The document provides solutions to calculating probabilities for a normally distributed random variable X with a mean of 80 and standard deviation of 5. (1) It finds the probability that X is greater than 95 by converting to the standard normal variable Z and using normal distribution tables. (2) It calculates the probability that X is less than 72 by again converting to Z and using tables. (3) It determines the probability that X is between 60.5 and 90 by breaking it into separate probabilities and adding the areas under the normal curve.

1. Normal
Distribution –Find the
area property
SUNDARA B. N.
Assistant
Professor

2. Problem
If the random variable X is normally distributed
with mean 80 and standard deviation 5, then
find
a) P(X>95)
b) P(X<72)
c) P(60.5<X<90)
d) P(85<X<97) and
e) P(64<X<76)

3. Given Information
Here we are given that
μ = 80
σ = 5
 X~N(80,25).
We know that is X~N(μ, σ²),then the S.N.V is given by
Z = X – μ
σ

4. a) P(X>95)
Consider X = 95 
P(X>95)=P(Z>3)
= 0.5-P(0<Z<3)
= 0.5-
0.4987(TV)
=0.0013
X=9
5
Z=3
X=8
0
Z=0

5. b) P(X<72)
Consider X =
72
 P(X<72)=P(Z<-1.6)
 = P(Z<1.6)
= 0.5-
P(0<Z<1.6)
= 0.5-
0.4452(TV)
=0.0548
X=8
0
Z=0
X=7
2
Z=-
1.6

6. c) P(60.5<X<90)
Consider X = 60.5 and X =
90
 P(60.5<X<90)=P(-
3.9<X<2)
= P(-3.9<X<0)+P(0<Z<2)
= P(0<X<3.9)+P(0<Z<2)
= 0.5000+0.4772 (TV)
=0.9772
X=8
0
Z=0
X=9
0
Z=2
X=60.5
Z=-
3.9

7. d) P(85<X<97
Consider X = 85 and X =
97

P(85<X<97)=P(1<X<3.
4)
= P(0<Z<3.4)-P(0<Z<1)
= 0.4997-0.34139(TV)
=0.1584
X=8
0
Z=0
X=8
5
Z=1
X=9
7
Z=3.
4

8. e) P(64<X<76)
Consider X = 64 and X =
76
 P(64<X<76)=P(-3.2<Z<-
0.8)
= P(0.8<Z<3.2)
= P(0<Z<3.2)-P(0<Z<0.8)
= 0.4993-0.2881(TV)
=0.2112
X=8
0
Z=0
X=7
6
Z=-
0.8
Z=-
3.2
X=6
4