11. Normal Distribution - CA Study · 2021. 2. 21. · 11. Normal Distribution 1. It is a continuos probability distribution where the variable ‘X’ can assume any value between

11. Normal Distribution

1. It is a continuos probability distribution

where the variable ‘X’ can assume any value

between – � to + �.

2. The Probability Density Function of a

Normal Distribution is given by

2

2

2σ

μ)(x–

e2πσ

1f(x)

��

�� , (– �� < x < ��)

where � = mean

��= SD

��and ��are the two parameters of Normal

Distribution and hence it is bi-parametric

in nature.

�� 3.1416 and e ��2.71828 which are

constant.

3. Replacing σ

μ–xby ‘z’ we obtain another

distribution called Standard Normal

Distribution with mean 0 and S.D. 1 and is

given by the density function f(z) =

2z

–2

e2π

1(– � < z < �)

Note : 1. N(��, ��2) implies Normal

Distribution with ��(mean) and ��2

(variance)

Note : 2. N(0,1) implies Standard Normal

Distribution with Mean = 0 and

S.D. = 1.

Note : 3. ‘z’ is called Standard Normal

Variate or Variable.

4. Caculation of Mean & S.D. of Z

(i) Calculation of Mean of Z

z = σ

μ–x

Mean of z = E(z)

= E ��

��

σ

μ–x

= σ1

[E(x) – E(�)]

= σ1

(��–��) = σ1

x 0 = 0

(ii) Calculation of Varince of z

Var (z) = E(z2) – [E(z)]2

= E(z2) – 02 [ ... E(z) = 0]

= E(z2)

= E

2

σ

μ–x��

��

= 2σ

1 E (x – �)2

=2

σ

1x �2

= 1 Var (z) = 1

hence S.D. (z) =1

... E (x – x

–)2

= E(x – �)2 = �2

Normal or Gaussian Distribution

J. K. SHAH CLASSES Normal Distribution

: 477 :


: 478 :

5. The probability of success under Normal Distribution in calculated by evaluating the area unde a

curve called Normal Freaquency curve which in shown in the following diagram –

Normal Curve

x = ��– 3� x = ��– 2� x = ��– � x = � x = ��+ � x = ��+ 2� x = ��+ 3�

– 3 – 2 – 1 z = 0 1 2 3

Mean = Median = Mode

Standard Normal Curve

Mean = Median = Mode = 0


: 479 :

6. CONVERSION OF X VALUES FROM NORMAL FREQUENCY CURVE TO

STANDARD NORMAL CURVE VALUES (Z - VALUES)

1. x = �� or, X – ��= 0 or, ��

��04–x

Z = 0

2. x = ��+�� x = ��–��

1–x ��

1––x ��

Z = 1 Z = – 1

3. x = ��+ 2�� x = ��– 2��

2–x ��

2––x ��

Z = 2 Z = – 2

4. x = ��+ 3�� x = ��– 3��

3–x ��

3––x ��

Z = 3 Z = – 3

PROPERTIES OF NORMAL CURVE AND NORMAL DISTRIBUTION

1. It is a bell shaped curve symmetrical about the line x = ��and assymptotic to the horizontal axis

(x = axis)

2. The two tails extend upto infinity at both the ends.

3. As the distance from the mean increases, The curve comes closer to the horizontal axis (x = axis)

4. The curve has a single peak at x = ��.

5. The two points of inflection of the normal curve are at x = ��– ��and x = ��+ ��respectively where

the normal curve changes its curvature.

6. The same points of inflection under standard normal curve are at z = – 1 and z = 1.

7. It is a continous prob. distribution where – �� < x < + ��

8. The distribution has two parameters ��and ��. Where ��= mean ��= standard deviation. Hence

normal is bi-parametric distribution.

9. The normal curve has a single peak. Hence it is unimodal and mean. Median and mode coincide.

at x = ��.

10. The maximum ordinate (i.e. y) lies at x = ��.

11. The distribution being symmetrical,

i) Mean = Median = Mode

ii) Skewness = 0

iii) Kurtosis = 0 (i.e. it is a mesorurtic distribution)


: 480 :

12. The two Quartiles are Q1 = ��– .675�� (Lower Quartile)

And Q3 = ��+ .675�� (Upper Quartile)

13. Quartile Deviation (Q. D.)

Q. D. = 2

)Q–Q( 13 , = 2

)675.–(–)675.( ��

= 2

675.–675. ��, =

2675.2 ��

, = .675��

14. Mean Deviation (M. D.) = .8��

15. Q. D. : MD : SD = 10 : 12 : 15

16. (i) The total area under the Normal or Standard Normal Curve = 1 (... Total Probability = 1),

Symbolically,

(i) ��

��

��–

1dx)x(f or (ii) ��

��

��–

1dz)z(f or (iii) �� b

a

1dx)x(f when a �� X �� b

(ii) f(x) �� 0 for all X

17. The curve being Symmetrical,

x = ��divides curve into two equal halves

such that (Area between – ��to ��)

= (Area between �� to + ��) = 0.5

18. Similarly, under standard normal curve,

(area between –�� to z = 0)

= (area between z = 0 to z = + ��) = 0.5

19. Symbilically

i) P(– �� < X ��) = P(�� X < + ��) = 0.5

ii) P(– �� < Z ��0) = P(0�� Z < + ��) = 0.5

20. The curve being symmetrical area of portions cut off from right and left of X = ��(or z = 0) are equal.

Symbolically, P (– a ��Z ��0) = P(0�� Z ��a).

Note : Here “Area” implies “Probability”


: 481 :

21. The probability that a normal variate Z will take a value less than or equal to a particular value (say

X = K) will be denoted by ��(K) = P( Z �� K)

Note : (i) The probability of success is calculated by evaluating the areas from the standard

normal curve, and the areas are obtained from normal table.

22. % Distribution of areas under Normal Curve / Standard Normal Curve

C-I

P(– 1 �� Z �� 0) = .3413,

P(0 �� Z �� 1) = .3413.

P(– 1 �� Z �� 1) = .6826.

68.26% of total area lies between Z = – 1 and Z = + 1 or X = ��– ��and Z = ��+ ��

C-II

P(– 2 �� Z �� 0) = .4772.

P(0 �� Z �� 2) = .4772.

P(– 2 �� Z �� 2) = .9544.

95.44% of total area lies between Z = – 2 and Z = + 2 or X = ��– 2��and X = ��+ 2��

C-III

P(– 3 �� Z �� 0) = .4987.

P(0 �� Z �� 3) = .4987.

P(– 3 �� Z �� 3) = .9974.

99.74% of total area lies between Z = – 3 and Z = + 3 or X = ��– 3��and X = ��+ 3��


: 482 :

23. Additive Property of Normal Distribution

If X & Y are independent normal variates with means ��1 & ��2 and standard deviation ��1 & ��2

respectively, then Z = X + Y will also follow a Normal Distribution with mean (��1 + ��2) and S.D.

= 22

21 �� symbolically,,

X ~ N(��1, ��1)

Y ~ N(��2, ��2), Z = X + Y ~ N(��1 + ��2,�� 21 �� 2

2 )

24. In continous probability Distribution, Probability is to be assigned to inervals and not to individual

values and accordingly the Probability that a Random Variable X will take any specific value will be

“0” i.e. P(X = C) = 0 when Distribution is continous.

25. Concept of Cumulative Distribution Function (C. D. F.)

Cumulative Distribution Function (C. D. F.) is defined as the Probability that a Random Variable X

takes a value less than or equal to A specified value x and is denoted by F(X)

F(x) = P (X �� x)... F(X) represents Probability; 0 �� F(X) �� 1

26. F(X) = P(X �� C) will imply the area under the probability curve to the left of vertical line at C.

27. Uniform Distribution (Continuous)

A. A continuous Random Variable is said to follow uniform distribution if the probabilities associated

with intervals of same width are always equal at all parts and for any range of values.

B. P. D. F. of uniform distribution is given by : f(x) = a–b

1 (a �� x �� b)

C. It is also known as “Rectangular Distribution”

D. Probability that X lies between any two specified values C and D within the range (“b and a”)

is given by : P(c �� x �� d) = a–bc–d

28. Areas under Standard Normal Curve

C-I

P(0 �� Z �� 1) = .3413, [Z = 0 & Z = any +VE value]


: 483 :

C-II

P(–1 �� Z �� 0)

= P(0 �� Z �� 1)

= .3413

[Z = –VE value to Z = 0] (Due to Symmetry)

C-III

P(Z ��1)

= .5 – P(0 �� Z �� 1)

= .5 – .3413

= .1587

C-IV

P(Z ��– 1)

= P(Z ��1) (Due to symmetry)

= .5 – P(0 �� Z �� 1)

= .5 – .3413

= .1587

C-V

P(Z ��1)

= .5 + P(0 �� Z �� 1)

= .5 + .3413

= .8413

[Area of the Region Greater than A +VE value]

[Area of the Region Less than A negative value]

Area of the Region less than A +VE value]


: 484 :

C-VI

[Area of the Region greater than A –VE value]

P(Z ��– 1)

= P(Z ��1) (Due to symmetry)

= .5 + P(0 �� Z �� 1)

= .5 + .3413

= .8413

C-VII

(Area between two +VE or Two –VE values of Z]

P(1 �� Z �� 2)

= P(0 ��Z �� 2)

– P(0 �� Z �� 1)

= .4772 – .3413

= .1359

Or P(– 2 �� Z �� – 1)

= P(1 �� Z �� 2) (Due to symmetry)

= P(0 �� Z �� 2) – P(0 �� Z �� 1)

= .4772 – .3413 = .1359


: 485 :

C-VIII

(AREA BETWEEN ONE -VE AND ONE + VE VALUE)

P(-2 �� Z ��)

= P (-2 �� Z ��) + (0 �� Z �� 1)

= P (0 �� Z ��) + (0 �� Z �� 1)

= .4772 + .3413

= .8185

NOTE:

1) If the -ve and +ve values happen to be identical .i.e P (-a �� z �� a) in such a case the total

area will be = 2P (0 �� z �� a)

2) When in the problem the magnitude of the given area is greater than “.5” it implies that

area from -� to that particular value of ‘z’ is provided, for evaluating the area from 0 to

that particular value of ‘z’ subtract .5 from ti

E.G. �� (2) = .9772

= P( Z ��) = .9772

= .5 + P (0 �� Z �� 2) = .9772

P (0 �� Z �� 2) = .9772 - .5

= .4772

�� (2) = .9772


: 486 :

29. Methods of fitting Normal Distribution or a Normal Curve

There Are Two Methods Of Fitting Normal Distribution

1) Ordinate Method

2) Area Method

30. Condition under which “Binomial” and “Possion” approaches “Normal Distribution”

Case I

Normal Distribution” is a limiting case of Binomial Distribution when

a) n, the number of trials is infinitely large I.e. n � ��

b) Neither P(or Q) is very small, i.e. p and q are fairly near equal

c) In other words, if neither p nor q is very small but n is sufficiently large Binomial

Distribution approaches Normal Distribution.

Case II

Poission Distribution tends to Normal Distribution with standardised Variable

Z = m

m–xWhere m = Mean = �� = Variance

m = S.d = ��as n increases indefinitely (i.e. as n � ��)


Normal and Standard Normal Distribution - Properties

1. For a normal distribution mean = 50, median = 52. Comment.

a) Statement is correct, as median of normal distribution is greater than its mean.

b) Statement is correct, as difference between median and mean of normal

distribution is always 2.

c) Statement is Incorrect.

d) No comments.

2. If the mean deviation of a normal variable is 16, what is its quartile deviation?

a) 10

b) 15

c) 13.33

d) 12.05

3. If the points of inflexion are 40 and 60 respectively, then its mean is:

a) 40

b) 45

c) 50

d) 60

4. If the quartile deviation of a normal curve is 4.05, then its mean deviation is:

a) 5.26

b) 6.24

c) 4.24

d) 4.80

5. If the 1st quartile and mean deviation about median of a normal distribution are 13.25

and 8 respectively, then the mode of the distribution is:

a) 10

b) 12

c) 15

d) 20

: 487 :


6. If the two quartiles of normal distribution are 14.6 and 25.4 respectively, what is the

standard deviation of the distribution?

a) 6

b) 8

c) 9

d) 10

7. If x and y are 2 independent normal variable with mean 10 and 12 and SD 3 and 4

respectively, then (x + y) is also a normal distribution with mean ____ and SD _____.

a) 22, 7

b) 22, 25

c) 22, 5

d) 22, 49

Area under Normal / Standard Normal Curve

Find the area under the standard normal curve for the following values of standard normal

variate z :

8. Between 0& 1.4z z= =

a) 0.5000

b) 0.4192

c) 0.1942

d) 0.2192

9. Between 0.56& 0z z= − =

a) 0.2323

b) 0.2159

c) 0.2594

d) 0.2123

: 488 :


10. Between 0.45& 2.45z z= − =

a) 0.5568

b) 0.6685

c) 0.6658

d) 0.6665

11. Between 0.75& 1.96z z= =

a) 0.2016

b) 0.2106

c) 0.2594

d) 0.2156

12. 0.8z < −

a) 0.2191

b) 0.2119

c) 0.2911

d) 0.3112

13. 1.76z >

a) 0.0245

b) 0.3256

c) 0.0392

d) 0.0540

14. 1.96z <

a) 0.9750

b) 0.9580

c) 0.9980

d) 0.9780

: 489 :


15. 0.8z > −

a) 0.8871

b) 0.1788

c) 0.8781

d) 0.7881

16. If x follows the Normal Distribution with Mean 12 and Variance 16, find ( 20)P x ≥ .

a) 0.22750

b) 0.25789

c) 0.02275

d) 0.03357

17. If a random variable x follows normal distribution with mean as 120 and standard

deviation as 40, what is P(x≤ 150 / x > 120)?

a) 0.85

b) 0.90

c) 0.95

d) None of the above

Use of Probability Curve to Calculate Probabilities

The mean height of 1000 student at a certain college is 165 cm and standard deviation of

height is 10 cm. Find the number of students whose height is:

18. Less than 172 cms.

a) 298

b) 278

c) 758

d) 246

19. Between 159 and 178 cms

a) 926

b) 629

c) 659

d) 596

: 490 :


20. More than 173.2 cms.

a) 602

b) 206

c) 504

d) 405

The weekly wages of 1000 workers are normally distributed around a mean of ` ` ` ` 70 and a

standard deviation of ` ` ` ` 5. Estimate the number of workers whose weekly wages will be:

21. Between ` ` ` ` 70 and `̀̀̀ 72

a) 155

b) 165

c) 175

d) 185

22. Between ` ` ` ` 69 and ` ` ` ` 72

a) 225

b) 235

c) 245

d) 265

23. More than ` ` ` ` 75

a) 149

b) 169

c) 159

d) 209

24. Less than ` ` ` ` 63

a) 18

b) 81

c) 91

d) 101

25. The 1. Q. s of army volunteers in a given year are normally distributed with

mean =110 and standard deviation =10. The army wants to give advance training to

20% of those recruits with the highest 1 Q. Find that lowest score acceptable for the

advanced training?

a) 118.4

b) 116.4

c) 108.4

d) 101.6 : 491 :


26. The income of a group of 10,000 persons was found to be normally distributed with

mean equal to ` ` ` ` 750 and standard deviation equal to ` ` ` ` 50 what was the lowest income among the richest 250?

a) ` ` ` ` 848

b) ` ` ` ` 648

c) ` ` ` ` 748

d) ` ` ` ` 878

Calculation of missing Parameters

27. Find the mean and standard deviation of a normal distribution of marks in an examination where 44% of the candidates obtained marks below 55 and 6% got above 80 marks.

a) 57.2, 14.66

b) 57.2, 15.98

c) 60.2, 15.98

d) 60.2, 14.66

28. The wages of a group of workers were found to be normally distributed with mean of

` ` ` ` 400 and standard deviation of ` ` ` ` 100. Estimate the total number of workers in this group in each of the following cases:

a) If there were 3174 workers getting below ` ` ` ` 300

b) If there were 6170 workers getting below 450

c) If there were 2996 workers getting wages between ` ` ` ` 300 and 350.

29. The wages of the workers are normally distributed with a standard deviation of 100.

The probability that workers get less than ` ` ` ` 300 is 33%. What is the mean of the distribution? (a) 344 (b) 347 (c) 352 (d) 350

30. The wages of the workers are normally distributed. 33% get less than ` ` ` ` 600 and

30.5% get more than ` ` ` ` 790. Find the mean and standard deviation of the distribution. (a) 700, 200 (b) 688, 200 (c) 650, 200 (d) None of these

: 492 :


Theoretical Aspects

31. Which of the following are continuous distribution?

i. Binomial Distribution

ii. Chi-square Distribution

iii. Normal Distribution

iv. Standard Normal Distribution

v. Student’s t – Distribution

vi. Snedecor’s F – Distribution

vii. Poisson Distribution

viii. Hypergeometric Distribution

ix. Multinomial Distribution

a) II, III, IV, V, VI above

b) All of I, II, III, IV, V, VI, VII, VIII, IX above

c) All but I, VII, VIII above

d) Only III, IV above

32. Probability density function of a continuous distribution satisfies which of the

following(s) condition(s):

i. ( ) 0f x >

ii. ( ) 0f x ≥

iii. ( ) 1

b

a

f x dx =∫ , where a x b≤ ≤

a) All of I, II & III above

b) All but II above

c) All but I above

d) None of I, II, III above

33. Probability density function is associated with which of the following variable?

a) Discrete

b) Continuous

c) Both Discrete and Continuous

d) Neither Discrete nor Continuous

: 493 :


34. Probability density function is always

a) Greater than 0

b) Greater than equal to 0

c) Less than 0

d) Less than equal to 0

35. The most important continuous probability distribution is known as :

a) Chi-square distribution

b) Normal distribution

c) Poisson distribution

d) Sampling distribution.

36. ______ Distribution is a continuous probability distribution and is defined by the

density function:

2

2

( )

21

( )2

x

f x e

µ

σ

σ π

−−

= , where x−∞ < < ∞ .

a) Normal

b) Gaussian

c) Rectangular

d) Both of a) and b) above

37. The probability distribution of z is called Standard Normal Distribution and is defined by the probability density function:

a) 1

( )2

f x eπ

= ; x−∞ < < ∞

b) 21

( )2

xf x eπ

−= ; x−∞ < < ∞

c)

2

21

( )2

z

p z eπ

−= ; z−∞ < < ∞

d) 21

( )2

zf z eπ

−= ; z−∞ < < ∞

38. The parameters of the normal distribution are:

a) &µ β

b) &µ σ

c) &β σ


: 494 :


39. What are the parameters of standard normal distribution?

a) 0 and 1

b) &µ β

c) &µ σ

d) The distribution has no parameters

40. If a random variable x is normally distributed with mean µ and standard deviation σ ,

then x

zµ

σ−

= is called:

a) Normal Variate

b) Standard Normal Variate

c) Chi-square Variate

d) Uniform Variate

41. The probability curve of normal distribution is known as:

a) Uniform Curve

b) Standard Curve

c) Normal Curve

d) Lorenz Curve

42. The curve of which of the following distribution is uni-modal and bell shaped with the

highest point over the mean

a) Poisson

b) Binomial

c) Normal

d) All of the above

43. The total area under the normal curve is:

a) 1.00

b) 0.50

c) 0.85

d) 1.25

: 495 :


44. The curve of which of the following distribution(s) has single peak?

a) Poisson

b) Binomial

c) Normal

d) Both a) and b) above

45. The normal curve is:

a) Positively Skewed

b) Symmetrical in nature

c) Negatively Skewed


46. The shape of the Normal Curve is:

a) Inverse J

b) U

c) J

d) Bell

47. In Normal distribution as the distance from the _______ increases, the curve comes

closer and closer to the horizontal axis.

a) Standard Deviation

b) Mean

c) Both a) and b) above

d) Neither a) nor b) above

48. The mean and mode of a Normal distribution

a) May be different.

b) May be equal

c) Are always equal

d) Either a) or b) above

: 496 :


49. For Standard Normal distribution, which of the following is correct?

a) Mean = 1; S.D. = 1

b) Mean = 1, S.D. = 0

c) Mean = 0, S.D. = 1

d) Mean = 0, S.D. = 0.

50. Because of the symmetry of Normal distribution the median and the mode have the

______ value as that of the mean.

a) Greater

b) Smaller

c) Same

d) Nothing can be said

51. In Normal distribution the probability has the maximum value at the

a) Mode

b) Median

c) Mean

d) All of the above

52. The mean deviation about median of a Standard Normal Variate is:

a) 0.675

b) 0.675 σ

c) 0.80 σ

d) 0.80

53. For a standard normal distribution, the points of inflexion are:

a) 1− to 1+

b) σ− to σ+

c) µ σ− to µ σ+

d) The distribution has no points of inflexion

: 497 :


54. The interval ( µ – 3σ , µ + 3σ ) covers

a) 96% area of a normal distribution.

b) 95% area of a normal distribution.

c) 99% area of a normal distribution.

d) All but 0.27% area of a normal distribution.

55. If neither p nor q is very small but n is sufficiently large, then Binomial distribution is very closely approximated by distribution.

a) Poisson

b) Normal

c) t

d) None.

56. Poisson distribution approaches a Normal distribution as n:

a) Increase infinitely

b) Increase moderately

c) Decrease


57. Let x be a normal variable for any point k . Let ( )kφ = ( )P x k≤ then ( ) ( )k kφ φ+ − =?

a) 1

b) – 1

c) 0.5


58. ( ) ( )F c P x c= ≤ is known as:

a) Probability Density Function

b) Continuous Density Function

c) Both of a) and b) above


59. The cumulative distribution function of random variable x is given by

a) F(x) = P (X ≤x)

b) F(X) = P (X = x)

c) F(x) = P (X≥x)

d) None of these

: 498 :


60. The number of methods of fitting the Normal Curve is

a) 1

b) 2

c) 3

d) 4

61. The symbol ∅ (a) indicates the area of standard normal curve between

a) 0 to a

b) a to ∞

c) - ∞ to a

d) -∞to∞

62. If the area of standard normal curve between z = 0 to z = 1 is 0.3413, the value of ∅(1) is

a) 0.5000

b) 0.8413

c) -0.5000

d) 1

63. For which distribution, whatever may be the parameter of distribution, it has same shape,

a) Normal

b) Binomial

c) Poisson

d) None

64. Distribution is asymptotic to the horizontal axis

a) Binomial

b) Normal

c) Poisson

d) t

65. f(x) =1

b a− where a ≤ x ≤ b, is the probability density function of

a) Uniform Distribution

b) Rectangular Distribution

c) Both of a) and b) above


: 499 :


THEORY ANSWERS:

31 a 41 c 51 d 61 c

32 c 42 c 52 d 62 b

33 b 43 a 53 a 63 a

34 b 44 c 54 d 64 b

35 b 45 b 55 b 65 c

36 d 46 d 56 a

37 c 47 b 57 a

38 b 48 c 58 d

39 d 49 c 59 a

40 b 50 c 60 b

: 500 :

Documents

11. Normal Distribution - CA Study · 2021. 2. 21. · 11. Normal Distribution 1. It is a continuos probability distribution where the variable ‘X’ can assume any value between