Question

Assume a normal random variable, X, with mean 90 and standard deviation 10. Find the probability that a randomly chosen value of X is less than 95. Find the probability that a randomly chosen value of X is between 60 and 100. Find the 97th percentile of the X distribution (i.e., find the value x such that P(X<x)=0.97). Find the probability that a randomly chosen value of X is greater than 100, given that it is greater than 90 (i.e., find P(X>100|X>90

Answer 1

(a)

= 90

= 10

To find P(X<95):

:Z = (95 - 90)/10 = 0.5

Table of Area Under Standard Normal Curve gives area = 0.1915

So,

P(X<95) = 0.5 + 0.1915 = 0.6915

(b)

= 90

= 10

To find P(60 < X<100):

Case 1: For X for 60 to miid value:

:Z = (60 - 90)/10 = - 3

Table of Area Under Standard Normal Curve gives area = 0.4987

Case 2: For X from mid value to 100:

Z = (100 - 90)/10 = 1

Table gives area = 0.3413

So,

P(60<X<100) = 0.4987 + 0.3413 = 0.8400

(c) 97th percentile is equivalent to area = 0.97 - 0.50 = 0.47 from mid value to Z on RHS.

Table gives Z = 1.88

So,

Z = 1.88 = (X - 90)/10

X = 90 + (1.88X 10) = 108.8

(d) P(X>100/P(X>90) = P(X>100)/P(X>90)

To findP(X>100):

Z = (100 - 90)/10 = 1

Table gives area = 0.3413

So,

P(X>100) = 0.5- 0.3413 = 0.1587

P(X>90) = 0.5, since 90 is the mean.

So, we get:
P(X>100/X>90) = 0.1587/0.5 = 0.3174