. Then find the corresponding value for Z, and use the Inverse Transformation Let X be normally distributed with mean 120 and standard deviation σ 20. a. Find P(X3 86). b. Find P(80 <X3100). c. Find x such that P(Xx) 0.40. d. Find x such that P(X> x) 0.90.

Answer 1

a)
P( X<86)  
  
I know that, z = (X-mean)/(sd)  
z1 = (86-120)/20)   -1.7000
hence,  
P( X<86)   =P(Z<-1.7)
NORMSDIST(-1.7) =   0.0446

b)
P(80 < X < 100)  
= P(X<100) - P(X<80)  
  
I know that, z = (X-mean)/(sd)  
z1 = (80-120)/20) =   -2.0000
z2 = (100-120)/20) =   -1.0000
  
hence,  
P(80 < X < 100)=   = P(Z<-1) - P(Z<-2)
= NORMSDIST(-1) - NORMSDIST(-2) =   13.59%

c)
z=   NORMSINV(0.4)
z=   -0.253347103
I know that, z = (X-mean)/sd  
(X-mean)/sd =   -0.2533
X=    -0.2533*20+120
X=    114.93

d)
P(Z<z)   10%
z=   NORMSINV(0.1)
z=   -1.281551566
I know that, z = (X-mean)/sd  
(X-mean)/sd =   -1.2816
X=    -1.2816*20+120
X=    94.37