Normal Distribution

UNIT 13 NORMAL DISTRIBUTION

Structure
13.1 Introduction
Objectives

13.2 Normal Distribution

13.3 Chief Characteristics of Normal Distribution
13.4 Moments of Normal Distribution
13.5 Mode and Median of Normal Distribution
13.6 Mean Deviation about Mean
13.7 Some Problems Based on Properties of Normal Distribution
13.8 Summary
13.9 Solutions/Answers

13.1 INTRODUCTION
In Units 9 to 12, we have studied standard discrete distributions. From this unit
onwards, we are going to discuss standard continuous univariate distributions.
This unit and the next unit deal with normal distribution. Normal distribution
has wide spread applications. It is being used in almost all data-based research
in the field of agriculture, trade, business, industry and the society. For
instance, normal distribution is a good approximation to the distribution of
heights of randomly selected large number of students studying at the same
level in a university.
The normal distribution has a unique position in probability theory, and it can
be used as approximation to most of the other distributions. Discrete
distributions occurring in practice including binomial, Poisson,
hypergeometric, etc. already studied in the previous block (Block 3) can also
be approximated by normal distribution. You will notice in the subsequent
courses that theory of estimation of population parameters and testing of
hypotheses on the basis of sample statistics have also been developed using
the concept of normal distribution as most of the sampling distributions tend to
normality for large samples. Therefore, study of normal distribution is very
important.
Due to various properties and applications of the normal distribution, we have
covered it in two units – Units 13 and 14. In the present unit, normal
distribution is introduced and explained in Sec. 13.2. Chief characteristics of
normal distribution are discussed in Sec. 13.3. Secs. 13.4, 13.5 and 13.6
describes the moments, mode, median and mean deviation about mean of the
distribution.
Objectives
After studing this unit, you would be able to:
 introduce and explain the normal distribution;

5
Continuous Probability
Distributions  know the conditions under which binomial and Poisson distributions
tend to normal distribution;
 state various characteristics of the normal distribution;
 compute the moments, mode, median and mean deviation about mean
of the distribution; and
 solve various practical problems based on the above properties of
normal distribution.

13.2 NORMAL DISTRIBUTION

The concept of normal distribution was initially discovered by English
mathematician Abraham De Moivre (1667-1754) in 1733. De Moivre obtained
this continuous distribution as a limiting case of binomial distribution. His
work was further refined by Pierre S. Laplace (1749-1827) in 1774. But the
contribution of Laplace remained unnoticed for long till it was given concrete
shape by Karl Gauss (1777-1855) who first made reference to it in 1809 as the
distribution of errors in Astronomy. That is why the normal distribution is
sometimes called Gaussian distribution. Though, normal distribution can be
used as approximation to most of the other distributions, here we are going to
discuss (without proof) its approximation to (i) binomial distribution and (ii)
Poisson distribution.
Normal Distribution as a Limiting Case of Binomial Distribution
Normal distribution is a limiting case of binomial distribution under the
following conditions:
i) n, the number of trials, is indefinitely large i.e. n  ;
ii) neither p (the probability of success) nor q (the probability of failure) is too
close to zero.
Under these conditions, the binomial distribution can be closely associated by
X  np
a normal distribution with standardized variable given by Z  . The
npq
approximation becomes better with increasing n. In practice, the
approximation is very good if both np and nq are greater than 5.
For binomial distribution, you have already studied [see Unit 9 of this course]
that
2 2
 2  npq  q  p   q  p 
1  33   3
 ,
2  npq  npq

 4 npq 1  3  n  2  pq  1  6pq

2  2
 2
 3 ,
2  npq  npq

q  p 1  2p
1  1   , and
npq npq

1  6pq
 2  2  3  .
npq
6
From the above results, it may be noticed that if n  , then moment Normal Distribution
coefficient of skewness ( 1 )  0 and the moment coefficient of kurtosis i.e.
 2  3 or  2  0 . Hence, as n  , the distribution becomes symmetrical
and the curve of the distribution becomes mesokurtic, which is the main
feature of normal distribution.
Normal Distribution as a Limiting Case of Poisson Distribution
You have already studied in Unit 10 of this course that Poisson distribution is a
limiting case of binomial distribution under the following conditions:
i) n, the number of trials is indefinitely large i.e. n 
ii) p, the constant probability of success for each trial is very small i.e. p  0.
iii) np is a finite quantity say ‘’.
As we have discussed above that there is a relation between the binomial and
normal distributions. It can, in fact, be shown that the Poisson distribution
approaches a normal distribution with standardized variable given by
X
Z as λ increases indefinitely.

For Poisson distribution, you have already studied in Unit 10 of the course that
32  2 1 1
1  3
 3   1  1  ; and
2   

 4 3 2 1 1
2  2
 2    3    2  2  3  .
2   
Like binomial distribution, here in case of Poisson distribution also it may be
noticed from the above results that the moment coefficient of skewness
( 1 )  0 and the moment coefficient of kurtosis i.e. 2  3 or  2  0 as
λ . Hence, as λ  , the distribution becomes symmetrical and the curve
of the distribution becomes mesokurtic, which is the main feature of normal
distribution.
Under the conditions discussed above, a random variable following a binomial
distribution or following a Poisson distribution approaches to follow normal
distribution, which is defined as follows:
Definition: A continuous random variable X is said to follow normal
distribution with parameters  (      ) and 2(>0) if it takes on any real
value and its probability density function is given by
2
1  x  
1   
f x  e 2  
,   x  ;
 2
which may also be written as

1  1  x   2 
= exp     ,   x  .
 2  2    

7
Continuous Probability
Distributions Remark
i) The probability function represented by f  x  may also be written as

f x; ,  2 .
ii) If a random variable X follows normal distribution with mean  and
variance 2, then we may write, “X is distributed to N(, 2)” and is
expressed as X  N(, 2).
iii) No continuous probability function and hence the normal distribution
can be used to obtain the probability of occurrence of a particular value
of the random variable. This is because such probability is very small,
so instead of specifying the probability of taking a particular value by
the random variable, we specify the probability of its lying within
interval. For detail discussion on the concept, Sec. 5.4 of Unit 5 may be
referred to.
X
iv) If X ~ N  ,  2  , then Z  is standard normal variate having

mean ‘0’ and variance ‘1’. The values of mean and variance of standard
normal variate are obtained as under, for which properties of
expectation and variance are used (see Unit 8 of this course).
 X 1
Mean of Z i.e. E  Z   E   =  E  X    
   
1
  E  X    

1
     = 0 [ E(X) = Mean of X =  ]

 X  
Variance of Z i.e. V(Z) = V  
  
1 1
  V  X      2  V  X  
2 
 
1 2

2
  [ variance of X is 2]

= 1.
X 
v) The probability density function of standard normal variate Z 

1  12 z2
is given by   z   e ,   z  .
2

This result can be obtained on replacing f  x  by   z  , x by z,  by 0

and  by 1 in the probability density function of normal variate X i.e. in
2
1  x  
1   
f x  e 2  
,   x  
 2

8
 vi) The graph of the normal probability function f  x  with respect to x is Normal Distribution

famous ‘bell-shaped’ curve. The top of the bell is directly above the
mean . For large value of , the curve tends to flatten out and for
small values of  , it has a sharp peak as shown in (Fig. 13.1):

Fig. 13.1

Normal distribution has various properties and large number of applications. It

can be used as approximation to most of the other distributions and hence is
most important probability distribution in statistical analysis. Theory of
estimation of population parameters and testing of hypotheses on the basis of
sample statistics (to be discussed in the next course MST-004) have also been
developed using the concept of normal distribution as most of the sampling
distributions tend to normality for large samples. Normal distribution has
become widely and uncritically accepted on the basis of much practical work.
As a result, it holds a central position in Statistics.
Let us now take some examples of writing the probability function of normal
distribution when mean and variance are specified, and vice-versa:
Example 1: (i) If X ~ N (40, 25) then write down the p.d.f. of X
(ii) If X ~ N (  36, 20) then write down the p.d.f. of X
(iii) If X ~ N (0, 2) then write down the p.d.f. of X
Solution: (i) Here we are given X ~ N (40, 25)
 in usual notations, we have

μ = 40, σ 2 = 25  σ = ± 25
5    0always 
Now, the p.d.f. of random variable X is given by
2
1  x  
1  
2 σ 

f(x) = e
σ 2π
2
1  x  40 
1  
2 5 

= e ,   x  
5 2π
9
Continuous Probability
Distributions
(ii) Here we are given X ~ N (  36, 20).
in usual notations, we have
μ  36, σ 2  20  σ  20
Now, the p.d.f. of random variable X is given by
2
1 x µ 
1  
2 σ 

f(x) = e
σ 2π
2
1  x  (  36)  1
1    1  (x +36) 2
2 20 
= e = e 40
20 2π 40π
1
1  (x +36) 2
= e 40 ,   x  
2 10π
(iii) Here we are given X ~ N (0, 2).
 in usual notations, we have

μ  0, σ 2  2  σ  2
Now, the p.d.f. of random variable X is given by
2 2
1 x µ  1 x 0 
1  
2 σ 
 1  
2 2 

f(x) = e = e
σ 2π 2 2π
1
1  x2
4
 e ,   x  
2 π
Example 2: Below, in each case, there is given the p.d.f. of a normally
distributed random variable. Obtain the parameters (mean and variance)
of the variable.
2
1  x  46 
1  
2 6 

(i) f(x) = e ,  x 
6 2π
1
1  (x  60) 2
(ii) f(x) = e 32 ,   x  
4 2π
2
1  x  46 
1  
2 6 

Solution: (i) f(x) = e ,  x
6 2π
Comparing it with,
2
1 x µ 
1  
2 σ 

f(x) = e
σ 2π
we have
µ  46,   6

 Mean    46, var iance   2  36

10
 1
1  (x  60) 2 Normal Distribution
(ii) f(x) = e 32 ,   x  
4 2π
2
1  X  60 
1  
2 4 

= e
4 2π
Comparing it with,
2
1 x µ 
1  
2 σ 

f(x) = e
σ 2π
we get
µ = 60,   4

 Mean    60, var iance   2  16

Here are some exercises for you.
E 1) Write down the p.d.f. of r. v. X in each of the following cases:
1 4
(i) X ~ N  , 
2 9
(ii) X ~ N (  40,16)
E 2) Below, in each case, is given the p.d.f. of a normally distributed
random variable. Obtain the parameters (mean and variance) of the
variable.
x2
1 
(i) f(x) = e 8 ,   x  
2 2π
1
1  (x  2) 2
4
(ii) f(x) = e ,  x 
2 π

Now, we are going to state some important properties of Normal distribution in

the next section.

13.3 CHIEF CHARACTERISTICS OF

NORMAL DISTRIBUTION
The normal probability distribution with mean  and variance 2 has the
following properties:
i) The curve of the normal distribution is bell-shaped as shown in Fig. 13.1
given in Remark (vi) of Sec. 13.2.
ii) The curve of the distribution is completely symmetrical about x =  i.e. if
we fold the curve at x  , both the parts of the curve are the mirror images
of each other.
iii) For normal distribution, Mean = Median = Mode

11
Continuous Probability
Distributions iv) f  x  , being the probability, can never be negative and hence no portion
of the curve lies below x-axis.
v) Though x-axis becomes closer and closer to the normal curve as the
magnitude of the value of x goes towards  or  , yet it never touches it.

vi) Normal curve has only one mode.

vii) Central moments of Normal distribution are
µ1  0,  2   2 , 3  0,  4  3 4 and

µ 32 
β1 = 3
 0, 2  24  3
µ2 2
i.e. the distribution is symmetrical and curve is always mesokurtic.
Note: Not only µ1 and µ 3 are 0 but all the odd order central moments
are zero for a normal distribution.

viii) For normal curve,

Q3 – Median = Median – Q1
i.e. First and third quartiles of normal distribution are equidistant from
median.
Q  Q1 2
ix) Quartile Deviation (Q.D.) = 3 is approximately equal to of the
2 3
standard deviation.
4
x) Mean deviation is approximately equal to of the standard deviation.
5
2 4
xi) Q.D. : M.D. : S.D. =  :  :  = 10 : 12 : 15
3 5
xii) The points of inflexion of the curve are
1
1 
x    , f  x   e 2
 2
xiii) If X1 , X 2 ,..., X n are independent normal variables with means
1 ,  2 ,...,  n and var iances 12 , 22 , ...,  n2 respectively then the linear
combination a1X1 + a 2 X 2 +...+ a n X n of X1 , X 2 ,..., X n is also a normal
variable with
mean a11  a 2 2  ...  a n  n and var iance a1212  a 22 22  ...  a n2  n2 .
xiv) Particularly, sum or difference of two independent normal variates is also
a normal variate. If X and Y are two independent normal variates with
means 1, 2 and variances 12 ,  2 2 , then

X + Y  N(1 + 2, 12 + 22) and X – Y  N(1  2, 12 + 22).

Also, if X1, X2, …, Xn are independent variates each distributed as
 2 
N( ,2), then their mean X ~ N  , .
 n 
12
 Normal Distribution

xv) Area property:



P     X       f  x  dx  0.6827,

1
Or P  1  Z  1     z  dz = 0.6827,
1
 2 

P    2  X    2   f  x  dx  0.9544,
 2 

2
Or P  2  Z  2     z  dz = 0.9544, and
2

 3 

P    3  X    3    f  x  dx  0.9973.
 3

3
Or P  3  Z  3     z  dz = 0.9973.
3

This property and its applications will be discussed in detail in Unit 14.
Let us now establish some of these properties.

13.4 MOMENTS OF NORMAL DISTRIBUTION

Before finding the moments, following is defined as gamma function [See Unit
16 of the Course also for detail discussion] which is used for computing the
even order central moments.
Gamma Function

If n > 0, the integral  x n 1e x dx is called a gamma function and is denoted by
0

n .
 
2 x 31  x
e.g.  x e dx   x e dx   3
0 0

  1
1/ 2  x
1
x 1
and 0 x e dx  0 x 2 e dx   2 
Some properties of the gamma function are

i) If n > 1,  n    n  1  n  1
ii) If n is a positive integer, then n  n  1

1
iii)     .
2
13
Continuous Probability
Distributions
Now, the first four central moments of normal distribution are obtained as
follows:
First Order Central Moment
As first order central moment (1) of any distribution is always zero [see Unit
3 of MST-002], therefore, first order central moment (1) of normal
distribution = 0.
Second Order Central Moment

2
2 =   x    f  x  dx

[See Unit 8 of MST-003]

2
 1  x  
2 1   
  x  
2  
 . e dx
  2
x 
Put  z  x    z

Differentiating
dx
 dz

 dx  dz
Also, when x  , we have z   and
and when x   , z  
 1
2 1  z2
 2    z
  2
e 2 dz

 1
2 2
 z2
2
 ze dz
2 

2
 z
2 2 2

2
  z e dz
2 0
z2

2
 on changing z to – z, the integrand i.e. z e does not get changed 2

i.e. it is an even function of z [see Unit 2 of MST-001]. Now, the

following property of definite integral can be used:
 

 f  z  dz  2  f  z  dz if f  z  is even function of z
 0

1
1  12
Now, put z 2  2t  z  2 t  2 t 2  dz  2 t dt
2

14
  1 Normal Distribution
2 1 2
 2  2  (2t)e t 1
dt   2 2  t 2 e  t dt
0  0
2t 2

2 2 32 1  t
  t e dt
 0

2 2  3 
   [By def. of gamma function]
 2

2 2 1 1
 [By Property (i) of gamma function]
 2 2

2


  [By Property (iii) of gamma function]

 2
Third Order Central Moment

3
3 =   x    f  x  dx


2
 1  x  
3 1   
  x  
2  
 e dx
  2
x 
Put  z  x    z  dx  dz

and hence

3 1  12 z 2
3 

  z  .
 2
e dz

 1
1  z2
 3  z 3e 2
dz
2 

1 1 1
 z2  z2  z2
3 3 3
Now, as integrand z e 2 changes to z e 2 on changing z to – z i.e. z e 2

is an odd function of z.
Therefore, using the following property of definite integral:
a

 f  z  dz  0 if f(z) is an odd function of z

a

we have,
1
3  3 0 = 0
2

15
Continuous Probability
Distributions Fourth Order Central Moment
2
  1  x  
4 4 1   
  x    f  x  dx    x   
2  
4 = e dx
   2
x 
Putting z

 dx  dz

4 1  12 z 2
 4    z 
  2
e dz

 1  1
4  z2 2 4  z2
  z 4e 2
dz  z 4
e 2
dz
2  2 0

z2
4 
 integrand z . e does not get changed on changing z to – z and hence it is
2

an even function of z [using the same property as used in case of 2].

z2
Put  t  z2 = 2t
2
 2zdz = 2dt
 z dz = dt
dt dt
 dz = 
z 2t

2 4 1
 4   (2t) 2 e  t dt
2 0 2t
 
24 .4 2 t 1 4 4 32  t
=  t e dt   t e dt
2 2 0 t  0

4 4 52 1  t
  t e dt
 0

4 4 5
 [By definition of gamma function]
 2

4 4 3 3
 [By Property (i) of gamma function]
 2 2

4 4 3 1 1
 [By Property (i) of gamma function]
 22 2

3 4 1
  [Using   (Property (iii) of gamma function)]
 2

 3 4

16
Thus, the first four central moments of normal distribution are Normal Distribution

1  0,  2   2 , 3  0,  4  3 4 .

 32 4 34 3 4
 1  = 0,  2   = 3
 23 22 2
2
4 
Therefore, moment coefficient of skewness ( 1 )  0

 the distribution is symmetrical.

The moment coefficient of kurtosis is 2  3 or  2  0.

 The curve of the normal distribution is mesokurtic.

Now, let us obtain the mode and median for normal distribution in the next
section.

13.5 MODE AND MEDIAN OF NORMAL

DISTRIBUTION
Mode
Let X ~ N  ,  2  , then p.d.f . of X is given by
2
1  x  
1  
 

f (x)  e 2 ...(1)
 2
,   x  
Taking logarithm on both sides of (1), we get
2
1 1 x   logmn  log m  log n 
logf(x) = log    log e  n 
 2 2    and log m  n log m 
1 1
 log  2
(x  )2 as loge = 1
σ 2π 2
Differentiating w.r.t.x
1 ' 1 (x  )
f (x)  0  2 2(x  )  
f(x) 2 2
(x  )
 f ' (x)   f (x) … (2)
2
For maximum or minimum
f ' (x)  0

(x  )
 f(x)  0
2

 x   0 as f (x)  0

17
Continuous Probability x
Distributions
Now differentiating (2) w.r.t. x, we have
 x   1
f  (x)   f '(x)  f (x)
 2
f () f ( )
f (x) at x   0 2
 2 0
 
 x =  is point where function has a maximum value.
 Mode of X is  .
Median
Let M denote the median of the normally distributed random variable X.
We know that median divides the distribution into two equal parts
M 
1
  f (x)dx   f (x)dx 
 M
2
M
1
  f (x)dx  2


2
µ 1  x   M
1  
 
 1
  e 2 dx   f (x)dx 
 σ 2π 
2

x 
In the first integral, let us put z

Therefore, dx   dz
Also when x    z  0 , and
when x    z   .
Thus, we have
0 M
1  12  z 2 1
 e dz   f (x)dx 
 2 
2
0 1 M
1  z2 1
  e 2 dz   f (x)dx 
 2 
2

 1  12 z2 
M  Z is s.n.v.with p.d.f. (z)  e 
1 1  2 
 +  f (x) dx 
2  2   0
1 
So  (z)dz  1   (z)dz  
   2 
M
  f(x)dx = 0
µ

M  as f (x)  0
18
 Median of X   Normal Distribution

From the above two results, we see that

Mean = Median = Mode = 

13.6 MEAN DEVIATION ABOUT THE MEAN

Mean deviation about mean for normal distribution is

  | x  Mean | f  x dx

[See Section 8.4 of Unit 8]

2
 1  x  
1  
2  

  x  e dx
  2
x 
Put  z  x    z

dx
  dz

 1
1  z2
 M.D. about mean =  | z | e 2 dz
  2
 1
  z2
2
  | z |e dz
2 

1
 z2
Now, | z | e 2 (integrand) is an even function z as it does not get changed on
changing z to –z,  by the property,
a a
“  f (x)dx  2 f (x)dx, if f  x  is an even function of x ”, we have
a 0

1 
  z2
M.D. about mean = 2  z e 2 dz
2 0
Now, as the range of z is from 0 to ∞ i.e. z takes non-negative values,
 z = z and hence
1 
2  z2
2
M.D. about mean   ze dz
2 0

z2
Put  t  z 2  2t  2zdz = 2dt  zdz = dt
2
 
2   e t  2 2
 M.D. about mean =  e t
dt  2      0  1 = 
2 0   1  0  

19
Continuous Probability
Distributions 2
In practice, instead of , its approximate value is mostly used and that is

4
.
5

2 2 7 7 4
    0.6364  0.7977  0.08 or (approx.)
 22 11 5
Let us now take up some problems based on properties of Normal Distribution
in the next section.

13.7 SOME PROBLEMS BASED ON PROPERTIES

OF NORMAL DISTRIBUTION
Example 3: If X1 and X2 are two independent variates each distributed as
N(0, 1), then write the distribution of (i) X1 + X2. (ii) X1 – X2.
Solution: We know that, if X1 and X2 are two independent normal variates s.t.

 
X1 ~ N 1 , 12 and X 2 ~ N  2 , 22 
then

 
X1  X 2 ~ N 1   2 , 12   22 , and


X1  X 2 ~ N 1   2 , 12   22  [See Property xiii (Section 13.3)]

Here, X1 ~ N  0,1 , X 2 ~ N  0,1

 i) X1  X 2 ~ N  0  0, 1  1

i.e. X1  X 2 ~ N  0, 2  , and

ii) X1  X 2 ~ N  0  0, 1  1

i.e. X1  X 2 ~ N  0, 2 

Example 4: If X ~ N  30, 25  , find the mean deviation about mean.

Solution: Here  = 30, 2 = 25   = 5.

2 2 2
 Mean deviation about mean =  = .5 = 5
  

Example 5: If X ~ N  0, 1 , what are its first four central moments?

Solution: Here  = 0, 2 = 1  σ = 1.
 first four central moments are:
1  0,  2   2  1,  3  0,  4  3 4  3.

Example 6: If X1 , X 2 are independent variates such that X1 ~ N (40, 25) ,

X 2 ~ N (60, 36) , then find mean and variance of (i) X = 2X1 +3X 2
(ii) Y  3X1  2X 2
20
Solution: Here X1 ~ N (40, 25), X 2 ~ N (60, 36) Normal Distribution

 Mean of X1 = E(X1 )  40

Variance of X1  Var(X1 )  25

Mean of X 2 = E(X 2 )  60

Variance of X 2  V ar(X 2 )  36

Now,
(i) Mean of X = E(X)  E(2X1  3X 2 )  E(2X1 )  E(3X 2 )

= 2E(X1 ) + 3E(X 2 ) = 2  40  3  60 = 80 + 180 = 260

Var (X) = Var(2X1  3X 2 )

 Var(2X1 )  Var(3X 2 )  X1 and X 2 are independent 

 4Var(X1 ) + 9Var(X 2 )

 4  25  9  36 = 100 + 324 = 424

(ii) Mean of Y = E(Y)  E(3X1  2X 2 )

 E(3X1 )  E( 2X 2 )

= 3E(X1 ) + (  2)E(X 2 )

= 3  40  2  (60) = 120 – 120 = 0

Var (Y) = Var (3X1  2X 2 )

= Var(3X1 )  Var( 2X 2 )

2
= (3) 2 Var  X1    2  Var  X 2 

= 9  25  4  36 = 225 + 144 = 369

You can now try some exercises based on the properties of normal distribution
which you have studied in the present unit.
E3) If X1 and X2 are two independent normal variates with means 30, 40
and variances 25, 35 respectively. Find the mean and variance of
i) X1 + X2
ii) X1 – X2
E4) If X  N(50, 225), find its Quartile deviation.
E5) If X1 and X2 are independent variates with each distributed as
X  X2
N (50, 64), what is the distribution of 1 ?
2
E6) For a normal distribution, the first moment about 5 is 30 and the fourth
moment about 35 is 768. Find the mean and standard deviation of the
distribution.

21
Continuous Probability
Distributions 13.8 SUMMARY
The following main points have been covered in this unit:
1) A continuous random variable X is said to follow normal distribution with
parameters  (      ) and 2(>0) if it takes on any real value and its
probability density function is given by
2
1  x  
1   
f x  e 2  
,   x  
 2
X
2) If X  
N ,  2 , then Z 

is standard normal variate.

3) The curve of the normal distribution is bell-shaped and is completely

symmetrical about x   .
4) For normal distribution, Mean = Median = Mode.
5) Q3 – Median = Median – Q1
Q3  Q1 2
6) Quartile Deviation (Q.D.) = is approximately equal to of the
2 3
standard deviation.
4
7) Mean deviation is approximately equal to of the standard deviation.
5
8) Central moments of Normal distribution are
1  0,  2   2 , 3  0,  4  34
9) Moment coefficient of skewness is zero and the curve is always mesokurtic.
10) Sum of independent normal variables is also a normal variable.

13.9 SOLUTIONS/ANSWERS
1 4
E 1) (i) Here we are given X ~ N  , 
2 9
 in usual notations, we have
1 4 2
  , 2  
2 9 3
Now, p.d.f. of r.v. X is given by
2
1  x  
1  
2  

f(x) = e ,   x  
 2
2
1  x 1/2 
1  
2  2/3 

= e
2
2π
3

22
 2
9  2x 1  Normal Distribution
3  
2 4 

= e ,   x  
2 2π
(ii) Here we are given X ~ N(40, 16)
 in usual notations, we have
   40, 2  16 4
Now, p.d.f. of r.v. X is given by
2
1  x  
1  
2  

f(x) = e ,   x 
σ 2π
2
1  x  (  40) 
1  
2 4


= e
4 2π
2
1  x + 40 
1  
2 4 

= e ,   x  
4 2π
x2
1 
E 2) (i) f(x) = e 8,   x  
2 2π
2
1x
1 
= e 24
2 2π
2
1  x 0 
1  
2 2 

= e ...(1) ,   x  
2 2π
Comparing (1) with,
2
1  x  
1  
2  

f(x) = e ,   x  
σ 2π
we get
  0, 2

 Mean    0 and var iance   2  (2) 2  4

1
1  (x  2) 2
4
(ii) f(x) = e ,   x  
2 π
1
1 
2×2
(x  2)2
= e
2× 2 π
2
1  x 2 
1  
2 2 

= e ...(1) ,   x  
2 2π
Comparing (1) with,
2
1  x  
1  
2  

f (x)  e ,   x  
 2
23
Continuous Probability
Distributions
we get

  2,  2

 Mean    2 and var iance   2  ( 2) 2  2

E3) i) X1 + X2  N(1 + 2, 12 + 22)
 X1 + X2  N(30 + 40, 25 + 35)
 X1 + X2  N(70, 60)
ii) X1 – X2  N(30  40, 25 + 35)
 X1 – X2  N(  10, 60)
E4) As 2 = 225
  = 15
2 2
and hence Q.D. =   15  10
3 3
E5) We know that if X1, X2, .., Xn are independent variates each distributed as
 2 
N(, 2), then X ~ N  , 
 n 
Here X1 and X2 are independent variates each distributed as N(50, 64),
X1  X 2  64 
 their mean i.e. X i.e. ~ N  50, 
2  2 

i.e X ~ N(50, 32).

E6) We know that 1  x  A [See Unit 3 of MST-002]

where 1 is the first moment about A.

 30 = x  5  x  35  Mean = 35
Given that fourth moment about 35 is 768. But mean is 35, and hence the
fourth moment about mean = 768.
 4 = 768
 34 = 768
768
 4 =   4  3 4 
3
 4 = 256 = (4)4   = 4.

24