B. S.

Parajuli
 Random Variable

A random variable is a variable whose possible values are the

numerical outcomes of a random experiment. Therefore, it is
a function which associates a unique numerical value with every
outcome of an experiment. Further, its value varies with every
trial of the experiment.
It can take any one of the various possible values each with a
definite probability. For example, in a throw of a die, if X denotes
the number obtained then X is a random variable, which can
take any one of the values 1, 2, 3, 4, 5 or 6 each with equal
probability 1/6. Similarly, in a toss of a coin, if X denotes the
number of heads, then X is a random variable which can take any
one of the two values 0 (no head i.e. tail) or 1 (i.e. head) each
with equal probability 1/2 .
Let us consider a random experiment of three tosses of a coin (or
three coins tossed simultaneously), then the exhaustive number
of cases (total number of outcomes) or sample space S consists
of 23 = 8 points as shown below.
S = (H,T) x (H,T) x (H,T) = (HH, HT, TH, TT) x (H,T)
= (HHH, HTH, THH, TTH, HHT, HTT, THT, TTT)
If we consider the variable X, which is the number of heads
obtained, then X is a random variable, which can take any one of
the values 0, 1, 2, 3. This can be presented as below.
Outcome: HHH HTH THH TTH HHT HTT THT TTT
Value of X: 3 2 2 1 2 1 1 0
Discrete Random Variables
Discrete random variables take on only a countable
number of distinct values. Usually, these variables
are counts (not necessarily though). If a random
variable can take only a finite number of distinct
values, then it is discrete.
Number of members in a family, number of
defective light bulbs in a box of 10 bulbs, etc.
You toss a coin 10 times. The random variable X is
the number of times you get a ‘tail’. X can only take
values 0, 1, 2, … , 10. Therefore, X is a discrete
random variable.
• Experiment: Toss 2 Coins. Let X = no. of heads.
4 possible outcomes

T T Probability Distribution
X Value Probability
0 1/4 = 0.25
H T
1 2/4 = 0.50
2 1/4 = 0.25
T H

H H
Continuous Random Variables
• Continuous random variables take up an infinite number of
possible values which are usually in a given range. Typically,
these are measurements like weight, height, temperature of a
solution, the time needed to finish a task, etc.
• To give you an example, the life of an individual in a
community is a continuous random variable. Let’s say that the
average lifespan of an individual in a community is 110 years.
• Therefore, a person can die immediately on birth (where life =
0 years) or after he attains an age of 110 years. Within this
range, he can die at any age. Therefore, the variable ‘Age’ can
take any value between 0 and 110.
• If a random variable (X) takes ‘k’ different
values, with the probability that X = xi is
defined as P(X = xi) =pi, then it must satisfy the
following:
• 0 < pi < 1 (for each ‘i’)
• p1 + p2 + p3 + … + pk = 1
Mathematical Expectation:
The mathematical expectation, also called the expected value of a random
variable is the weighted arithmetic mean of the variable; the weights are
being the respective probabilities of the values that the random variable can
possibly assume.
Thus, if X is a discrete random variable, which can take the values x1, x2,
……..xn with respective probabilities p1, p2, ……..pn, where , then the
mathematical expectation of X denoted by E(X) is defined as
E(X) = p1X1 + p2X2 + …………+ pnXn = σ 𝑝𝑖 𝑥𝑖
Where, σ 𝑝𝑖 = 𝑝1 + 𝑝2 + 𝑝3 +……..+ 𝑝𝑛 = 1
More precisely, if X is a random variable with probability distribution {x, p(x)},
then E(X) = σ 𝒙 . 𝒑(𝒙)

Variance of X: Var(X)= Σ[xi – E(xi ) ]2 * P(xi)=

where: E(X) = Expected value of the discrete random variable X
Xi = the ith outcome of X
P(Xi) = Probability of the ith occurrence of X

Standard deviation = 𝒗𝒂𝒓𝒊𝒂𝒏𝒄𝒆

Example Find the expected sales of Toyota car in Kathmandu city
in a week from the following information.
Day Sun Mon Tue Wed Thu Fri
Sales(Rs.) 90 60 90 50 75 55
Probability 0.25 0.18 0.12 0.05 0.20 0.20
Expected Sales of Toyota car in Kathmandu is
E(X) =
=

= 22.5+10.8 + 10.8 + 2.5 + 15 + 11 = 72.6

Example A random variable X has the following probability
distribution.
X: 0 1 2 3 4
P(X): 0.35 0.28 0.15 0.12 0.10
Compute the expected value.
Compute the standard deviation.
X P(X) X.P(X) X2.P(X)
0 0.35 0 0
1 0.28 0.28 0.28
2 0.15 0.30 0.60
3 0.12 0.36 1.08
4 0.10 0.40 1.60
1.34 3.56
 Actual vs Theoretical Distributions

• The empirical (observed or experimental)

frequency distributions are based on the
actual sample observations (drawn from
population).
• The theoretical probability distribution is
generally used to estimate the future values
through the way of limited information (or
theoretical assumptions), and to make
decision logically.
 Probability Distributions
The concept of probability distribution is analogous to that of
frequency distribution. As frequency distribution is the
summarization of data into tabular form, a theoretical
probability distribution is also a systematic arrangement of
probability value associated with the outcome of the
experiment. Probability distribution tells us how total probability
of 1 is distributed among different values, which the random
variable can take.

Discrete Continuous
Probability Probability
Distributions Distributions

Binomial Distribution Normal Distribution

Poisson Distribution
 Binomial Probability Distribution
• Binomial distribution is one of the most widely used discrete
probability distribution. It is based on "Bernoulli process" developed
by Swiss mathematician Jacob Bernoulli (17th century).

• The Binomial Distribution is a widely used probability distribution of

a discrete random variable. It plays a major role in quality control
and quality assurance function.
• A manufacturing plant labels items as either defective or acceptable
• Binomial distribution is also being used in service organizations like
banks, and insurance corporations to get an idea of the proportion
customers who are satisfied with the service quality.
• A firm bidding for contracts will either get a contract or not
• A marketing research firm receives survey responses of “yes I will
buy” or “no I will not”
• New job applicants either accept the offer or reject it
 Assumptions of Binomial Distribution

1. The number of experiment is finite (n)

2. Each experiment has only two possible outcome
(dichotomous), e.g., success (S) or failure (F),
heads (H) or tails (T), Yes (1) or No (0).
3. "The outcome of any trial is independent of the
outcome of any other trials". i.e. the trials are
statistically independent.
4.The probability of success (happening of an event)
or failure in any trial is constant (same) for each
trial.
 Binomial Distribution Formula
• if X denotes the number of successes in n trials of a
binomial experiment satisfying the above conditions,
then X is a random variable, which can take any one of
the (n+1) integer values 0, 1, 2,………n, then the
probability of r successes out of n trials is given by
P(x) = P(X = x) = ncx pxqn-x,
r = 0, 1, 2, 3, ………..n
• Where, q = 1 – p; the probability of failure on each
trial. The binomial probability distribution has two
parameters n and p.
• The mean of the binomial distribution E(x) = 𝑛𝑝
• variance V(x) = 𝑛𝑝𝑞
• Standard deviation (𝜎)= 𝑛𝑝𝑞
Example : Suppose the probability of purchasing a defective
computer is 0.07. What is the probability of purchasing 2
defective computers in a group of 10?
x = 2, n = 10, and p = 0.02
n!
P(X = 2) = 𝑝x (1 − 𝑝)n−x
x!(n − x)!
10!
= (0.02)2 (1 − 0.02)10−2
2!(10 − 2)!
= (45)(0.0004)(0.8508)
= 0.01531

1
(i) If p = q = the binomial distribution is
2
symmetrical.
1
(ii) If p > , binomial distribution is negatively
2
skewed.
1
(iii) If p < , it is positively skewed.
2
 Fitting of Binomial Distribution:
If a random experiment consisting of n trials is repeated N times satisfying
the conditions of binomial distribution, then the frequency of r successes is
given by f (x) = N. P(x) = N. ncx pxqn-x
Putting r = 0, 1, 2, ………n, we can get the expected or theoretical
frequencies of the binomial distribution.

No. of successes(r) f(x) = N. p (x) = N. ncx pxqn-x

0 f(0) = N. p (0) = N. nc0 p0qn-0 = N.qn
1 f(1) = N. p (1) = N. nc1 p1qn-1
2 f(2) = N. p (2) = N. nc2 p2qn-2
. .
n f(n)= N. p (n) = N. ncn pnqn-n = N. pn

• However, if p is not known and if we want to fit a binomial distribution to

a given frequency distribution, first we find the mean of the given
σ 𝑓𝑋
ത
frequency distribution by the formula 𝑋= and equate it to np, which
𝑁
is the mean of the binomial probability distribution. Hence, p can be
𝑋ത
estimated by the following relation 𝑋 = np  p = , then q = 1-p
ത
𝑛
 Fits the following binomial Data
X: 0 1 2 3 4
f: 28 62 46 10 4
since p is unknown, its value can be calculated by using following formula,
σ 𝑓𝑥
np = n=4 , N = 150
𝑁
0×28 + 1×62 + 2×46 + 3×10 +(4×4)
4.p = , p = 1/3, q = 1- 1/3 = 2/3
150

r 1 2
E = N. p (x) = N. ncx pxqn-x = 150 . 4C𝐱 . (3)𝒙 (3)4−𝒙

0 1 2
150 . 4C0 . ( )0 ( )4−0 = 29.63 ≈ 30
3 3
1 4 1 2
150 . C1 . (3)1 (3)4−1 = 59.26 ≈ 59
2 4 1 2
150 . C2 . (3)2 (3)4−2 = 44.44 ≈ 44
3 1 2
150 . 4C3 . (3)3 (3)4−3 = 14.81 ≈ 15
4 1 2
150 . 4C4 . ( )4 ( )4−4 = 1.85 ≈ 2
3 3

Hence the required expected frequencies are 30,

59, 44, 15, 2.
• The shape of the binomial distribution depends on the values of
p and n
 Poisson Distribution
• Poisson distribution was derived in 1837 by a French
Mathematician Simeon D. Poisson. Poisson distribution is a
limiting form of binomial probability distribution may be
used under the following conditions:
• It is used when a random event happens at a constant
average.
• n, the number of trials is indefinitely large i.e. n→
• p, the probability of success or happening of an event for
each trial is indefinitely small i.e. p→0
• np = 𝜆 (say) is finite.
• In other words, if the number of trials (n) is indefinitely
large (generally, n ≥ 20) and the probability of success for
each trial (p) is indefinitely small (generally, p ≤ 0.05)
Poisson distribution will be more appropriate to use. This is
poissson approximation to binomial distribution.
• Poisson distribution is used to describe the behavior of rare
events where probability of success is low. Therefore, it is also
called "Law of improbable events“
Examples of Poisson Distribution
• The number of accidents in a busy road
• Number of defective items in a lot containing large number of
manufactured items
• Number of printing mistakes in a book
• Number of death reported in a city due to heart-attacks,
cancer, tumor, suicides etc.
• Number of wrong telephone calls in a busy hour of the day
• Number of vehicles passing per minute from the station.
• Number of customers arriving at super market per hour, etc.
 Poisson Distribution Formula

𝑒 −𝜆 𝜆𝑥
𝑃(𝑋 = 𝑥) =
𝑥!

where:
x = number of events or no. of success in an area of
opportunity = 0, 1, 2, 3,…………………
 = mean of the distribution or average no of
occurrence
e = base of the natural logarithm system (2.71828...)
Mean, = 𝝀
Variance = 𝝀
Standard deviation = 𝝀
Example : The quality control manager of Marlyin’s Cookies is
inspecting a batch of chocolate-chip that has just been baked. If
the production process is in control, the average number of chip
parts per cookies is 6. What is the probability that in any
particular cookie being inspected..
i. Fewer than five chip parts will be found?
ii. Exactly five chip parts will be found?
iii. Five or more chip parts will be found?
iv. Four or five chip parts will be found?
Solution:
𝑒 − 𝑥
 = 6 , we have P(X=x) =
𝑥!
(i) P(X< 5)= P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
=……
 𝑒 −6 60 𝑒 −6 61 𝑒 −6 62 𝑒 −6 63 𝑒 −6 64
= 0!
+
1!
+
2!
+
3!
+
4!
0 61 62 63 64
−6 6
= 𝑒 [ + + + + ]
0! 1! 2! 3! 4!
= 0.00248[1+6+18+36+54] = 0.00248× 115 = 0.28581
𝑒 −6 65
(ii) P(X = 5) =
5!
(iii) P(X ≥ 5) = 1 – P(X<5) = 1 – 0.2851 =0.715
(iv) P(X = 4 or 5) = P(X = 4) + P(X = 5) =
Fitting of Poisson Distribution:
• If N is the total number of frequencies, then the expected or
theoretical frequencies of the Poisson distribution are given by
𝑒 −  𝑥
f ( x ) = N. p( x ) = N. P (X=x) = N.
𝑥!

If 𝜆 is unknown we computed by using relationship

σ 𝑓𝑋
𝑋ത = =𝜆
𝑁
Example The number of telephone calls received per day over a
period of 100 days is given below
Number of calls 0 1 2 3
Number of days 45 25 20 10
Find the expected frequencies.
σ 𝑓𝑥 95
Solution: N = 100 , 𝑋ത = = = 0.95 , 𝜆 = 0.95
𝑁 100
e−λ . λx
Expected frequency (E) = N. P r = N x
x!
e−0.95 . 0.95x
= 100 x
x!
x e−0.95 . 0.95𝐱
𝐄=𝐟 𝐱 = 100 x
𝐱!

e−0.95 . 0.950
0 100 x 0!
= 39

e−0.95 . 0.951
1 100 x 1!
=37

e−0.95 . 0.952
2 100 x 2!
=17

e−0.95 . 0.953
3 100 x 3!
=6

Hence the expected frequencies are 39,37,17 and 6.

 e−λ λx e−0.50 (0.50)2
P(X = 2) = = = 0.0758
x! 2!
 Normal Distribution
• Normal distribution is one of the most used and best-known
continuous theoretical distributions in statistics. Several
mathematicians were instrumental in its development,
including the eighteenth-century mathematician-astronomer
Karl Gauss. In honor of his work, the normal probability
distribution is often called the Gaussian distribution.
• There are two basic reasons why the normal distribution
occupies such a prominent place in statistics. First, it has some
properties that make it applicable to a great many situations
in which it is necessary to make inferences by taking samples.
We will find that the normal distribution is a useful sampling
distribution. Second, the normal distribution comes close to
fitting the actual observed frequency distributions of many
phenomena relating to economics, business, social and
physical science, including human characteristics such as
weights, heights and IQs and so on.
A continuous random variable X is said to be normally
distributed with mean and standard deviation if it has a
probability density function of X represented by the
equation 𝟏 𝑿−𝝁 𝟐
𝟏 − ( )
f(x) = 𝒆 𝟐 𝝈
𝝈 𝟐𝝅
Where
e = the mathematical constant approximated by 2.71828
π = the mathematical constant approximated by 3.14159
μ = the population mean
σ = the population standard deviation
X = any value of the continuous variable
The mean  and  are called the parameters of the normal
distribution. Symbolically, x follows normal variable with mean(𝜇)
and standard deviation(𝜎) (or variance 𝜎 2 ) and is written as
X ~N(𝜇 , 𝜎 2 ).
By varying the parameters μ and σ, we obtain different
normal distributions
 Properties of Normal Distributions

• It is perfectly symmetrical about the mean and is bell-shaped curve.

• The random variable has an infinite theoretical range:
f(X)
+  to − 
• In a normal distribution ,
Mean = Median = Mode
• It is uni-modal i.e. it has only one
mode. (having single peak) X
μ
• The ordinate at the mean of the
Mean = Median = Mode
normal curve divides the total are under the normal curve into two
equal parts and each part is of area 0.5.
• The two tails of the normal distribution extend indefinitely both
sides and never touch the horizontal axis.
• Normal distribution is a limiting case of binomial and Poisson
distribution.
Area Property:
If the mean and S.D. of a normal distribution are 𝜇 and 𝜎 ,Then area under the normal
probability curve between the ordinate
• P(  - < 𝑿 <  +  ) = 0.6826 i.e.    occupies 68.26% of the observations.
• P( - 2 < 𝑿 <  + 2) = 0.9544 i.e.   2 occupies 95.44% of the observations.
• P(  - 3 < 𝑿 <  + 3 ) = 0.9973 i.e.   3 occupies 99.73% of the observations.
The curve are shown in the figure below:

68.26%

95.44%

𝑋 𝑆𝑐𝑎𝑙𝑒
 − 3  − 2  −  𝜇  +   + 2  + 3

-3 -2 -1 0 1 2 3 𝑍 𝑆𝑐𝑎𝑙𝑒
99.74%
Standard Normal Distribution

• If x is a random variable following the normal distribution with

mean  and standard deviation , then the random variable z
defined as Z = 𝑋−𝜇 is called the standard normal variate.
𝜎

• The standard normal distribution always has zero mean and

standard deviation unity and is denoted by N(0,1)
• The formula for the standardized normal probability density
function is 1 − (1/2)Z 2
f(Z) = e
2π
Where e = the mathematical constant approximated by 2.71828
π = the mathematical constant approximated by 3.14159
Z = any value of the standardized normal distribution
• It is also known as Z Distribution.
If X is distributed normally with mean of 100 and standard
deviation of 50, the Z value for X = 200 is
X −μ 200 − 100
Z= = = 2.0
σ 50

This says that X = 200 is two standard deviations (2

increments of 50 units) above the mean of 100.
Probability is measured by the area under the curve
f(X)
P(a ≤ X ≤ b)

X
a b
 Area Under Normal Curve
• When the values of Z is known
The Cumulative Standardized Normal table in the textbook
(Appendix table ) gives the probability less than a desired value
of Z (i.e., from negative infinity to Z)

P(Z < 2.00) = 0.9772

If areas under standard normal distribution between mean and
positive values of Z (o to 𝑍)
0.5

2.0
P(Z < 2.00) = 0.5 + 0.4772
 The Column gives the value of Z to the
second decimal point
Z 0.00 0.01 0.02 …
The row shows the
value of Z to the 0.0 The value within the table
first decimal 0.1 gives the probability from Z = −
point .  up to the desired Z value
.
.
P(Z < 2.00) = 0.9772
2.0 .9772

Z 0.00 0.01 0.02 …

0.0 The value within the table
0.1 gives the probability from Z = 0
. up to the desired Z value
.
.

2.0 .4772
P(Z < 2.00) = 0.5 + 0.4772= 0.9772
 • When the Scores is known/given
To find P(a < X < b) when X is distributed normally:
• Draw the normal curve for the problem in terms of X
• Translate X-values to Z-values
• Use the Standardized Normal Table
Suppose X is normal with mean 8.0 and standard deviation 5.0.
(a) Find P(8 < X < 8.6)
𝑋 −𝜇
We have, Z =
𝜎
8 −8
When X = 8, Z= =0
5
8.6 −8
When X = 8, Z= = 0.12
5
8 8.6
P(8 < X < 8.6) = P(0 < Z < 0.12) = 0.0478 0 0.12
OR = P(Z < 0.12) – P(Z ≤ 0) = 0.5478 - .5000 = 0.0478
(b) P(X > 8.6) = P(Z > 0.12) = 1.0 - P(Z ≤ 0.12) = 1.0 - 0.5478 = 0.4522
OR = 0.5 – P(0 < Z < 0.12) = 0.5 – 0.0478 = 0.4522
Example: The inside mean diameter of 500 washers produced by a machine is
5.02 mm and the standard deviation is 0.05 mm. The purpose for which these
washers are intended allows a maximum tolerance in the diameter of 4.96 to 5.08
mm., otherwise the washers are considered defective. Determine the percentage
of defective washers produced by the machine assuming the diameters are
normally distributed.
Solution:
N= 500, 𝜇= 5.02 mm., 𝜎 = 0.05 mm.
𝑋 −𝜇
We have, Z =
𝜎
𝑋 −𝜇 4.96 −5.02
When, X= 4.96, Z = = = -1.2 X
𝜎 0.05 1.2
-1.2
𝑋 −𝜇 5.08 −5.02 5.02
Again, when, x=5.08, Z = = = 1.2
𝜎 0.05
Now, P(4.96<x<5.08) = P(-1.2<z<1.2)
= P(-1.2<z<0) + P(0<z<1.2) = 2 . P(0<z<1.2) [by symmetry]
= 0.3849 + 0.3849 = 0.7698
The probability of non defective washers = 0.7698
The probability of defective washers = 1 – 0.7698 = 0.2302
The % of defective washers = 0.2302 100 = 23.02%
Example: A banker claimed that the life of a regular saving account opened in his
bank average 18 months with a standard deviation of 6.45 months. What is the
probability that:
i. There will be still money in a saving account between 20 to 22 months by a
depositor.
ii. The account will be closed (no money in the deposit) after 2 years?
Solution: With the usual notation, =18 months, 6.45 months
(i) Probability that there will be still money in the saving account between 20 to
𝑋 −𝜇
22 months is P(20 <X <22) We have, Z =
𝜎
20 −18 22 −18
When X = 20 , Z1 = = 0.31 Again, When X = 22 , Z1 = = 0.62
6.45 6.45
P(20<X<22) = P(0.31<z<0.62) = P(0<z<0.62) – P(0<z<0.31)
= 0.2324-0.1217 = 0.1107
(ii) Probability that the account will be closed (no money in the deposit) after two
24 −18
years(24 months) is P(X >24) When X = 22 , Z = = 0.93
6.45
P(x>24) = P(z>0.93) = P(0<z< )-P(0<z<0.93) = 0.5-0.3238 = 0.1762
Finding the scores(values) when the area (probability) is known

Steps to find the X value for a known probability:

1. Find the Z value for the known probability
2. Convert to X units using the formula: X = 𝝁 + Z 𝝈

Example : Given a normal distribution with 𝝁 = 50 and 𝝈 = 10 find

the value of X that has (i) 13% of the area to its left and
(ii) 4% of the area to its right
Solution:
0.13 0.04

P(-∞ <Z< −𝟏.13) = 0.13 0.37 0.46

OR
X
P(0< Z < 1.13) = 0.37 ? 8.0 ?
-1.13 0 1.75 Z
 Z …….. .02 .03 .04 If we use the table of
areas between 0
1.0 ……….. .3461 .3485 0.3508 to values of Z
1.1 …………. .3686 .3708 0.3729

1.2 …………. .3888 .3907 0.3925

Z ………. .01 .02 .03 If we use the table of

.04
areas between
-1.2 ……… 0.1131 0.1112 0.1093 0.1075 -∞ to values of Z

-1.1 ……… 0.1335 0.1314 0.1292 0.1271

-1.0 ………. 0.1562 0.1539 0.1515 0.1452

(i) X = μ + Zσ (ii) X = μ + Zσ
Similarly
= 50 + ( −1.13 )10 = 50 + (1.75 )10

= 38.7 = 67.5
Problem: The marks of 500 candidates in an examination are
normally distributed with a mean of 45 and standard deviation of 20
marks.
a. Given that the pass mark is 40, estimate the number of
candidates who passed the examination.
b. If 5% of the candidates obtained a distinction by scoring x marks
or more, estimate the value of x.
c. If 400 candidates to be passed, what should be the lowest mark
for passing?
Solution:
Let the random variable X denotes the marks obtained by the
candidates. Then X follows normal distribution with mean and
standard deviation . Given, N = 500, 𝜇 = 45, and 𝜎 = 20
𝑋−𝜇 40 − 45
(a) We have Z = , When X = 40, Z = = -0.25
𝜎 20
If the pass mark is 40, the area in the normal curve indicates the area above x = 40,
which can be written as
P(40≤ 𝑋 ≤ ∞ ) = P(-0.25≤ Z≤ ∞) = P(-0.25≤Z≤0) + P(0≤ 𝑍 ≤ ∞)
= 0.0987 + 0.5 = 0.5987
The expected number of students who passed the examination
=500 × 0.5987= 299.71 ≈ 300
(b) Let 𝑥1 denotes the lowest mark of 5% candidates, who have scored distinction
marks, then P(X≥ 𝑥1 )= 0.05
𝑋−𝜇
We have Z = 0.05
𝜎
𝑋−𝜇 𝑥1 −45 0.5 0.45
let Z = = = z1 ………………..(i)
𝜎 20
P(Z≥z1)= 0.05
𝒙𝟏 =?
P(0≤Z ≤ ∞) – P(0≤Z ≤ z1) = 0.05
0.5 - P(0≤Z ≤ z1) = 0.05 z1
P(0≤Z ≤ z1) = 0.45
From the normal table, the value of z corresponding to the probability 0.45 is 1.645
i.e. z1=1.645
Hence from (i) 𝑥1 = 𝜇+ z1 𝜎 = 45 + 1.645× 20 = 45 + 32.9 = 77.9
Hence, the lowest score of top 5% is 77.9 marks.
 400
(c) The percentage of candidates to be passed = ×100% = 80%
500
i.e 20% of the candidates are to be failed. Let 𝑥2 denotes the highest score of the
lowest 20% of the candidates.
Then, P(X ≤ 𝑥2 ) = 20% = 0.20 0.20
𝑋−𝜇 𝑥2 −45 0.30
let Z = = = z2 ………………..(ii)
𝜎 20 0.5
P(Z ≤ z2 ) = 0.20
P(−∞ ≤ Z ≤ 0) – P(- z2≤Z ≤0) = 0.20 𝒙𝟐 =?
0.5 - P(- z2≤Z ≤0) = 0.20 z2
P(- z2≤Z ≤0) = 0.30
From the normal table, the value of z closure to 0.30 is 0.2995 at z = 0.84.
i.e. z2= -0.84
from (ii) 𝑥2 = 𝜇+ z2 𝜎 = 45 + -0.84× 20 = 45 -16.8 = 28.2
Hence if 400 candidates are to be passed out of 500, the lowest mark for passing
= 28.2 mark.
Example: In a certain examination 20% of the students, who appeared in
Statistics paper, got less than 30 marks and 97% of the students got less than 62
marks. Assuming the distribution to be normal, find the mean and standard
deviation of the distribution.
Solution: Let x be a random variable of marks obtained by the students, which is
normally distributed.
According to the question, we have,
P(x<30) = 0.20 ………..(i)
P(x<62) = 0.97 ……….(ii)
The standard normal variate is
30−𝜇
When x =30, then, Z = 𝜎
= - z1 …………(iii)
62−𝜇
When x = 62, then, Z= 𝜎
= - z2 …………(iv)
Now, P(x<30) = 0.20 or P(z < -z1) = 0.20
Or, P(-∞ < 𝑍 < 0) – P(- 𝑧1 < 𝑍 < 0) = 0.20
Or, 0.5 – P(0< 𝑍 < 𝑧1 ) = 0.20
Or, P(0< 𝑍 < 𝑧1 ) = 0.30
From the normal table, the value of Z closer to 0.30 is 0.2995 at Z = 0.84
Z = z1 = - 0.84
 30−𝜇
From equation (iii) we get, = - 0.84 or, 𝟑𝟎 − 𝝁 = -0.84 𝝈 ………….(v)
𝜎
Again, P(x<62) = 0.97 or P(z < z2) = 0.97
Or, P(-∞ < 𝑍 < 0) + P(0< 𝑍 < 𝑧2 ) = 0.97
Or, 0.5 + P(0< 𝑍 < 𝑧2 ) = 0.97
Or, P(0< 𝑍 < 𝑧2 ) = 0.47
From normal table, the value of Z closure to 0.47 is 0.4699 at Z = 1.88
62−𝜇
From equation (iv), we get, = 1.88 or, 𝟔𝟐 − 𝝁 = 1.88 𝝈 ………………..(vi)
𝜎
Subtracting (vi) from (v), we have,
62 − 𝜇 = 1.88 𝜎
30 − 𝜇 = -0.84 𝜎
- + +
………………………………………..
32 = 2.72 𝜎 or, 𝜎 = 32/2.72 = 11.76
Substituting the value of in eqn. (v), we get, 30 - 𝜇= - 0.84 × 11.76
or 𝜇= 30 + 9.88 = 39.88
Hence mean(𝜇) = 39.88 and standard deviation(𝜎) = 11.76