Probability Distribution
PROBABILITY DISTRIBUTION
A series of the values that a random a random experiment takes together with their probabilities
is called a probability distribution. There are mainly three types of probability distribution. They
are
1. Binomial probability Distribution
2. Poisson Distribution
3. Normal Distribution
1. BINOMIAL DISTRIBUTION
Binomial means two outcomes of an experiment. So, the probability distribution associated with
a random experiment which results into two mutually disjoint outcomes is known as a Binomial
distribution. For example, a coin has only two outcomes head and tail. Similarly, the even and
odd faces of a die are also associated with binomial distribution being it consists of two mutually
disjoint outcomes.
In a binomial Probability Distribution, the two basic features are:
i. The trial is repeated for a finite number of times.
ii. There must be only two types of outcomes of the trial. (Test of Binomial Distribution)
Notation:
p = Prob. of success in every trial (desired outcomes)
q = Prob. of failure in every trial (unwanted outcomes)
n = no. of repetition of a trial.
x = no. of success.
Note that 𝒑 + 𝒒 = 𝟏.
Formula:
𝑃(𝑥 𝑠𝑢𝑐𝑐𝑒𝑠𝑠) = 𝐶(𝑛, 𝑥)𝑝 𝑞
.
Examples
D.B.Thapa, DMC
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Probability Distribution
1. A box contains 50 items of which 20 are defectives. If one item is selected at random
from the box, what is the probability that it is a non-defective item? [TU 2020]
Solution: Here,
𝑝 = P(success) = P(Non-defective) =
= .
𝑞 =1−𝑝=1− = .
𝑛 = 1.
P(1 success) = ?
We know,
𝑃(𝑥 = 1) = 𝐶(𝑛, 𝑥)𝑝 𝑞
.
(
= 𝐶(1,1)
)
.
= 1 × × 1.
= . Ans
Alternative way
Here,
No. of defective items = 20.
∴
No. of non-defective items = 50-20 = 30.
Then, 𝑃(𝑁𝑜𝑛 − 𝑑𝑒𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑖𝑡𝑒𝑚𝑠) =
= . Ans
2. Three coins are tossed simultaneously. Find the probability of getting
a. no head
b. exactly 2 heads.
c. at least one head.
d. not more than 3 heads. Head is our success and tail is our failure.
Solution: Here,
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Probability Distribution
𝑝 = 𝑃(ℎ𝑒𝑎𝑑) = .
𝑞 = 𝑃(𝑡𝑎𝑖𝑙) = 1 − 𝑝 = 1 − = .
𝑛 = 3.
a. 𝑃(𝑥 = 0) =?
We know, 𝑃(𝑥 = 0) = 𝐶(𝑛, 𝑥)𝑝 𝑞
.
= 𝐶(3, 0)
=1×1× .
= . Ans
𝑏. 𝑃(𝑥 = 2) =?.
Solution: We know,
.
𝑃(𝑥 = 2) = 𝐶(𝑛, 𝑥)𝑝 𝑞
= 𝐶(3,2)
.
×
=3× .
= . Ans
𝑐. 𝑃(𝐴𝑡 𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑒 ℎ𝑒𝑎𝑑) = 𝑃(𝑥 ≥ 1).
= 𝑃(𝑥 = 1) + 𝑃(𝑥 = 2) + 𝑃(𝑥 = 3) + ⋯ + 𝑃(𝑥 = 10).
= 1 − [𝑃(𝑥 = 0) + 𝑃(𝑥 = 1)].
= 1 − {𝐶(𝑛, 0)𝑝 𝑞
+ 𝐶(10,1)𝑝 𝑞
= 1 − 𝐶(10,0)
=1−
+ 10 ×
+ 𝐶(10,1)
.
.
. Ans
Mean and Variance of Binomial Distribution
Consider a binomial distribution with probability distribution as
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Probability Distribution
No. of Success
Prob.
𝑥
𝑝
𝑥
𝑝
𝑥
𝑝
𝑥
𝑝
𝑥
𝑝
𝑥
𝑝
Then,
Mean of the distribution = no. of repetition of trial × Prob. of success in each trial.
i.e;
Mean = 𝑛𝑝.
And, variance of the binomial distribution is given by
Variance = no. of repetition of trial × Prob. of success in each trial × Prob. of failure
i.e;
Variance = 𝑛𝑝𝑞.
3. If the mean and variance of a binomial distribution are 0.4 and 0.36 respectively, find the
probability of at least one success.
Solution: Here, given that
𝑛𝑝 = 𝑜. 4,
(i)
And,
(ii)
𝑛𝑝𝑞 = 0.36.
Dividing Eqn. (ii) by Eqn. (i), we get
𝑞=
.
.
= 0.9.
Then,
𝑝 = 1 − 𝑞 = 1 − 0.9 = 0.1.
Also, from Eqn. (ii), we have
𝑛=
.
=
.
. × .
= 4.
No. of successes (x) are : 0, 1, 2, 3, 4.
Now, P(At least one success) = 𝑃(𝑥 ≥ 1), where x represents the no. of success.
= 𝑃(𝑥 = 1) + 𝑃(𝑥 = 2) + 𝑃(𝑥 = 3) + 𝑃(𝑥 = 4).
= 1 − 𝑃(𝑥 = 0).
= 1 − 𝐶(𝑛, 0)𝑝 𝑞
D.B.Thapa, DMC
.
Page 4
Probability Distribution
= 1 − 𝐶(4,0) × (0.1) × (0.9)
.
= 1 − 1 × (0.9) .
= 1 − 0.6561.
= 0.3439. Ans
4. In a binomial distribution with 6 independent trails for the probabilities of 3 and 4
successes are found to be 0.2457 and 0.0819. Find the parameter p of the binomial
distribution.
Solution: Here,
n = 6.
𝑃(𝑥 = 3) = 0.245.
and,
𝑃(𝑥 = 4) = 0.0819.
We know,
. [∵ 𝑃(𝑥 𝑠𝑢𝑐𝑐𝑒𝑠𝑠) = 𝐶(𝑛, 𝑥)𝑝 𝑞
𝑃(𝑥 = 3) = 𝐶(𝑛, 3)𝑝 𝑞
0.2457 = 𝐶(6,3)𝑝 𝑞
].
.
Or,
0.2457 = 20 × 𝑝 𝑞 .
Or,
𝑝 𝑞 = 0.012285.
Again,
𝑃(𝑥 = 4) = 0.0819.
Or,
𝐶(𝑛, 4)𝑝 𝑞
= 0.0819.
Or,
𝐶(6,4)𝑝 𝑞
= 0.0819.
Or,
15𝑝 𝑞 = 0.0819.
Or,
𝑝 𝑞 = 0.00546.
(i)
(ii)
Now, dividing equation (ii) by (i), we get
=
Or,
.
.
.
= 0.44.
∴
D.B.Thapa, DMC
𝑝 = 0.44𝑞.
(iii)
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Probability Distribution
Using equation (iii) in equation (i), we get
(0.44𝑞) 𝑞 = 0.012285.
Or,
0.085184 𝑞 = 0.012285.
Or,
𝑞 = 0.1442.
∴
𝑞 = √0.1442 = 0.724.
Then, from equation (iii), we get
𝑝 = 0.44 × 0.724 = 0.318. Ans
Fitting of Binomial Distribution (Theoretical Frequency of a particular success)
If n independent trials satisfying binomial distribution are repeated N no. of times, then the
expected or theoretical frequency of any no. of success say x successes is given by
𝑓(𝑥) = 𝑁 × 𝑃(𝑥 𝑠𝑢𝑐𝑐𝑒𝑠𝑠) = 𝑁 × 𝐶(𝑛, 𝑥)𝑝 𝑞
, where 𝑥 = 0,1,2, ….
5. Calculate theoretical frequencies from the following by using binomial distribution law.
No. of Success
5
4
3
2
Frequency(observed) 190 500 900 960
1
0
500 150
Solution: Here, total no. of repetition of the experiment (N) = 3200.
Table:
No. of success (x)
5
4
3
2
1
0
Total
Then,
Mean =
∑
∑
=
f
fx
190
500
900
960
500
150
∑𝑓 = 3200
950
2000
2700
1920
500
0
∑𝑓𝑥 = 8070
= 2.52.
But, for the binomial distribution,
Mean = 𝑛𝑝.
Or,
2.52 = 5 × 𝑝.
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Probability Distribution
∴
Then,
.
𝑝=
= 0.5
𝑞 = 1 − 𝑝 = 1 − 0.5 = 0.5.
Table:
No. of Success
Frequency(observ
ed)
P(x
success)
𝑪(𝒏, 𝒙)𝒑𝒙 𝒒𝒏 𝒙
=
Theoretical
(Expected)
Freq. =
𝑵 × 𝑪(𝒏, 𝒙)𝒑𝒙 𝒒𝒏 𝒙.
5
190
4
500
3
900
2
960
1
500
0
150
𝐶(5,5)𝑝 𝑞
= (0.5) (0.5)
= 0.03125
𝐶(5,4)𝑝 𝑞
= 5. (0.5) (0.5) .
= 5. (0.5) .
= 0.1562.
𝑁 × 0.1562
= 3200×0.1562
= 500.
𝐶(5,3)𝑝 𝑞
= 10. (0.5) (0.5)
= 10 × (0.5)
= 0.3125
𝑁 × 0.3125
= 3200×0.3125
= 1000.
𝐶(5,2)𝑝 𝑞
= 10. (0.5) (0.5)
= 10 × (0.5)
= 0.3125
𝑁 × 0.3125
= 3200×0.3125
= 1000.
𝐶(5,1)𝑝 𝑞
= 5. (0.5) (0.5)
= 5 × (0.5)
= 0.1562
𝑁 × 0.1562
= 3200×0.1562
= 500.
𝐶(5,0)𝑝 𝑞
= (0.5) (0.5)
= (0.5)
= 0.03125
𝑁 × 0.3125
= 3200×0.03125
= 100.
𝑁 × 0.03125
= 3200×0.03125
= 100.
14. Fit a binomial distribution to the following data:
x 0
1
f
28 62
Solution: Here, no. of repetition of trial (n) = 4.
2 3 4
46 10 4
To find p and q,
Mean of the distribution =
∴
Or,
∑
∑
×
=
×
×
×
×
= 1.28.
𝑛𝑝 = 1.28.
4𝑝 = 1.28
∴
𝑝 = 0.7.
Then, 𝑞 = 1 − 𝑝 = 1 − 0.7 = 0.3
Also, total no. of repetition of trials (N) = 28 + 62 + 46 + 10 + 4 = 150.
Table:
x
f
𝑷(𝒙 𝒔𝒖𝒄𝒄𝒆𝒔𝒔)
Exp. Frequency
D.B.Thapa, DMC
0
28
𝐶(4,0)𝑝 𝑞 .
= 0.0081.
150 × 0.0081.
= 1.
1
62
𝐶(4,1)𝑝 𝑞 .
= 0.076.
0.076 × 150.
= 11.
2
46
𝐶(4,2)𝑝 𝑞 .
= 0.2646.
0.2646 × 150.
= 40.
3
10
𝐶(4,3)𝑝 𝑞 .
= 0.4116.
0.4116 × 150.
= 62.
4
4
𝐶(4,4)𝑝 𝑞 .
0.2401.
0.2401 × 150.
= 36.
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Probability Distribution
2. POISSON DISTRIBUTION
It mainly concerns with the discrete random variable for which the probability of getting success
is very small and total number of possible outcomes is very large. That means, Poisson
distribution is the limiting case of binomial distribution in which
(i)
𝑝 is very small, and
(ii)
𝑛 is indefinitely large.
Thus, a Binomial distribution turns into a Poisson distribution when 𝑝 → 0 and 𝑛 → ∞.
The major assumptions of Poisson distribution are as follows.
a. The events i.e. outcomes are independent.
b. The number of trials (𝑛) is indefinitely large. i.e.; 𝑛 → ∞.
c. The probability of success if very small. i.e.; 𝑝 → 0.
d. The parameters of Poisson distribution are 𝑛 and 𝑝. We denote their product to be 𝜆.
Thus, 𝜆 = 𝑛𝑝. In this respect, the parameter of the Poisson distribution is λ only.
e.
The parameter λ is called mean of the Poisson distribution. That is,
Mean of Poisson Distribution = λ.
f. The variance of the Poisson Distribution is also given by the same parameter λ. That is,
Variance of Poisson Distribution = λ.
g. The standard deviation of Poisson Distribution is √𝜆. That is;
SD = √𝜆.
Characteristics of Poisson Distribution
a. It is a discrete probability distribution, where the random variable X can takes the integer
values 0,1,2,3,…. etc.
b. It is a uni-parametric distribution. The parameter is λ.
c. In this distribution, mean = Variance = λ, and SD = √𝜆.
d. It may be uni-modal or bimodal depending upon the value of λ.
Probability of x successes in Poisson Distribution
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Probability Distribution
The probability of x successes in Poisson Distribution is given by
𝑃(𝑥 𝑠𝑢𝑐𝑐𝑒𝑠𝑠) =
.
!
Examples
1. Assume a Poisson distribution with λ = 5.0. What is the probability that
a. 𝒙 = 𝟏
b. 𝒙 ≤ 𝟑
c. 𝒙 > 2.
Solution: a. Here, the random variable X may take values 0,1,2,3,….
𝑃(𝑥 = 1) =
b.
!
=𝑒
×
!
= 0.03. Ans
𝑃(𝑥 ≤ 3) = 𝑃(𝑥 = 0) + 𝑃(𝑥 = 1) + 𝑃(𝑥 = 2) + 𝑃(𝑥 = 3).
=
!
+
!
+
!
+
!
.
= 0.0067 + 0.033 + 0.08422 + 0.1403.
= 0.2643. Ans
c.
𝑃(𝑥 > 2) = 1 − [𝑃(𝑥 = 0) + 𝑃(𝑥 = 1) + 𝑃(𝑥 = 2)].
=1−
!
+
!
+
!
.
= 0.8763. Ans
2. The number of accidents that occurs on an assembly line with an average of three
accidents per week. What is the probability that
a. A particular week will be accident free.
b. Exactly five accidents occur in a week.
c. At least three accidents occur in a week.
Solution: Let X be a random variable denoting the number of accidents occur per week. Then, X
may take the values 0,1,2,3,….
Also mean of the distribution (λ) = 3. Now, we have
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Probability Distribution
a. P(no accidents per week) = 𝑃(𝑥 = 0).
=
.
!
×
=
.
!
= 0.0497.
b. P(5 accidents per week) = (𝑃(𝑥 = 5).
=
!
.
= 0.1.
c. P(At least 3 accidents per week) = 𝑃(𝑥 ≥ 3).
= 1 − [𝑃(𝑥 = 2) + 𝑃(𝑥 = 1) + 𝑃(𝑥 = 0)].
=1−
!
+
!
+
!
.
= 0.58. Ans
2. Calculate mean and variance of the Poisson distribution X if 𝑷(𝒙 = 𝟒) = 𝑷(𝒙 = 𝟓).
Solution: Given,
𝑃(𝑥 = 4) = 𝑃(𝑥 = 5).
Or,
!
=
!
.
!
Or,
𝜆= .
!
∴
𝜆 = 5.
∴ Mean = Variance = 5. Ans
4. Given a Binomial Distribution with 𝒏 = 𝟐𝟖 trials and 𝒑 = 𝟎. 𝟎𝟐𝟓. Use the Poisson
distribution to the Binomial to find
a. 𝑷(𝒙 = 𝟎).
b. 𝑷(𝒙 = 𝟓).
c. Estimate the theoretical frequency of 0 and 5 successes if the trial is repeated for
10500 times.
D.B.Thapa, DMC
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Probability Distribution
Solution: Here, mean = 𝑛𝑝 = 28 × 0.025 = 0.7
1. 𝑃(𝑥 = 0) =
2. 𝑃(𝑥 = 5) =
!
!
. ×
=
.
!
. ×
=
!
.
= 0.496.
= 0.00069.
3. The expected frequency of 0 success = 𝑁 × 𝑃(𝑥 = 0) = 10500 × 0.496 = 5208.
Also, expected frequency of 5 successes = 𝑁 × 𝑃(𝑥 = 5) = 10500 × 0.00069 = 724. Ans
5. Find the probability that at most 5 defective bolts will be found in a box of 200 bolts if it
is known that 2% bolts are expected to be defective.
Solution: Here,
no. of defective bolts (n) = 5
𝑝 =P( a defective bolt) = 2% = 0.02.
And,
∴
Mean (λ) = 𝑛𝑝 = 5 × 0.02 = 0.1
Now,
P(At most 5 defective bolts) = 𝑃(𝑥 ≤ 5).
= 𝑃(𝑥 = 0) + 𝑃(𝑥 = 1) + ⋯ + 𝑃(𝑥 = 5).
6. The standard deviation of a Poisson Distribution is 2. Find the probability that 𝑿 = 𝟑.
Solution: Here,
SD=2.
i.e;
𝑉𝑎𝑟(𝑋) = 2.
Or,
√𝜆 = 2.
∴
𝜆 = 4.
Then,
𝑃(𝑥 = 3) =
=
D.B.Thapa, DMC
!
!
.
.
Page 11
Probability Distribution
= 0.195. Ans
7. A manufacturer of pins knows that on an average 5% of his product is defective. He
sells pins in boxes of 100 and guarantees that not more than 4 pins will be defective.
What is the probability that
a. a box will meet the guaranteed quality.
b. A box will not meet the guaranteed quality.
Solution: Here,
𝑝 = P(a pin is defective) =
= 0.05.
𝑛 = No. of pins in a box = 100.
Then,
Mean of the distribution (λ) = 𝑛𝑝 = 100 × 0.05 = 5.
a. For a box to meet the required quality, it should not contain more than 4 defective pins.
So, we have
P(required quality) = 𝑃(𝑥 ≤ 4).
= 𝑃(𝑥 = 0) + 𝑃(𝑥 = 1) + 𝑃(𝑥 = 2) + 𝑃(𝑥 = 3) +
𝑃(𝑥 = 4).
=
!
+
!
+
!
+
!
+
!
.
= 0.0067 + 0.033 + 0.084 + 0.140 + 0.175.
= 0.4387. Ans
Fitting of Poisson Distribution
The finding of theoretical (expected) frequencies of a discrete random variable satisfying a
Poisson distribution. we use the relation,
Expected frequency of x successes = 𝑁 ×
!
,
where,
𝑁 = Total no of independent trials repeated.
8. Fit a Poisson distribution to the following data.
D.B.Thapa, DMC
Page 12
Probability Distribution
Number of 0
1
Death
Frequencies 122 60
2
3
4
Total
15
2
1
200
×
×
Solution: Here,
Mean number of death (λ) =
×
×
×
= 0.5.
Table:
No. of deaths 0
(x)
F
122
P(x-success) =
. (
. )
.
!
= 0.6065.
!
Expected Freq 200 ×
0.6065.
=𝑁×
!
= 121.
1
2
3
4
60
15
2
1
. (
. )
.
!
= 0.3032.
200 ×
0.3032.
= 61.
. (
. (
. )
.
!
= 0.0758.
200 ×
0.0758.
= 15.
. )
.
!
= 0.0126.
200 ×
0.0126.
= 3.
. (
. )
.
= 0.00158.
200 ×
0.00158.
= 0.
!
9. Fit a Poisson distribution to the following data.
Number
of 0
1
2
typing mistakes
Frequencies
142 156 69
3
4
5
27 5
1
Solution: Here,
Mean number of typing mistakes (λ) =
×
×
×
×
×
×
= 1.
Table:
No.
of 0
mistakes (x)
f
142
P(x-success)
=
𝒆 𝝀 𝝀𝒙
.
= 0.3679.
400 × 0.3679.
= 147.
!
𝒙!
Expected Freq =
𝑵×
( )
𝒆 𝝀 𝝀𝒙
1
2
3
4
5
156
69
27
5
1
( )
.
= 0.3679.
400 × 0.3679.
= 147.
!
( )
.
= 0.1839.
400 × 0.1839.
= 74.
!
( )
.
= 0.0613.
400 × 0.0613.
= 25.
!
( )
.
= 0.0153.
400 × 0.0153.
= 6.
!
( )
.
= 0.0031
400 × 0.0031.
= 1.
!
𝒙!
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Page 13
Probability Distribution
3. NORMAL DISTRIBUTION
It is a continuous probability distribution as a continuous random variable only can satisfies this
type of distribution. Most of the data relating to economic, business, statistics or even social and
physical sciences are continuous in nature.
A continuous random variable X is said to follow a normal distribution with mean 𝜇 and
standard deviation 𝜎 if it's probability density function is given by,
𝑝(𝑥) =
√
, −∞ < 𝑥 < ∞.
𝑒
Standard Normal Variate
Let X denotes a continuous random variable satisfying a normal distribution with mean μ and
standard deviation σ. Then the random variable that transform X into another one having mean 0
and standard deviation 1 is called a Z variate, and it is defined as,
,
𝑧=
where μ is the mean and σ is the SD of the given distribution. The Z-variate transform the
variable X into the variable Z such that the mean of the variable Z is 0 and standard deviation 1.
Example: Transform the X variate into the Z variate from the following.
x:
Sol: Here, mean 𝜇 =
D.B.Thapa, DMC
2
4
6
8
10,
= 6, and
Page 14
Probability Distribution
(
𝜎=
)
(
)
(
)
(
)
(
)
.
=
=
.
= √8,
= 2.8.
The z transformation is 𝑧 =
So,
For x=2, 𝑧 =
For x=4, 𝑧 =
For x=6, 𝑧 =
For x=8, 𝑧 =
,
= −1.4,
.
= −0.7,
.
= 0,
.
= 0.7,
.
For x=10, 𝑧 =
= 1.4.
.
Then,
Mean of z series =
SD =
(
.
)
(
(
.
. ) (
. )
)
( .
.
)
.
( .
= 0, and
)
= 1.
Q. Why to transform X into Z?
Since a continuous random variable X satisfies a normal distribution. Two such variables are
comparable if they have similar variances. The comparison of two variables having different
variances doesn't make complete sense. They are exactly comparable if they have the similar
variabilities. Let us discuss this matter with an example.
Suppose there are two students A and B. A gets 65 marks in Maths and B gets 80 in English.
Now, if we are asked to tell who performed better with respect to others, we cannot say that A
did better than B by just looking at the scores as, their variability may be different. That is to say
that the way students performed in English may be different from the way they performed in
Maths exam. So, direct comparison by just looking at the scores will not work.
D.B.Thapa, DMC
Page 15
Probability Distribution
We have the following further information:
Maths marks follow Normal distribution with mean 60 and sd 4
English marks also follow Normal distribution with mean 79 and sd 2.
Here, we can see that the variance is different. So, in order to enable comparison we need to
unitize the deviations, that is we have to express the deviation from the mean per unit sd. By this,
we are calculating a quantity called Z score by scaling the deviations. The resulting deviation per
unit sd therefore has sd =1 and mean = 0 ,which is usual if the data comes from normal
distribution. So, now we can say with certainty in the former example that A performed better
than B with respect to others which is totally opposite of what we thought initially as B's marks
are way more than that of A.
So to make conclusions like this we need to convert Normal distribution into Standard Normal as
comparison without a common base will serve no purpose.
To find Area under Normal Curve:
Case a. If 𝑧 < 0 and 𝑧 > 0, we find the area 𝑃(𝑧 ≤ 𝑍 ≤ 𝑧 ) as,
𝑃(𝑧 < 𝑍 < 𝑧 ) = 𝑃(𝑧 < 𝑍 < 0) + 𝑃(0 < 𝑍 < 𝑧 ).
= 𝑃(0 < 𝑍 < −𝑧 ) + 𝑃(0 < 𝑍 < 𝑧 ).
Case b. If 𝑧 < 𝑧 < 0, we find the area 𝑃(𝑧 ≤ 𝑧 ≤ 𝑧 ) as,
𝑃(𝑧 < 𝑧 < 𝑧 ) = 𝑃(𝑧 < 𝑧 < 0) − 𝑃(𝑧 < 𝑧 < 0).
Case c. If 0 < 𝑧 < 𝑧 , we find the area 𝑃(𝑧 ≤ 𝑧 ≤ 𝑧 ) as,
𝑃(𝑧 < 𝑍 < 𝑧 ) = 𝑃(0 < 𝑍 < 𝑧 ) − 𝑃(0 < 𝑍 < 𝑧 ).
Case d. If 𝑧 < 0, we find the area 𝑃(𝑧 ≤ 𝑧 ) as,
𝑃(𝑍 < 𝑧 ) = 0.5 − 𝑃(𝑧 < 𝑍 < 0).
Case e. If 𝑧 > 0, we find the area 𝑃(𝑧 > 𝑧 ) as
𝑃(𝑍 > 𝑧 ) = 0.5 − 𝑃(𝑍 > 𝑧 ).
D.B.Thapa, DMC
Page 16
Probability Distribution
Notation: We write 𝑋~𝑁(𝜇 , 𝜎) to represents that "X follows a normal distribution with mean μ
and standard deviation σ". If 𝑋~𝑁(𝜇 , 𝜎), then 𝑍~𝑁(0,1).
Examples:
1. Let 𝒁~𝑵(𝟎, 𝟏), then find 𝑷(−𝟎. 𝟓 ≤ 𝒁 ≤ 𝟎. 𝟏) and 𝑷(𝟏 ≤ 𝒁 ≤ 𝟐).
Solution: Here,
(i)
𝑃(−0.5 ≤ 𝑍 ≤ 0.1) = 𝑃(−0.5 ≤ 𝑍 ≤ 0) + 𝑃(0 ≤ 𝑍 ≤ 0.1).
= 𝑃(0 ≤ 𝑍 ≤ 0.5) + 𝑃(0 ≤ 𝑍 ≤ 0.1).
= 0.1915 + 0.0398.
= 0.2313. Ans
(ii)
𝑃(1 ≤ 𝑍 ≤ 2) = 𝑃(0 ≤ 𝑍 ≤ 2) − 𝑃(0 ≤ 𝑍 ≤ 1).
= 0.4772 − 0.3413.
= 0.1359. Ans
2. A continuous random variable X satisfies a normal distribution with mean 750 and
standard deviation 100. Find the probability that X lies between 554 and 946.
Solution: Here,
Mean (𝜇) = 750, and
SD (𝜎) = 100.
First we need to transform X into Z by the relation,
𝑍=
(
)
=
(
)
.
We have to find,
𝑃(554 ≤ 𝑋 ≤ 946) = ?
Now transforming 𝑥 = 554 and 𝑋 = 946 into Z, we get
When X = 554, 𝑍 =
When X= 946, 𝑍 =
D.B.Thapa, DMC
(
)
=
=−
= −1.96, and
= 1.96.
Page 17
Probability Distribution
Then,
𝑃(554 ≤ 𝑋 ≤ 946) = 𝑃(−1.96 ≤ 𝑍 ≤ 1.96).
= 𝑃(−1.96 ≤ 𝑍 ≤ 0) + 𝑃(0 ≤ 𝑍 ≤ 1.96).
= 𝑃(0 ≤ 𝑍 ≤ 1.96) + 𝑃(0 ≤ 𝑍 ≤ 1.96).
= 𝟐 × 𝑃(0 ≤ 𝑍 ≤ 1.96).
= 2 × 0.4750.
= 0.95. Ans
3. Given a normal distribution with mean 100 and SD 10. What is the probability that
a. 𝑿 > 75?
b. 𝑿 < 70?
c. 𝟕𝟓 < 𝑋 < 85?
d. 𝑿 < 80 or 𝑿 > 110?
e. 𝟕𝟎% of the values are less than what value of X? What is the largest value of first
70% items?
f. 𝟕𝟎% of the values are more than what value of X?
g. 𝟖𝟎% of the values are in the middle of what two values values of X?
Solution: Here,
Mean = 100, and
Standard Deviation = 10.
Now, first we need to transform X into Z by the relation,
𝑍=
(
)
=
(
)
.
a. We have to find,
𝑃(𝑋 > 75) = ?
Now transforming𝑋 = 75 into Z, we get
𝑍=
D.B.Thapa, DMC
(
)
= −2.5, and
Page 18
Probability Distribution
Then,
𝑃(𝑋 > 75) = 𝑃(𝑍 > −2.5).
= 𝑃(−2.5 < 𝑍 ≤ 0) + 0.5.
= 𝑃(0 ≤ 𝑍 < 2.5) + 0.5.
= 0.4938 + 0.5.
= 0.9938. Ans
b. Next, we find 𝑃(𝑋 < 70) =?.
Now, transforming𝑋 = 70 into Z, we get
𝑍=
(
)
= −3.0, and
Then,
𝑃(𝑋 < 70) = 𝑃(𝑍 < −3.0).
= 0.5 − 𝑃(−3.0 < 𝑍 ≤ 0).
= 0.5 − 𝑃(0 ≤ 𝑍 < 3.0).
= 0.5 − 0.49865.
= 0.00135. Ans
c. 𝑃(75 < 𝑋 < 85) =?.
Now, transforming𝑋 = 75 and 𝑋 = 85 into Z, we get
When X=75,
𝑍=
(
)
= −2.5, and
When X = 85,
𝑍=
= −1.5.
Then,
𝑃(75 < 𝑋 < 85) = 𝑃(−2.5 < 𝑍 < −1.5).
= 𝑃(−2.5 < 𝑍 ≤ 0) − 𝑃(−1.5 < 𝑍 < 0).
D.B.Thapa, DMC
Page 19
Probability Distribution
= 𝑃(0 < 𝑍 ≤ 2.5) − 𝑃(0 < 𝑍 < 1.5).
= 0.4938 − 0.4332.
= 0.0606. Ans
d. Here, 𝑃(𝑋 < 80 or 𝑋 > 110) =?
Now, transforming𝑋 = 80 and 𝑋 = 110 into Z, we get
When X = 80,
𝑍=
(
)
= −2.0, and
When X = 110,
𝑍=
= 1.
Then,
𝑃(𝑋 < 80 or 𝑋 > 110) = 𝑃(𝑍 < −2.0) + 𝑃(𝑍 > 1).
= [0.5 − 𝑃(−2.0 < 𝑍 < 0)] + [0.5 − 𝑃(0 ≤ 𝑍 ≤ 1)].
= [0.5 − 𝑃(0 < 𝑍 < 2.0)] + [0.5 − 𝑃(0 < 𝑍 < 1)].
= (0.5 − 0.4772) + (0.5 − 0.3413).
= 0.0228 + 0.1587.
= 0.1815. Ans
e. Let x1 be the value of X below which 70% of the values are found. Then,
𝑃(𝑋 < 𝑥 ) = 0.70.
Now, transforming𝑋 = 𝑥 into Z, we get
When X = x1,
𝑧 =
(
)
.
Then, the above equation becomes,
𝑃(𝑍 < 𝑧 ) = 0.70.
Or,
D.B.Thapa, DMC
0.5 + 𝑃(0 ≤ 𝑍 ≤ 𝑧 ) = 0.70.
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Probability Distribution
Or,
𝑃(0 ≤ 𝑍 ≤ 𝑧 ) = 0.20.
Or,
𝑃(0 ≤ 𝑍 ≤ 𝑧 ) = 𝑃(0 ≤ 𝑍 ≤ 0.52).
⇒
𝑧 = 0.52.
(
Or,
)
Or,
= 0.52.
𝑥 = 100 + 5.2 = 105.2. Ans
f. Let x1 be the value of X above which 70% of the values are found. Then,
𝑃(𝑋 > 𝑥 ) = 0.70.
Now, transforming𝑋 = 𝑥 into Z, we get
When X = x1,
𝑧 =
(
)
.
Then, the above equation becomes,
𝑃(𝑍 > 𝑧 ) = 0.70.
Or,
𝑃(𝑧 ≤ 𝑍 ≤ 0) + 0.5 = 0.70.
Or,
𝑃(𝑧 ≤ 𝑍 ≤ 0) = 0.20.
Or,
𝑃(0 ≤ 𝑍 ≤ −𝑧 ) = 0.20.
Or,
𝑃(0 ≤ 𝑍 ≤ −𝑧 ) = 𝑃(0 ≤ 𝑍 ≤ 0.52).
⇒
−𝑧 = 0.52.
(
)
Or,
−
Or,
−(𝑥 − 100) = 5.2.
Or,
𝑥 = 100 − 5.2 = 94.8. Ans
= 0.52.
g. Let 𝑥 and 𝑥 be two values of X between which middle 80% of the values are lying.
Then,
𝑃(𝑥 ≤ 𝑋 ≤ 𝑥 ) = 0.80.
(a) Now, transforming 𝑋 = 𝑥 and 𝑋 = 𝑥 into Z, we get
When, 𝑋 = 𝑥 ,
D.B.Thapa, DMC
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Probability Distribution
, and
𝑧 =
When, 𝑋 = 𝑥 ,
.
𝑧 =
Then, from Eqn. (a), we get
𝑃(𝑧 ≤ 𝑍 ≤ 𝑧 ) = 0.80.
Or,
𝑃(𝑧 ≤ 𝑍 ≤ 0) + 𝑃(0 ≤ 𝑍 ≤ 𝑧 ) = 0.80.
But, being the middle 80% items, we must have |𝑧 | = |𝑧 |. Therefore, we have
2 × 𝑃(0 ≤ 𝑍 ≤ 𝑧 ) = 0.80.
Or,
𝑃(0 ≤ 𝑍 ≤ 𝑧 ) = 0.40.
Or,
𝑃(0 ≤ 𝑍 ≤ 𝑧 ) = 𝑃(0 ≤ 𝑍 ≤ 1.28).
∴
𝑧 = 1.28.
Or,
= 1.28.
∴
𝑥 = 112.8.
Similarly, we have
𝑧 = −1.28.
Or,
= −1.28.
∴
𝑥 = 87.2.
Hence, the range of middle 80% values lie between 87.2 to 112.8. Ans
4. A particular product has a declared weight of 60 gm. Assume the process mean is 61.8
gm and standard deviation 1.5 gm. Assume the weight is normally distributed.
b. What proportion of product will be above the declared weight?
c. Determine the proportion of product below 95% of the declared weight.
d. Determine the proportion of product between 60 gm and 63 gm.
Solution: Here,
D.B.Thapa, DMC
Page 22
Probability Distribution
Mean = 61.8 and standard deviation = 1.5
Declared weight = 60 gm.
Let us transform the variable X into Z by the relation,
𝑍=
.
=
.
.
a. 𝑃(𝑋 > 60) =?.
To find it, transforming𝑋 = 60 into Z, we get
𝑍=
(
. )
.
= −1.2.
Then,
𝑃(𝑋 > 60) = 𝑃(𝑍 > −1.2).
= 𝑃(−1.2 < 𝑍 < 0) + 0.5.
= 𝑃(0 ≤ 𝑍 ≤ 1.2) + 0.5.
= 0.3849 + 0.5.
= 0.8849.
Hence, nearly 88.5% values are above the declared weight.
b. Here, 95% of the declared weight = 60 ×
= 57.
So, we want to find 𝑃(𝑋 < 57) =?.
To find it, transforming𝑋 = 57 into Z, we get
𝑍=
(
. )
.
= −3.2.
Then,
𝑃(𝑋 > 60) = 𝑃(𝑍 > −3.2).
= 𝑃(−3.2 < 𝑍 < 0) + 0.5.
= 𝑃(0 ≤ 𝑍 ≤ 3.2) + 0.5.
= 0.4993. Ans
c. 𝑃(60 < 𝑋 < 63) =?.
D.B.Thapa, DMC
Page 23
Probability Distribution
For this, when 𝑋 = 60,
.
𝑍=
= −1.2.
.
When 𝑋 = 63,
.
𝑍=
Then,
= 0.8.
.
𝑃(60 < 𝑋 < 63) = 𝑃(−1.2 < 𝑍 < 0.8).
= 𝑃(−1.2 < 𝑍 < 0) + 𝑃(0 < 𝑍 < 0.8).
= 𝑃(0 < 𝑍 < 1.2) + 𝑃(0 < 𝑍 < 0.8).
= 0.0793 + 0.2882.
= 0.3675. Ans
5. In final examination of statistics, the marks obtained by students was observed to be
normally distributed with mean 73 and standard deviation 8.
a. What is the probability of a student getting marks more than 60?
b. What % of students scored marks between 60 and 90?
c. What is the minimum marks of top 5% marks scorer ?
Solution: Here,
Mean = 73 and standard deviation = 8.
Let us transform the variable X into Z by the relation,
𝑍=
.
=
a. 𝑃(𝑋 > 60) =?.
To find it, transforming𝑋 = 60 into Z, we get
𝑍=
(
)
= −1.625.
Then,
𝑃(𝑋 > 60) = 𝑃(𝑍 > −1.625).
= 𝑃(−1.625 < 𝑍 < 0) + 0.5.
D.B.Thapa, DMC
Page 24
Probability Distribution
= 𝑃(0 ≤ 𝑍 ≤ 1.625) + 0.5.
= 0.4474 + 0.5.
= 0.9474. Ans
b. 𝑃(60 < 𝑋 < 90) =?
When 𝑋 = 60,
𝑍=
(
)
(
)
= −1.625.
When 𝑋 = 90,
𝑍=
= 2.125.
Then,
𝑃(60 < 𝑋 < 90) = 𝑃(−1.625 < 𝑍 < 2.125).
= 𝑃(−1.625 < 𝑍 < 0) + 𝑃(0 < 𝑍 < 2.125).
= 𝑃(0 < 𝑍 < 1.625) + 𝑃(0 < 𝑍 < 2.215).
= 0.4474 + 0.4864.
= 0.9338. Ans
c. Let the minimum marks of highest 5% marks scorer be 𝑥 . Then,
𝑧 =
.
By question,
𝑃(𝑋 > 𝑥 ) = 0.05.
Or,
𝑃(𝑍 > 𝑧 ) = 0.05.
Or,
0.5 − 𝑃(0 < 𝑍 < 𝑧 ) = 0.05.
Or,
𝑃(0 < 𝑍 < 𝑧 ) = 0.45.
Or,
𝑃(0 < 𝑍 < 𝑧 ) = 𝑃(0 < 𝑍 < 1.65).
⇒
Or,
D.B.Thapa, DMC
𝑧 = 1.65.
= 1.65.
Page 25
Probability Distribution
Or,
𝑥 = 86.2.
Therefore, the minimum marks of the top 5% scorer is 86.2. Ans
6. The income of a group of 10,000 people were found to be normally distributed with mean
Rs. 520 and standard deviation Rs 60. Find the limit of the middle 40% of the income. Also,
find the number of people having middle 40% income.
Solution: Here,
Mean = Rs. 520 and standard deviation = Rs. 60.
Let us transform the variable X into Z by the relation,
𝑍=
.
=
Let 𝑥 and 𝑥 be the limit of the middle 40% income. Then, their corresponding Z variates
are
𝑧 =
and 𝑧 =
.
Now, we have,
𝑃(𝑥 < 𝑋 < 𝑥 ) = 0.4.
Or,
𝑃(𝑧 < 𝑍 < 𝑧 ) = 0.4.
Or,
𝑃(𝑧 < 𝑍 < 0) + 𝑃(0 < 𝑍 < 𝑧 ) = 0.4.
Or,
𝑃(0 < 𝑍 < −𝑧 ) + 𝑃(0 < 𝑍 < 𝑧 ) = 0.4.
Or,
2 × 𝑃(0 < 𝑍 < −𝑧 ) = 0.4.
Or,
𝑃(0 < 𝑍 < −𝑧 ) = 0.2.
Or,
𝑃(0 < 𝑍 < −𝑧 ) = 𝑃(0 < 𝑍 < 0.0793).
⇒
Or,
−𝑧 = 0.0793.
−
∴
And,
= 0.0793.
𝑥 = 515.242.
= 0.0793.
⇒
D.B.Thapa, DMC
𝑥 = 524.758.
Page 26
Probability Distribution
Hence, the limit of middle 40% of the income is (Rs. 512.242 to Rs. 524.758). Ans
Next, the number of people having middle 40% income = 10,000 × 0.4 = 4000. Ans
7. The weekly wages of 1000 workers are normally distributed with mean Rs. 72 and
standard deviation Rs. 5. Estimate the number of workers when weekly wages will be
a. Between Rs. 70 to Rs. 72.
b. Between Rs. 69 to Rs. 75.
c. More than Rs. 75.
d. Less than Rs. 63.
Solution: Here,
Mean = Rs. 72 and standard deviation = Rs. 5.
Let us transform the variable X into Z by the relation,
𝑍=
=
.
a. Here, first we find 𝑃(70 < 𝑋 < 72). For this, we first change these values into Z values,
The corresponding Z values are
𝑧 =
= −0.4 and 𝑧 =
= 0.
Now, we have,
𝑃(70 < 𝑋 < 72) = 𝑃(𝑧 < 𝑍 < 0).
= 𝑃(−0.4 < 𝑍 < 0).
= 𝑃(0 < 𝑍 < 0.4).
= 0.1554.
Therefore, the number of people having income Rs. 70 to Rs. 72 = 0.1554 × 1000 =
155.4 ≈ 155.
b. We first find 𝑃(69 < 𝑋 < 75). For this, we first change these values into Z values, The
corresponding Z values are
𝑧 =
= −0.6 and 𝑧 =
= 0.6.
Now, we have,
D.B.Thapa, DMC
Page 27
Probability Distribution
𝑃(69 < 𝑋 < 75) = 𝑃(−0.6 < 𝑍 < 0.6).
= 𝑃(−0.6 < 𝑍 < 0) + 𝑃(0 < 𝑍 < 0.6).
= 2 × 𝑃(0 < 𝑍 < 0.6)..
= 2 × 0.2257.
= 0.4510.
Therefore, the number of people having income Rs. 69 to Rs. 75 = 0.4510 × 1000 =
451.
c. We find 𝑃(𝑋 > 75). For this, we first change these values into Z values, The
corresponding Z value is
𝑧=
= 0.6.
Now, we have,
𝑃(𝑍 > 0.6) = 0.5 − 𝑃(0 < 𝑍 < 0.6).
= 0.5 − 0.2257.
= 0.2743.
Therefore, the number of people having income more than Rs. 75 = 0.2743 × 1000 =
274.3 ≈
274.
d. We find 𝑃(𝑋 < 63). For this, we first change these values into Z values, The
corresponding Z value is
𝑧=
= −1.8.
Now, we have,
𝑃(𝑍 < −1.8) = 0.5 − 𝑃(−1.8 < 𝑍 < 0).
= 0.5 − 𝑃(0 < 𝑍 < 1.8).
= 0.5 − 0.4641.
= 0.0359.
Therefore, the number of people having income less than Rs. 63 = 0.0359 × 1000 = 35.9 ≈ 36.
D.B.Thapa, DMC
Page 28
Probability Distribution
8. The mean height and variance of height of 500 students were found to be 165 cm and 25
cm2. Find the range of middle 80% of the students. T.U Question (2022)
Solution: Given
Mean Height (μ) = 165
Variance (σ2) = 25
Range of height of middle 80% students =?
Let 𝑋 and 𝑋 denote the points representing the middle 80% items.
By question,
𝑃(𝑋 ≤ 𝑋 ≤ 𝑋 ) = 0.8.
Now, Let us transform the variable X into Z by the relation,
𝑍=
.
=
Let 𝑧 and 𝑧 be the corresponding values of 𝑥 and 𝑥 respectively. Then,
and 𝑧 =
𝑧 =
.
Also, since the region represents the middle 80% area, we have
𝑧 = −𝑧 .
Now, from the question,
𝑃(𝑧 ≤ 𝑍 ≤ 𝑧 ) = 0.8.
Or,
2 × 𝑃(𝑜 ≤ 𝑍 ≤ 𝑧 ) = 0.8. [∵ 𝑧 = −𝑧 ].
Or,
𝑃(0 ≤ 𝑍 ≤ 𝑧 ) = 0.4.
Or,
𝑃(0 ≤ 𝑍 ≤ 𝑧 ) = 𝑃(0 ≤ 𝑍 ≤ 1.28).
⇒
Or,
𝑧 = 1.28.
= 1.28.
Or,
𝑥 = 171.4.
Also,
𝑧 = 1.28.
D.B.Thapa, DMC
Page 29
Probability Distribution
−𝑧 = 1.28.
Or,
Or,
Or,
𝑧 = −1.28.
= −1.28.
𝑥 = 158.6.
Therefore, the range of middle 80% items = 171.4 − 158.6 = 12.8 Ans
D.B.Thapa, DMC
Page 30