Question

Suppose a random variable X is normally distributed with mean 66 and standard deviation 7.6. Answer the following questions:

a) P(46.24 <  X < 71.32) =

b) P(X ? 75.12) =

c) P(X = 71.32) =

d) Suppose a is such that: P(X ? a) = 0.56. Then a =

e) What is the IQR (inter-quartle range) of X?

Answer 1

Mean =66

S.D = 7.6

a)

Z at x = 46.24

Z = (X - ?) / ?
Z = (46.24 - 66) / 7.6

Z = -2.6

Z at x= 71.32

Z = (X - ?) / ?
Z = (71.32 - 66) / 7.6

Z = 0.7

From Z score table

P(46.24 <  X < 71.32) = P( -2.6 < Z<0.7)= 0.7534

B) P(X<= 75.12)

Z at x =75.12

Z = (X - ?) / ?
Z = (75.12 - 66) / 7.6

Z = 1.2

From Z-Score Table

P(X<= 75.12) = P(Z<= 1.2) = 0.8849

C)

P(X=71.32) = 0

As X is continuous random varibale sp there is 0 probablity of x=71.32

D)

P(X<=a) =0.56

Z Score at P Value of 0.56

Z = 0.151

Z = (X - ?) / ?
0.151= (X - 66) / 7.6
X-66 = 1.1476

X = 67.1476

P(X<= 67.1476) =0.56

a= 67.1476

e)

IQR is the difference of x of middle 50 % (leaving 25 % each side)

$P(a\leq x\leq b)=0.5$

IQR = b-a

For b

$P(x\leq b)=0.5+0.25=0.75$

Z Score at P Value of 0.75

Z = 0.674

Z = (X - ?) / ?
0.67449= (b - 66) / 7.6
b-66 = 5.126124

b = 71.126

For a

$P(x\leq a)=0.5-0.25=0.25$

Z Score at P Value of 0.75

Z = -0.67449

Z = (X - ?) / ?
-0.67449= (a - 66) / 7.6
a-66 = -5.126124

a = 60.874

IQR = b-a = 71.126 - 60.874 = 10.252