Probability

Distributions
Discrete Continuous
Probability Probability
Distributions Distributions

Binomial Normal

Poisson Uniform

Hypergeometric Exponential
 Binomial Distribution
If there are only two possible outcomes of an experiment, say success and failure, then the
probability of ‘r’ successes in ‘n’ trials is given by:
P(X=r) = nCr pr (1-p)n-r ~ B(n, p)
where ‘p’ is the probability of success (which remains constant), r ≤ n and, r = 0, 1, 2, 3………n and
n = 1, 2, 3, 4, …………, ∞
✓ Sometimes it is also written as P(X=r) = nCr pr qn-r, where ‘q’ is the probability of failure and, p+q =
1.
✓ Mean and variance of a Binomial distribution are ‘np’ and ‘npq’ respectively.
Ex.1: Probability that a delivery boy delivers a correct order is 0.9. What is the probability that he
correctly delivers 8 out of 10 orders? What is the probability that he correctly delivers 8 or more
times?
Soln: Prob. of correct delivery (success) = 0.9; Prob. of wrong delivery (failure) = 1-0.9 = 0.1
n = 10, r = 8, p = 0.9, (1-p) = 1-0.9 = q
P(X=8) = 10C8 (0.9)8 (1-0.9)10-8 = 45.(0.4305).(0.01) = 0.1937
Hint: P(X≥8) = P(X=8) + P(X=9) + P(X=10); P(X=9) = 0.3874, P(X=10) = 0.3487; Answer: 0.9298
 Calculation of nCr

nC = 𝑛! 𝑛 𝑛−1 𝑛−2 …..3.2.1

r 𝑟! 𝑛−𝑟 ! =
𝑟 𝑟−1 𝑟−2 …..3.2.1 𝑛−𝑟 𝑛−𝑟−1 ….3.2.1
n=5, r=2
5C 5! 5.4.3.2.1
2 = 2! 5−2 ! =
2.1.(3.2.1)
= 10

0! = 1 and 1! = 1
Ex.2: Nitrogen gas is filled in automobile tyres to improve the ride quality. A filling station experiences
that 30% of the customers get nitrogen gas filled in their vehicle’s tyres. If 10 customers arrive at a
gas station then what is the probability that,
(i) All of them would fill nitrogen gas?
(ii) None of them would fill nitrogen gas?
(iii) Half of them would fill nitrogen gas?
(iv) Less than 8 customers would fill nitrogen gas?
(v) Find the mean no. of customers who would fill nitrogen gas.
Soln: n = 10, p = 0.3, 1-p = 1-0.3 = 0.7
P(X=r) = nCr pr (1-p)n-r
(i) P(X=10) = 10C10 (0.3)10 (1-0.3)10-10 = 1.(0.000006).(1) = Almost zero
(ii) P(X=0) = 10C0 (0.3)0 (1-0.3)10-0 = 1.(1).(0.0282) = 0.0282
10!
(iii) P(X=5) = 10C5 (0.3)5 (1-0.3)10-5 = . (0.3)5(1-0.3)10-5 = 252.(0.0024).(0.1681) = 0.0835
5! 10−5 !
(iv) P(X<8) = 1-P(X≥8) = 1-[P(X=8) + P(X=9) + P(X=10)] = 1-(0.0015+0.0001+0.000006) = 0.9984
(v) Mean no. of customers who fill nitrogen gas = np = 10.(0.3) = 3
Ex.3: Fifty percent of the people believe that the country is in recession. For a sample of 20
people, make the following calculations:
(i) Probability that 12 people believe the country is in recession. (0.1201)
(ii) Probability that at least 18 people believe the country is in recession. (0.0002)
(iii) Probability that more than three people believe the country is in recession.
P(X>3)=1-P(X≤3) = 1-[P(X=0)+P(X=1)+P(X=2)+P(X=3)] = (0.9987)
(iv) Expected no. of people who would say that the country is in recession. (10)
(v) Compute variance and std. deviation of the no. of people who believe the country is in
recession. (5, 2.236)

Ex.4: Twenty three percent of the vehicles are not covered by any insurance. On a special
checking day, 30 vehicles are checked randomly. What is the probability that more than 27
vehicles are insured? (here, p=0.77, n=30)
What is the expected no. of vehicles not covered by any insurance? What is the variance and
std. deviation? (here, n=30, p=0.23, q=0.77)
 Poisson Distribution
It is used to estimate the probability of no. of occurrences over a specified period of time. If the mean no. of
occurrences over a specified period of time is ‘λ’ then the probability of ‘r’ occurrences is given by,
−λ λ𝑟
P(X=r) = 𝑒 . ~ 𝑃(λ)
𝑟!
where r = 0, 1, 2, 3, …………∞ and e = 2.718.
✓ Both mean and variance of a Poisson distribution are ‘λ’.
✓ The value of ‘λ’ should be adjusted according to time-interval.
✓ Poisson distribution approximates the binomial distribution for large n and small p.
Ex.1: In a drive-thru window of a burger shop, average 10 customers arrive in a 30-minute interval. What is the
probability that exactly 5 customers arrive in 30 minutes?
What is the probability that 5 customers arrive in 15-minutes interval?
−λ λ𝑟
Soln: P(X=r) = 𝑒 . 𝑟!
For 30-minutes duration, λ = 10 and r = 5.
−10 105
P(X=5) = 𝑒 . = (0.000045).(100000/120) = (0.000045).(833.33) = 0.0378
5!
For 15-minutes duration, λ = 5 and r = 5.
−5 55
P(X=5) = 𝑒 . = (0.0067).(3125/120) = 0.1754
5!
Ex.2: An average of 15 aircraft accidents occur each year. Compute
(i) The mean no. of accidents per month.
(ii) Prob. of no accidents during a month.
(iii) Prob. of exactly one accident per month.
(iv) Prob. of more than one accidents per month.
−λ λ
𝑟
Soln: P(X=r) = 𝑒 .
𝑟!
For 12-months duration λ = 15, for one-month duration λ = 1.25.
(i) Mean no. of accidents per month = 15/12 = 1.25
−1.25 (1.25)0
(ii) For r=0, P(X=0) = 𝑒 . = (0.2865).(1/1) = 0.2865
0!
−1.25 1.25 1
(iii) For r=1, P(X=1) = 𝑒 . = (0.2865).(1.25/1) = 0.3581
1!
(iv) For r>1, P(X>1) = 1- P(X≤1) = 1-[P(X=0)+P(X=1)] = 1-(0.2865+0.3581) = 0.3554

Ex.3: Phone calls arrive at the rate of 48 per hour in a call centre. Compute
(i) The mean no. of calls in 5 minutes duration. (4)
(ii) The prob. of receiving three calls in 5 minutes. (0.1952)
(iii) Prob. of receiving exactly 10 calls in 15 minutes. (0.1048)
(iv) Prob. of receiving at least one call in 10 minutes. (0.9997)
Ex.4: It is estimated that 0.5 percent of the callers to the customer service department will receive a busy
signal. What is the probability that out of 1200 callers, at least 3 will receive a busy signal?
Soln: n = 1200, p = 0.5/100 = 0.005
Since the problem has large n and small p, hence we assume λ =np
Mean np = (1200).(0.005) = 6 = λ (as n is large and p is very small)
−λ λ
𝑟
P(X=r) = 𝑒 .
𝑟!
For r≥3, P(X ≥3) = 1-P(X<3) = 1-[P(X=0)+P(X=1)+P(X=2)]
−6 (6)0
P(X=0) = 𝑒 . = 0.0025, P(X=1) = 0.0149, P(X=2) = 0.0446
0!
P(X≥3) = 1-(0.0025+0.0149+0.0446) = 0.938

Ex.5: Patients arrive at a hospital at the rate of 6 per hour. Find the probability that in a 90-minute duration
(i) exactly 7 patients arrive in the hospital.
(ii) between 7 and 10 patients arrive in the hospital. P(7≤X≤10) = P(X=7)+P(X=8)+P(X=9)+P(X=10)
(iii) If a patient arrives at 11:30am then what is the probability that other patients arrive before 11:45am?
λ= 1.5, P(X≥1)= 1-P(X<1) = 1-P(X=0) = 1-e-1.5 = 1- 0.2232 = 0.7768
 Continuous Probability Distributions

• A continuous random variable is a variable that can assume any

value on a given range (can assume an uncountable number of
values)
• thickness of an item
• time required to complete a task
• temperature of a solution
• height, in inches

• These can potentially take on any value depending only on the

ability to precisely and accurately measure
 The Normal Distribution

• Bell Shaped
• Symmetrical f(X)
• Mean, Median and Mode
are equal
Location is determined by the σ
mean, μ X
μ
Spread is determined by the
standard deviation, σ
Mean = Median = Mode

The random variable has an infinite

theoretical range:
-  to + 
 The Normal Distribution Function

The formula for the normal probability density function is

2
1  (X−μ) 
1 −  
2  
f(X) = e
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
 Many Normal Distributions

By varying the parameters μ and σ, we obtain different normal distributions

 The Normal Distribution Shape

f(X) Changing μ shifts the distribution

left or right.

Changing σ increases or decreases

the spread.
σ

X=μ X
 The Standardized Normal Distribution
• Any normal distribution (with any mean and standard deviation
combination) can be transformed into the standardized normal
distribution (Z distribution).
• Need to transform X units into Z units.
• The standardized normal distribution (Z) has a mean of 0 and a
standard deviation of 1.
• Translate from X to the standard normal variate ‘Z’ by
subtracting the mean of X and dividing by its standard deviation:
X −μ
Z=
σ
The Z distribution always has mean = 0 and standard deviation = 1
The Standardized Normal Probability Density Function

• The formula for the standardized normal probability

density function is

1 −(1/2)Z2
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
 The Standardized Normal Distribution

• Also known as the ‘Z-distribution’

• Mean is 0
• Standard Deviation is 1
f(Z)

Z
Z=0

Values on right of Z=0 have positive Z-values and values on left of Z=0 have negative Z-values
 Example: Transforming X into Z

• 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.
 Comparing X and Z values

X=100 X= 200 X (μ = 100, σ = 50)

Z=0 Z=2.0 Z (μ = 0, σ = 1)

Note that the shape of the distribution is the same, only the scale has changed.
We can express the problem in original units (X) or in standardized units (Z)
 Finding Normal Probabilities

Probability is measured by the area under

the curve
f(X)
P (a ≤ X ≤ b )
= P (a < X < b )
(Note that the probability
of any individual value is
zero)

X
a b
Probability as Area Under the Curve

The total area under the curve is 1.0, and the curve is symmetric, so half is on
the right of mean and half is on the left.

f(X)
P( −  X  μ) = 0.5 P(μ  X  ) = 0.5

0.5 0.5

μ X

P( −  X  ) = 1.0
 The Standardized Normal Table

Standard normal table in the textbook (Appendix table

A.1, page-856) gives the probability equal to or less
than a desired value of Z (from 0 to Z). The probabilities
are same for negative values of Z.

0.50 0.4772
Example:
P(0<Z<2) = 0.4772
P(Z<2.00) = 0.50 + 0.4772
= 0.9772
Z=0 Z=2 Z
P(Z>2) = 0.5-0.4772 = 0.0228
 Finding Normal Probabilities

To find P(a < X < b) when X is distributed normally:

• Translate X-values to Z-values

• Draw a normal curve for the problem in terms of Z

• Use the Standard normal table

 Finding Normal Probabilities

• Let X represent the time it takes to download

an image file from the internet.
• Suppose X is normally distributed with mean
8.0 and standard deviation 5.0.
• Find P(X < 8.6)

X
8.0
8.6
 Finding Normal Probabilities
• Let X represent the time it takes to download an image file from the internet.
• Suppose X is normal with mean 8.0 and standard deviation 5.0. Find P(X < 8.6)

X − μ 8.6 − 8.0
Z= = = 0.12
σ 5.0

μ=8 μ=0
σ=5 σ=1

8.0 8.6 X 0 0.12 Z

P(X < 8.6) P(Z < 0.12)

 Solution: Finding P(Z < 0.12)

P(X < 8.6) = P(Z < 0.12)

Standard Normal Probability Table (Portion) = 0.5 + 0.0478
= 0.5478

Z 0.00 0.01 0.02

0.50 0.0478
0.0 0.0000 0.0040 0.0080

0.1 0.0398 0.0438 0.0478

0.2 0.0793 0.0832 0.0871

Z
0.3 0.1179 0.1217 0.1255 0.00
0.12
 Finding Normal Probabilities

• Suppose X is normal with mean 8.0 and

standard deviation 5.0.
• Now Find P(X > 8.6)

X
8.0
8.6
 Finding Normal Upper Tail Probabilities

Now Find P(X > 8.6)…

P(X > 8.6) = P(Z > 0.12) = 0.5 - P(0 ≤ Z ≤ 0.12)
= 0.5 - 0.0478 = 0.4522

0.0478
0.5 0.5 (blue area)
0.5 - 0.0478 = 0.4522

Z Z
Z=0 Z=0
Z=0.12
Finding a normal probability between two values

Suppose X is normal with mean 8.0 and standard

deviation 5.0. Find P(8 < X < 8.6)

Calculate Z-values:

X −μ 8 −8
Z= = =0
σ 5
8 8.6 X
X − μ 8.6 − 8 0 0.12 Z
Z= = = 0.12
σ 5 P(8 < X < 8.6)
= P(0 < Z < 0.12)
= 0.0478
 Solution: Finding P(0 < Z < 0.12)

P(8 < X < 8.6)

Standardized Normal Probability Table (Portion) = P(0 < Z < 0.12)
= P(Z < 0.12) – P(Z ≤ 0)
Z 0.00 0.01 0.02 = 0.5478 - 0.5000 = 0.0478

0.0 0.0000 0.0040 0.0080 0.0478

0.5000

0.1 0.0398 0.0438 0.0478

0.2 0.0793 0.0832 0.0871

0.3 0.1179 0.1217 0.1255 Z

0.00
0.12
 Probabilities in the Lower Tail

• Suppose X is normal with mean 8.0 and

standard deviation 5.0.
• Now Find P(7.4 < X < 8)

X
8.0
7.4
 Probabilities in the Lower Tail

Now Find P(7.4 < X < 8)…

P(7.4 < X < 8)
= P(-0.12 < Z < 0)
0.0478
= 0.0478

0.4522

The Normal distribution is symmetric, so this

probability is the same as P(0 < Z < 0.12)
7.4 8.0 X
-0.12 0 Z
 Practice Exercises
1. In a city, it is estimated that the maximum temperature is normally distributed with a mean of 23°C and
a standard deviation of 5°C. Calculate the number of days in this month in which it is expected to reach a
maximum of between 21°C and 27°C.

2. The mean weight of 500 college students is 70 kg and the standard deviation is 3 kg. Assuming that the
weight is normally distributed, determine how many students weigh:
a. between 60 kg and 75 kg
b. more than 90 kg
c. less than 64 kg
d. exactly 64 kg
e. 64 kg or less

3. For borrowers with good credit scores, the mean debt amount is $15,000. Assuming the debt amounts
to be normally distributed with standard deviation $3000, calculate the probability that
a. debt for a borrower is more than $18,000
b. debt for a borrower is less than $10,000
c. Debt for a borrower is between $12,000 and $18,000
 Empirical Rules

What can we say about the distribution of values around

the mean? For any normal distribution:

f(X)

μ ± σ encloses about 68.26%

of X’s
σ σ

X
μ-σ μ μ+σ

68.26%
 The Empirical Rule

• μ ± 2σ covers about 95.44% of X values

• μ ± 3σ covers about 99.73% of X values

2σ 2σ 3σ 3σ

μ x μ x

95.44% 99.73%