Distributions

Binomial Distribution: BIPARAMETRIC

Mean and Variance:
E(X)= ∑𝒏𝒙=𝟏 𝒙𝑷(𝒙)
𝑛
= ∑𝒏𝒙=𝟏 𝒙 𝐶𝑥 𝑝 𝑥 𝑞 𝑛−𝑥
𝑛
= ∑𝒏𝒙=𝟏 𝒙 𝐶𝑥 𝑝 𝑥 𝑞 𝑛−𝑥
𝑛!
= ∑𝒏𝒙=𝟏 𝒙 𝑝 𝑥 𝑞 𝑛−𝑥
𝑥! (𝑛−𝑥)!

(𝑛−1)!
= np ∑𝒏𝒙=𝟏 𝑝 𝑥−1 𝑞 𝑛−𝑥
(𝑥−1)! (𝑛−𝑥)!

Substitute, x-1=s
(𝑛−1)!
We get, E(X)= np ∑𝒏−𝟏
𝒔=𝟎 𝑝 𝑠 𝑞 𝑛−𝑠−1
𝑠! (𝑛−𝑠−1)!

(𝑛−1)!
= np ∑𝒏−𝟏
𝒔=𝟎 𝑝 𝑠 𝑞 𝑛−1−𝑠
𝑠! (𝑛−𝑠−1)!

= np (𝑝 + 𝑞)𝑛−1
= np
V(X)= 𝐸(𝑥 2 ) − [𝐸(𝑥)]2= 𝐸(𝑥 2 ) – (𝑛𝑝)2
𝑛
Where, 𝐸(𝑥 2 ) = ∑𝒏𝒙=𝟏(𝒙𝟐 − 𝒙 + 𝒙) 𝐶𝑥 𝑝 𝑥 𝑞 𝑛−𝑥
𝑛 𝑛
= ∑𝒏𝒙=𝟏(𝒙𝟐 − 𝒙) 𝐶𝑥 𝑝 𝑥 𝑞 𝑛−𝑥 + ∑𝒏𝒙=𝟏 𝒙 𝐶𝑥 𝑝 𝑥 𝑞 𝑛−𝑥
𝑛!
= ∑𝒏𝒙=𝟐 𝒙 (𝒙 − 𝟏) 𝑝 𝑥 𝑞 𝑛−𝑥 + np
𝑥! (𝑛−𝑥)!

(𝑛−2)!
= 𝑛(𝑛 − 1)𝑝2 ∑𝒏𝒙=𝟐 𝑝 𝑥−2 𝑞 𝑛−𝑥 + np
(𝑥−2)! (𝑛−𝑥)!

Substitute, x-2=s
(𝑛−2)!
= 𝑛(𝑛 − 1)𝑝2 ∑𝒏−𝟐
𝒔=𝟎 𝑝 𝑠 𝑞 𝑛−𝑠−2 + np
𝑠! (𝑛−𝑠−2)!

= 𝑛(𝑛 − 1)𝑝2 (𝑝 + 𝑞)𝑛−2 + np

= 𝑛(𝑛 − 1)𝑝2 + np
V(X)= 𝐸(𝑥 2 ) − [𝐸(𝑥)]2 = 𝑛(𝑛 − 1)𝑝2 + np – (𝑛𝑝)2
V(X)= npq

Problems:
1. Six coins are tossed. Find the probability of getting:
a) Exactly 3 heads --------P(X=3)--------5/16
b) At least 3 heads -------P(𝑋 ≥ 3)-----21/32
c) At most 3 heads -------P(𝑋 ≤ 3)-- --21/32
d) At least 1 head ---------P(𝑋 ≥ 1) = 1-P(zero heads)
Ans: n=6 X: number of heads
p: getting head = ½
q: getting tail = ½
2. What is the probability of getting a 6 at least once in 2 throws of a fair die.
Ans: n=2 X: number of times 6 obtained
P(𝑋 ≥ 1) ------------11/36
 p: getting six = 1/6
q: not getting six = 5/6
3. A fair die is thrown 180 times. What is the expected number of sixes.
Ans: n=180 X: number of times 6 obtained
E(x) =np=30
4. A die is thrown 8 times. Find the probability that 3 appears.
i. Two times……….P(X=2)= 0.2604
ii. At least 7 times ……….P(X≥ 7)= 0.0000244
iii. Exactly one time ……..P(X=1)= 0.372
5. Two percent fuses manufactured by company are defective. Find the probability that a box containing 200
fuses contains.
i. NO defective …………P(X=0)= 0.01758
ii. 3 or more defective……….P(3 or more)=1- P(0,1,2)= 0.7649
6. Find the probability in a family of 4 children there will be at least one boy by assuming that probability of
male birth is ½……….P(At least 1)= 1- P(NO boy)
7. A family has 6 children. Find the probability that there are fewer boys than girls.
Ans: n=6 and X- number of boys can take the values 0, 1, or 2
8. The sum and product of mean and variance of binomial distribution are 24 and 128. Find the distribution.
Solution: np+npq=24
np.npq=128
q= 2 and 0.5
2 not possible therefore q=0.5 hence p=0.5
We have np.npq=128 therefore n=32
The binomial distribution = 32Cx (0.5)𝑥 (0.5)32−𝑥
9. Numbers are selected at random one at a time from the two digit numbers 00,01,…, 99 with replacement. An
event occurs iff the product of the two digits of selected number is 18. If 4 numbers are selected find the
probability that the event occurs at least 3 times.
Solution: n= 4
p= 4/100 and q= 24/25
P(𝑋 ≥ 3)= 97/254
10. A perfect die is tossed 100 times in sets of 8 the occurrence of 5 and 6 is called a success. How many times
do you expect to get 3 success.
Solution:
p=1/3 and q=2/3 and n=8
P (3 success)= 8C3 (1/3)3 (2/3)5= 0.2731
E (3 success)=100 x 0.2731= 27.31
11. Suppose that the probability for A to win a game of tennis against B is 0.4. A has an option of playing either a
best of 3 games or a best of 5 games. Which option A should choose so that his probability of winning is
greater.
Solution:
X: number of games A wins against B
P= 0.4 and q= o.6
When n= 3
P(𝑋 ≥ 2)= 0.352
When n= 5
P(𝑋 ≥ 3)= 0.31744
12. An aircraft knows that 5% of the people making reservation on a certain flight will not show up.
Consequently, their policy is to sell 52 tickets for the flight that can only hold 50 passengers. What is the
probability that there will be a seat available for every passenger who turns up?
Solution: n=52
X: number of passengers who won’t turn up
p= passenger will not turn up= 0.05
q= 0.95
P(𝑋 ≥ 2)= 1- [P(X=0)+P(X=1)]= 0.7405
Poisson’s Distribution: UNIPARAMETRIC

Theorem: Let X be a binomial random variable with parameters n, p and pdf P(X=x)= nCx 𝑝𝑥 𝑞𝑛−𝑥 . Suppose
𝑒 −𝛼 𝛼𝑥
n→∞; np=𝛼 then lim 𝑃(𝑋 = 𝑥) = is a Poisson’s distribution with parameter 𝛼.
𝑛→∞ 𝑥!

Proof: General expression for binomial distribution is P(X)= nCx 𝑝 𝑥 𝑞 𝑛−𝑥

= nCx 𝑝 𝑥 (1 − 𝑝)𝑛−𝑥
𝑛(𝑛−1)(𝑛−2)…(𝑛−𝑥+1)
= 𝑝 𝑥 (1 − 𝑝)𝑛−𝑥
𝑥!

𝑛−𝑥
𝑛(𝑛−1)(𝑛−2)…(𝑛−𝑥+1) 𝛼 𝑥 𝛼
Let 𝑛𝑝 = 𝛼 = (𝑛) (1 − 𝑛)
𝑥!

𝛼 𝑛
𝛼𝑥 1 2 {𝑥+1} (1− )
𝑛
= [1. (1 − 𝑛) (1 − 𝑛) … (1 − )] 𝛼 𝑥
𝑥! 𝑛 (1− )
𝑛

Let n→∞ and 𝛼 = 𝑛𝑝

𝛼 𝑛
(1 − 𝑛) )
𝑥 {𝑥 + 1}
𝛼 1
lim 𝑃(𝑋 = 𝑥) = lim ( [1. (1 − ) … (1 − )]
𝑛→∞ 𝑛→∞ 𝑥! 𝑛 𝑛 𝛼 𝑥
(1 − 𝑛 )

𝛼 𝑥 𝑒 −𝛼
lim 𝑃(𝑋 = 𝑥) =
𝑛→∞ 𝑥!

Mean and Variance:

E(X)= ∑∞
𝒙=𝟎 𝒙𝑷(𝒙)
𝑒 −𝛼 𝛼𝑥
= ∑∞
𝒙=𝟏 𝒙 𝑥!
∞ 𝑒 −𝛼 𝛼𝑥
= ∑𝒙=𝟏
(𝑥−1)!
𝛼 (𝑥−1)
= 𝛼𝑒 −𝛼 ∑∞
𝒙=𝟏 (𝑥−1)!
Substitute, x-1=s
𝛼𝑠
= 𝛼𝑒 −𝛼 ∑∞𝒔=𝟎 𝑠!
= 𝛼𝑒 −𝛼 𝑒 𝛼 = 𝛼
Therefore,
E(X)= 𝛼
V(X)= 𝐸(𝑥 2 ) − [𝐸(𝑥)]2 = 𝐸(𝑥 2 ) − [𝛼]2
𝑒 −𝛼 𝛼𝑥
𝐸(𝑥 2 ) = ∑∞
𝒙=𝟏 𝒙
𝟐
𝑥!
∞ 𝑒 −𝛼 𝛼𝑥−1
= 𝛼 ∑𝒙=𝟏 𝒙 (𝑥−1)!
 Substitute, x-1=s
∞
2)
𝑒 −𝛼 𝛼 𝑠
𝐸(𝑥 = 𝛼 ∑(𝒔 + 𝟏)
𝑠!
𝒔=𝟎
𝑒 −𝛼 𝛼𝑠 −𝛼 ∑∞ 𝛼
𝑠
= 𝛼 ∑∞
𝒔=𝟏 𝒔 + 𝛼𝑒 𝒔=𝟎 𝑠!
𝑠!
2 −𝛼 ∑∞ 𝛼𝑠−1 −𝛼 𝛼
= 𝛼 𝑒 𝒔=𝟏 (𝑠−1)! + 𝛼 𝑒 𝑒
2
=𝛼 + 𝛼
Therefore,
V(X)= 𝐸(𝑥 2 ) − [𝐸(𝑥)]2 = 𝛼

Problems:
a) 2% of fuses manufactured by a company are defective. Find the probability that a box having 200 fuses
contains.
i. No defective fuse.
ii. 3 or more defective fuses.
Solution:
𝛼 = 𝑛𝑝
n=200, p=0.02, q= 0.98. Therefore, 𝛼 = 4.
i. P(X=0) = 0.0183
ii. 𝑃(𝑋 ≥ 3) = 0.769

b) A pot has 10% of defective items. What should be the number of items such that the probability of finding at
least 1 defective item is at least 0.95.
Solution:
n=? p= 0.1 and q= 0.9
P (𝑥 ≥ 1) ≥ 0.95
[1-P(x<1)] ≥ 0.95
P(x<1) ≤ 0.05
P(x=0) ≤ 0.05
𝑒 −𝛼 𝛼 0 ≤ 0.05
𝑒 −𝑛𝑝 ≤ 0.05
𝑛 ≥ 29.95
c) Suppose that a container contains 10,000 particles. The probability that such a particle escapes from the
container equals 0.0004. What is the probability that more than 5 such escape occurs.
Solution:
X: number of particle escapes
n= 10,000 p= 0.0004 and 𝛼 = 4
𝑒 −4 4𝑥
P (𝑋 > 5)= 1 − 𝑃(𝑋 ≤ 5) = 1 − ∑5𝑥=0 𝑥!
d) X is a Poisson’s variable and it’s found that the probability that X=2 is two third of the probability that X=1.
Find the probability that X=0 and X=3. What is the probability that X exceeds 3?
Solution: P(X=2) = 2/3 P(X=1)
4
𝛼=
3
P(X=0) =0.2635
P(X=3) = ……….
P(X>3) =…………
e) An insurance company has discovered that only about 0.1% of the population is limited in a certain type of
accidents each year. If its 10000 policy holders were randomly selected from the population. What is the
probability that not more than 5 of the clients are involved in such accidents each year.
Solution:
 X: number of clients involved in accidents
p=0.001 n=1000 𝛼 = 10
P(𝑋 ≥ 5) = 0.8686
f) Suppose that a book of 585 pages contains 43 typographical errors. If these errors are randomly distributed
throughout the book, what is the probability that 10 pages selected at random will be errorfree.
Solution:
X: number of errors
n=10
p= 43/585
𝛼 = 0.735

P(X=0)=0.4795

g) Probability that an individual suffers a bad reaction from an injection is 0.001. Find the probability that out of
2000 individuals (i) exactly 3 (ii) not more than 2 suffer from bad reaction.

Continuous Distributions
Exponential Distribution: UNIPARAMETRIC
This family of distributions is characterized by a single parameter λ, which is called the rate. Intuitively, λ
can be thought of as the instantaneous “failure rate” of a “device” at any time t, given that the device has
survived up to t. The exponential distribution is typically used to model time intervals between “random
events”.
Examples:
a. The length of time between telephone calls
b. The length of time between arrivals at a service station
c. The lifetime of electronic components, i.e., an inter failure time.

Mean and Variance:

∞

𝐸(𝑋) = ∫ 𝑥 𝑓(𝑥)𝑑𝑥
−∞
∞

𝐸(𝑋) = ∫ 𝑥 𝜆𝑒 −𝜆𝑥 𝑑𝑥
0
𝑒 −𝜆𝑥 𝑒 −𝜆𝑥 1
= 𝜆 {𝑥 − 2 }=
−𝜆 𝜆 𝜆
1
𝑆𝑖𝑚𝑖𝑙𝑎𝑟𝑙𝑦, 𝑉(𝑋) = 2
𝜆
Gamma Distribution
A continuous random variable X is said to have a gamma distribution with parameters 𝛼 > 0 and r > 0 ,
shown as X∼G(𝛼 , r) , if its PDF is given by

Note: When we sub r=1 in gamma distribution we get exponential distribution.

Mean and Variance:

∞

𝐸(𝑋) = ∫ 𝑥 𝑓(𝑥)𝑑𝑥
−∞
∞
𝑟−1
𝑥 𝑒 −𝛼𝑥 𝛼 𝑟
𝐸(𝑋) = ∫ 𝑥 { } 𝑑𝑥
Γ(𝑟)
0
Multiply and divide by "𝛼 𝑟"
𝑟 ∞ 𝑥 𝑟 𝑒 −𝛼𝑥 𝛼 𝑟+1 ∞ 𝑥 𝑟 𝑒 −𝛼𝑥 𝛼 𝑟+1
𝐸(𝑋) = ∫
𝛼 0
{ Γ(𝑟+1)
} 𝑑𝑥 where ∫0 { Γ(𝑟+1)
} 𝑑𝑥 = 1 𝑏𝑒𝑖𝑛𝑔 𝑎 𝑝𝑑𝑓 𝑜𝑓 𝐺𝑎𝑚𝑚𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
𝑟
𝐸(𝑋) =
𝛼
V(X)= 𝐸(𝑥 2 ) − [𝐸(𝑥)]2
∞

𝐸(𝑥 2 ) = ∫ 𝑥2 𝑓(𝑥)𝑑𝑥
−∞
∞
2) 2
𝑥𝑟−1 𝑒−𝛼𝑥 𝛼𝑟
𝐸(𝑥 = ∫𝑥 { } 𝑑𝑥
Γ(𝑟)
0
Multiply and divide by "𝛼 2 𝑟(𝑟 + 1)"

𝑟(𝑟+1) ∞ 𝑥𝑟+1 𝑒−𝛼𝑥 𝛼𝑟+2 𝑟(𝑟+1)

𝐸(𝑥 2 ) = ∫0 { } 𝑑𝑥= 𝛼2
𝛼2 Γ(𝑟+2)
∞ 𝑥 𝑟+1 𝑒 −𝛼𝑥 𝛼 𝑟+2
where ∫0 { Γ(𝑟+2)
} 𝑑𝑥 = 1 𝑏𝑒𝑖𝑛𝑔 𝑎 𝑝𝑑𝑓 𝑜𝑓 𝐺𝑎𝑚𝑚𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
Therefore,

𝑟
V(X)= 𝛼2

To check the pdf defined is valid:

𝑓(𝑥) ≥ 0
∞ ∞ ∞ 𝑥 𝑟−1 𝑒 −𝛼𝑥 𝛼 𝑟
∫−∞ 𝑓(𝑥)𝑑𝑥 = ∫0 𝑓(𝑥)𝑑𝑥 = ∫0 { Γ(𝑟)
} 𝑑𝑥
𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒, 𝛼𝑥 = 𝑣
∞
𝛼𝑟 𝑣 𝑟−1 −𝑣
= ∫ { } 𝑒 (1/𝛼)𝑑𝑣
Γ(𝑟) 𝛼
0

1 ∞
=
Γ(𝑟)
∫0 {𝑣}𝑟−1 𝑒 −𝑣 𝑑𝑣
=1
 Problems:
1. The number of road accidents per day in a city is following a gamma distribution with an average of 6 and
variance of 18. Find the probability that there will be (i) more than 8 accidents (ii) between 5 to 8 in a
day.
𝑟 𝑟 1
Solution: 𝛼 =6 and 𝛼2
= 18 → r=2 and 𝛼 = 3
1
−( )𝑥 1
8 (1/3) 𝑒 3 ( 𝑥)
P(X>8) = ∫0 { Γ(1)
3
} 𝑑𝑥 = 1/9 [11𝑒 −1/3 − 1]

2. The daily consumption of electric power is in million Kw in a certain city is a random variable x having the
𝑥
1
pdf f(x)= {9 𝑥 𝑒− 3 0<𝑥
. Find the probability that the power supply is inadequate on
0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
any given date if the daily capacity of the power plant is 12 million Kw.
𝑥
∞ 1
Solution: P(x>12) = ∫12 {9 𝑥 𝑒 − 3 } 𝑑𝑥
𝑟 2
Note: V(X)= =
𝛼 2 (1)2
= 18
3
𝑥
1
3. If X has the pdf f(x)= {4 𝑥 𝑒− 2 0<𝑥< ∞ . Find the mean and variance.
0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
4. Solution:
𝑟 𝑟
𝐸(𝑋) = 𝛼 and V(X)= 𝛼2
𝑥 1
We have, r-1=1 i.e r=2 and −𝛼𝑥 = − 2 𝑖. 𝑒 𝛼 = 2
𝑟 𝑟
𝐸(𝑋) = 𝛼
= 4 and V(X)= 𝛼2 = 8

5. The amount of time required to repair a TV is exponentially distributed with mean 2. Find (i) the
probability that the required time exceeds 2 hours. (ii) The conational probability that the required
time taken at least 10 hours given that already 9 hours have been spent on repairing the TV.
1
Solution: E(X)= = 2 λ = 0.5
λ
x
1 −
f(x)= {2 e 0<x
2

0 otherwise
(i) P(X>2)=1/e
1
𝑃(𝑋≥10∩ |𝑥>9 )
(ii) 𝑃(𝑋 ≥ 10 |𝑥 > 9 ) =
𝑃(𝑥>9 )
= 𝑒 −2
6. If 𝑋 ~ 𝐸(𝜆) with P(𝑋 ≤ 1) = P(x>1) then find V(X).
1 1
Solution: 𝛼 = ln 2 therefore 𝑉(𝑋) = =
𝜆2 (ln 2)2

Chi-square Distribution:
𝒏 𝟏
Special case of Gamma distribution: 𝒓 = 𝟐 and 𝜶 = 𝟐 in 𝚪 function we get 𝝌𝟐 distribution.
A continuous random variable X is said to have a chi-square distribution if its PDF is given by.

Mean and Variance:

𝑟 𝑟
𝐸(𝑋) = = n and V(X)= = 2n
𝛼 𝛼2
Normal Distribution: (Gaussian Distribution)

To check the pdf defined is valid:

𝑓(𝑥) ≥ 0
∞ ∞ 2
−(𝑥−μ )
1 2
∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑒 2σ 𝑑𝑥
−∞ −∞ σ√2𝝅
𝑥−𝜇
Sub, 𝜎
=𝑧
∞ ∞ 2
1 −(𝑧 )
∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑒 2 𝑑𝑧
−∞
√2𝝅 −∞
∞ ∞ 2
2 −(𝑧 )
∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑒 2 𝑑𝑧
−∞
√2𝝅 0
∞ ∞ 2
√2 −(𝑧 )
∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑒 2 𝑑𝑧
√𝝅
−∞ 0

𝑧2
Sub, 2 = t

∞ √2 ∞ 𝑒 −𝑡
∫−∞ 𝑓(𝑥)𝑑𝑥 = ∫0 𝑑𝑡=1
√𝝅 √2𝑡

Mean and Variance: Prove the following.

E(X)=𝜇 and V(X)= 𝜎 2
Problem on Chebyshev’s Inequality:
The number of patients requiring ICU in a hospital is a random variable with mean 18 and S.D 2.5. Determine
minimum probability- that number of patients are between 8 and 28.
1
Solution: 𝑃(8 ≤ 𝑋 ≤ 28) > 1 − 𝑘2
1
𝑃(𝜇 + 𝑘𝜎 ≤ 𝑋 ≤ 𝜇 − 𝑘𝜎) > 1 −
𝑘2
𝜇 + 𝑘𝜎 = 8 therefore k = 4
15
𝑃(8 ≤ 𝑋 ≤ 28) >
16
Standard normal distribution
The standard normal distribution, also called the z-distribution, is a special normal distribution where
the mean(𝜇) is 0 and the standard deviation(𝜎) is 1. The curve is symmetric about x=0. Denoted by 𝑍 ∼
𝑁(0, 1).
Its PDF is,
−z2
e 2
ϕ(z) = , -∞<z<∞
√2π
𝑥−𝜇
where, 𝑧 = when 𝜇 is 0 and and 𝜎 is 1 z=x.
𝜎

Properties:
1. Area under the curve is one.
2. P(a<x<b) is area under the curve from a to b.
3. Cdf of 𝝓(𝒂) = 𝑷(𝒁 ≤ 𝒂)
4. P(𝒂 ≤Z≤ 𝒃)= 𝝓(𝒃) − 𝝓(𝒂)
5. 𝝓(−𝒂) = 𝑷(𝒁 ≤ −𝒂) = 𝑷(𝒁 ≥ 𝒂)= 1-P(Z<a)= 1- 𝝓(𝒂)
𝑎−𝜇 𝑏−𝜇
6. 𝑃(𝑎 ≤ 𝑥 ≤ 𝑏)= 𝑃( 𝜎 ≤ 𝑧 ≤ 𝜎 )

Problems:
1. Suppose 𝑿~𝑵(𝟕𝟓, 𝟏𝟎𝟎). 𝑭𝒊𝒏𝒅
(i) P(X<60)
(ii) P (70< X<100)
(iii) P(X<65)
Solution: 𝜇 = 75 and 𝜎 2 = 100.
𝑋−75
Z= 10 and 𝑍 ∼ 𝑁(0, 1)
𝑋−75
(i) P(X<60) = P (Z< 10 )
= P(Z<-1.5)
= 𝜙(−1.5)
= 1 − 𝜙(1.5) = 1 − 0.9332= 0.0668
(ii) P(70<X<100) = P(-0.5< z <2.5)
= 𝜙(2.5) − 𝜙(−0.5) = 𝜙(2.5) − [1 − 𝜙(0.5)]
= 0.9938-1+0.6915
= 0.6853
(iii) P(X<65)= 0.1587

2. Suppose 𝑋~𝑁(2, 0.16). 𝐹𝑖𝑛𝑑.

(i) 𝑃(𝑋 ≥ 2.3)→ 0.2266
(ii) 𝑃(1.8 ≤ 𝑋 ≤ 2.1) → 0.2902

3. Diameter of an electric cable is normally distributed with mean 0.8 and variance 0.0004. What is the
probability that the diameter exceeds 0.81 inches.
Solution:
𝜇 = 0.8 𝑎𝑛𝑑 𝜎 = 0.02
0.81−𝜇
𝑃(𝑋 > 0.81) = 𝑃 (𝑍 > )= 0.3085
𝜎
4. 𝑋~𝑁(1, 4), 𝐹𝑖𝑛𝑑 𝑃(|𝑥| > 4). 𝐴𝑛𝑠 = 0.073

5. 𝑋~𝑁(75, 25). 𝐹𝑖𝑛𝑑 𝑃(𝑋 > 80|𝑋 > 77). 𝐴𝑛𝑠: 0.4605

6. The height of 500 soldiers is found to have normal distribution. Of them 258 are found to be within 2c.m
of the mean height of 170c.m. Find the standard deviation of X.

Solution: 𝜇 = 170
X: height of soldiers
𝑋~𝑁(170, 𝜎2)
258
P(168 <X<172)= 500 = 0.516
168−𝜇 172−𝜇
P( 𝜎
<Z< 𝜎 )= 0.516
2
2𝜙 ( , ) = 1.516
𝜎
2/𝜎 = 𝜙 −1 (0.758)
2
= 0.7
𝜎
𝜎 = 2.857

7. In normal distribution 31% of item are <45 and 8% are over 64. Find mean and Standard distribution.
Solution: P(X<45)=0.31
P(X>64) = 0.08
Simplifying, 45- 𝜇= -0.5𝜎
64 − 𝜇 = 1.41𝜎
Solving, 𝜇 = 49.97 𝑎𝑛𝑑 𝜎 = 9.94

8. Suppose X has 𝑁(3, 4). Find “c” such that P(𝑋 > 𝑐)= 2𝑃(𝑋 ≤ 𝑐).
Ans: c= 2.14

9. Suppose that the life span of two electronic device A and B have distribution N(40, 36) and N(45, 9). If
the electronic device is to be used for 45 hours period which device is to be preferred. If it is used for 48
hours which device is to be preferred.
Solution:
Device A Device B
N ( 40 , 36) N (45 , 9)
Mean = 40 Mean = 45
SD = 6 SD = 3
device A is to be used for a 45h period and device B is to be used for a 45 –h period.
above. 𝑃(𝑋 ≥ 45) = 0.5
𝑃(𝑋 ≥ 45) = 0.2025 0.5 > 0.2025 Hence Device B is better for 45
device A is to be used for a 48h period. hours.
𝑃(𝑋 ≥ 48) = 0.0915
Device B is to be used for a 48 –h period.
𝑃(𝑋 ≥ 48) = 0.1587
0.1587 > 0.0915
Hence Device B is better for 48 hours.
in Both cases Device B is better

10. In a normal distribution, 7% of the item are under 35 and 89% of the item are under 63. Find the mean
and variance of the distribution.
Solution: P(X<35) = 0.07 and P(X<65)= 0.89 gives 𝑋~𝑁(50, 100)
11. Obtain the percentage of students who are graded A, B, C, D, E and F.

Solution:
P(𝑋 > 𝜇 + 𝜎)= 0.1587
P(𝜇 < 𝑋 < 𝜇 + 𝜎)=0.3413
𝑃(𝜇 − 𝜎 < 𝑋 < 𝜇 )= 0.3413
𝑃(𝜇 − 2𝜎 < 𝑋 < 𝜇 − 𝜎 ) = 0.1359
𝑃(𝑋 < 𝜇 − 2𝜎) = 0.0228

12. The monthly income of a group of 10,000 person were found to be normally distributed with mean 750
rupees and SD rupees 50. Show that of this group about 95% had income exceeding rupees 668 and
only 5% had income exceeding rupees 832. What was the lowest income among the richest 100?
Solution:
X: Monthly income of a group
To Show P(X>668) =0.95 and P(X>832) =0.05
Consider, P(X>668) and P(X>832) and solve.
To find “C” such that P(X>𝜇+C) = 100/10000 =0.01
C= 116.5
Therefore, lowest income among the richest 100 = 𝜇 + 𝐶= 866.5

13. The annual rainfall at a certain locality is known to be normally distributed random variable with mean
29.5inches and SD 2.5inches. How many inches of rain annually exceeds about 5% of the time?
Solution:
X: annual rainfall at certain locality
𝜇 = 29.5 𝑎𝑛𝑑 𝜎 = 2.5
𝑃(𝑋 > 𝜇 + 𝑐) = 0.05
𝑐
1 − 𝑃 (𝑍 ≤ ) = 0.95
𝜎
C= 4.125
Rain exceeds about 5% of the time is 𝜇 + 𝑐= 33.625 inches.