Probability-1.pptx

• Introduction to Random Variables (Ch 10.1)
• Discrete (CH 10.2)
• continuous
• Expected value and Variance (Ch 10.3)
• Binomial (Ch 10.5)
• Poisson (Ch 10.6)
• Normal (Ch 11.2)
• Exponential Distributions (Ch 11.4)

• A random variable is a numerical description of the outcome of an
experiment.
• The probability distribution for a random variable describes how
probabilities are distributed over the values of the random variable.
• table, graph, or formula
The required conditions for a discrete probability
function f(x) are:
f(x) > 0
f(x) = 1

Discrete random variable
Experiment Random Variable
(x)
Possible values for
random variable
Contact five
customers
No: of customers
who place an order
0,1,2,3,4,5
Inspect shipment of
50 radios
No: of defective
radios
0,1,2,3…..49,50
Operate restaurant for
one day
No: of customers 0,1,2,3……..
Sell an automobile Gender of customer 0 for male , 1 for
female

Probability distribution for number obtained on a roll of
dice
No obtained Probability of x = f(x)
1 1/6 = 0.1667
2 1/6 = 0.1667
3 1/6 = 0.1667
4 1/6 = 0.1667
5 1/6 = 0.1667
6 1/6 = 0.1667
∑ f(x) = 1

Continuous random variable
Experiment Random Variable (x) Possible values for
random variable
Operate a bank Time between customer
arrivals in minutes
X≥0
Fill a soft drink can ( max
=12.1 ounces)
No: of ounces 0≤x≤12.1
Construct a new library % of project complete
after 6 months
0≤x≤100
Test a new chemical
process
Temp observed when
reaction takes place (150-
212 F )
150≤x≤212

Number
Units Sold of Days
0 80
1 50
2 40
3 10
4 20
200
Discrete Probability Distributions
x f(x)
0 .40
1 .25
2 .20
3 .05
4 .10
1.00
a tabular representation of the probability distribution for automobile
sales was developed.

.10
.20
.30
.40
.50
0 1 2 3 4
Values of Random Variable x (TV sales)
Probability
Discrete Probability Distributions

Problem 7 ( Pg 326)
• The probability distribution for the random variable x follows
a) Is this probability distribution valid? Explain
b) What is the probability that x =30?
c) What is the probability that x≤ 25?
d) What is probability that x> 30?
X f
20 100
25 75
30 125
35 200

Solution
a. f (x)  0 for all values of x.
 f (x) = 1 Therefore, it is a proper probability distribution.
b. Probability x = 30 is f (30) = .25
c. Probability x  25 is f (20) + f (25) = .20 + .15 = .35
d. Probability x > 30 is f (35) = .40
Practice Q 8-14 pg 326-328
X f f(x)
20 100 0.20
25 75 0.15
30 125 0.25
35 200 0.40

Expected value and variance
• The expected value is a weighted
average of the values the random
variable may assume. The weights are
the probabilities.
• The expected value does not have to be
a value the random variable can assume.
• The expected value, or mean, of a
random variable is a measure of its
central location.
• E(x) =  = xf(x)
• The variance is a weighted average of
the squared deviations of a random
variable from its mean. The weights
are the probabilities.
• The variance summarizes the
variability in the values of a random
variable.
• Var(x) =  2 = (x - )2f(x)
• = √Var

x Frequency f(x) E(x) =
x*f(x)
𝒙 −  𝟐
𝝈𝟐
= 𝒙 −
0 54
1 117
2 72
3 42
4 12
5 3
∑f(x)
No: of cakes sold in a bakery =1.5

x f f(x) x*f(x)
0 54 0.18 0*0.18=0
1 117 0.39 1*0.39=0.39
2 72 0.24 2*0.24=0.48
3 42 0.14 3*0.14=0.42
4 12 0.04 4*0.04=0.16
5 3 0.01 5*0.01=0.05
∑ =300 ∑ f(x) = 1 E(x) =μ= ∑

x f(x) x*f(x) (x-μ) (x-μ)^2 (x-μ)^2 *f(x)
0 0.18 0 0-1.50=-1.5 2.25 2.25(.18) =0.4050
1 0.39 0.39 1-1.50= -0.50 0.25 0.25(0.39) =0.0975
2 0.24 0.48 2-1.5= 0.50 0.25 0.25(0.24) =0.0600
3 0.14 0.42 1.5 2.25 2.25(0.14) =0.3150
4 0.04 0.16 2.5 6.25 6.25(0.04)=0.2500
5 0.01 0.05 3.5 12.25 12.25(0.01) =0.1225
∑ f(x) = 1 E(x) =μ= ∑
xf(x)=1.50
𝜎2
= ∑ 𝑥 − 𝜇 2
𝑓 𝑥) = 1.25
SD = 𝜎 = √1.25 =1.118
(μ = 1.50)

Calculate Problem 7 with expected value and
variance
X f(x)
20 0.20
25 0.15
30 0.25
35 0.40
E(x) =  = xf(x)
Var(x) =  2 = (x - )2f(x)

Calculate Problem 7 with expected value and
variance – solution
X f(x) x*f(x) x- μ (x- μ)^2 (x- μ)^2* f(x)
20 0.2 4 -9.25 85.5625 17.1125
25 0.15 3.75 -4.25 18.0625 2.709375
30 0.25 7.5 0.75 0.5625 0.140625
35 0.4 14 5.75 33.0625 13.225
29.3 33.1875
E(x) =  = xf(x)= 29.3
Var(x) =  2 = (x - )2f(x) = 33.1875
Practice problems 15-24 pg 330-333

Binomial Distribution
• Applied to single variable discrete data where results are the numbers of
“successful outcomes” in a given scenario.
• Statistically independent – one does not affect the other.
• Knowledge of outcomes either success or failure
• q+p=1
• e.g.:
• no. of times the lights are red in 20 sets of traffic lights,
• no. of students with green eyes in a class of 40
• no. of plants with diseased leaves from a sample of 50 plants

Properties of a Binomial Experiment
• The experiment consists of a sequence of n identical trials.
• Two outcomes, success and failure, are possible on each trial.
• The probability of a success, denoted by p, does not change from trial to
trial.
• The trials are independent

Binomial Probability Distribution
𝑛
𝐶𝑟𝑝𝑥
𝑞𝑛−𝑥
• We let x denote the number of successes occurring in the n trials.
𝑓 𝑥 =
𝑛!
𝑥! 𝑛 − 𝑥 !
𝑝𝑥𝑞𝑛−𝑥
where:
x = the number of successes
p = the probability of a success on one trial
n = the number of trials
f(x) = the probability of x successes in n trials
n! = n(n – 1)(n – 2) ….. (2)(1)
𝑁𝑜𝑡𝑒 𝑛
𝐶0=1 ;
𝑛
𝐶1=n;
𝑛
𝐶𝑟 𝑤ℎ𝑒𝑟𝑒
9
𝐶9 =1 ; 𝑝0
= 1

Binomial – sample problem – solution
=
20!
3! ∗ 17!
0.153 ∗ 0.8517
=1140*0.003375* 0.06311

Binomial Probability Distribution
• Expected Value
𝐸 𝑥 =  = 𝑛𝑝
• Variance
𝑉𝑎𝑟 𝑥 = 𝜎2 = 𝑛𝑝 1 − 𝑝)
• Standard Deviation
𝜎 = 𝑛𝑝 1 − 𝑝
0r 𝑛𝑝𝑞

Sample problem - Binomial
• Evans Electronics is concerned about a low retention
rate for its employees. In recent years, management has
seen a turnover of 10% of the hourly employees
annually. Thus, for any hourly employee chosen at
random, management estimates a probability of 0.1 that
the person will not be with the company next year.
Choosing 3 hourly employees at random, what is the
probability that 1 of them will leave the company this
year?
Let: p = .10, n = 3, x = 1

Solution
Let: p = .10, n = 3, x = 1
f x
n
x n x
p p
x n x
( )
!
!( )!
( )( )
=

 
1
1 2
3!
(1) (0.1) (0.9) 3(.1)(.81) .243
1!(3 1)!
f = = =


•E(x) = np = 3(.1) = .3 employees out of 3
•Var(x) = np(1 – p) = 3(.1)(.9) = .27
•𝜎 = 3 ∗ 0.1 ∗ 0.9 = 0.52 employees

Experimental
Outcome
(S, F, F)
(F, S, F)
(F, F, S)
Probability of
Experimental Outcome
p(1 – p)(1 – p) = (.1)(.9)(.9) = .081
(1 – p)p(1 – p) = (.9)(.1)(.9) = .081
(1 – p)(1 – p)p = (.9)(.9)(.1) = .081
Total = .243
With a .10 probability of an employee leaving on any
one trial, the probability of an employee leaving on
the first trial and not on the second and third trials is
given by
(.10)(.90)(.90) = (.10)(.90)2 = .081

Binomial Probability Distribution – tree diagram
1st Worker 2nd Worker 3rd Worker x Prob.
Leaves
(.1)
Stays
(.9)
3
2
0
2
2
Leaves (.1)
Leaves (.1)
S (.9)
Stays (.9)
Stays (.9)
S (.9)
S (.9)
S (.9)
L (.1)
L (.1)
L (.1)
L (.1) .0010
.0090
.0090
.7290
.0090
1
1
.0810
.0810
.0810
1

Binomial probability table
n x .05 .10 .15 .20 .25 .30 .35 .40 .45 .50
3 0 .8574 .7290 .6141 .5120 .4219 .3430 .2746 .2160 .1664 .1250
1 .1354 .2430 .3251 .3840 .4219 .4410 .4436 .4320 .4084 .3750
2 .0071 .0270 .0574 .0960 .1406 .1890 .2389 .2880 .3341 .3750
3 .0001 .0010 .0034 .0080 .0156 .0270 .0429 .0640 .0911 .1250
p

Practice Problem 31 pg 350
• Consider a binomial experiment with two trials and
p=0.4
a) Draw a tree diagram for this experiment
b) Compute the probability of one success
c) Compute f(0)
d) Compute f(2)
e) Compute probability of at least one success
f) Compute Expected value, Variance, SD

Practice Problem 31 Solution
1 1
2 2!
(1) (.4) (.6) (.4)(.6) .48
1 1!1!
f
 
= = =
 
 
0 2
2 2!
(0) (.4) (.6) (1)(.36) .36
0 0!2!
f
 
= = =
 
 
2 0
2 2!
(2) (.4) (.6) (.16)(1) .16
2 2!0!
f
 
= = =
 
 
P(x  1) = f (1) + f (2) = .48 + .16 = .64
Or
P(x  1) = 1- f (0) = 1-0.36 = 0.64
E(x) = n p = 2 (.4) = .8
Var(x) = n p (1 - p) = 2 (.4) (.6) = .48
 = .48 =0.6928

Problem 34 pg 351
• A Harris interactive survey for intercontinental hotels and resorts asked
respondents,” when travelling internationally, do you generally venture
out on your own to experience culture or stick with your tour group and
itineraries?” The survey found that 23% of the respondents stick with
their tour group
a) In a sample of six international travelers, what is the probability that
two will stick with their tour group?
b) If n=6,find probability that at least two will stick with the tour group
c) If n=10, find probability that none will stick with the group

Problem 34 solution
Practice problems 351-352 for binomial

Poisson
• A Poisson distributed random variable is often useful in
estimating the number of occurrences over a specified
interval of time or space.
• It is a discrete random variable that may assume an
infinite sequence of values (x = 0, 1, 2, . . . ).
• Eg : the number of vehicles arriving at a toll booth in
one hour
• Eg: the arrival of phone calls.

• Properties of a Poisson Experiment
• The probability of an occurrence is the same for any two
intervals of equal length.
• The occurrence or nonoccurrence in any interval is independent
of the occurrence or nonoccurrence in any other interval
• A property of the Poisson distribution is that the mean and
variance are equal.
• Since there is no stated upper limit for the number of
occurrences, the probability function f(x) is applicable for
values x = 0, 1, 2, … without limit.
• In practical applications, x will eventually become large
enough so that f(x) is approximately zero and the probability
of any larger values of x becomes negligible.

Poisson Probability Function
f x
e
x
x
( )
!
=

 
where:
x = the number of occurrences in an interval
f(x) = the probability of x occurrences in an interval
 = mean number of occurrences in an interval
e = 2.71828
x! = x(x – 1)(x – 2) . . . (2)(1)
A property of the Poisson distribution is that
the mean and variance are equal.
 =  2
e - Euler's number - compounding
factor less than annual - non-
terminating series

Sample problem – Poisson
• Patients arrive at the emergency room of Mercy Hospital at the average
rate of 6 per hour on weekend evenings. What is the probability of 4
arrivals in 30 minutes on a weekend evening?

Sample problem – Poisson
• Patients arrive at the emergency room of Mercy Hospital at the average
rate of 6 per hour on weekend evenings. What is the probability of 4
arrivals in 30 minutes on a weekend evening?
 = 6/hour = 3/half-hour, x = 4
4 3
3 (2.71828)
(4) .1680
4!
f

= =
Variance for Number of Arrivals during 30-Minute Periods
 =  2 = 3

Problem 44 – pg 355
• Consider a poisson distribution with  =3.
• Write the appropriate Poisson probability function
• Compute f (2)
• Compute f (1)
• Compute P(x  2)

Poisson – problem 46 pg 355
• Phone calls arrive at the rate of 48 per hour at the reservation desk for
regional airways
a. Compute the probability of receiving three calls in a 5 minute interval
b. Compute the probability of receiving exactly 10 calls in 15 minutes
c. Suppose no calls are currently on hold. If the agent takes 5 minutes to
complete the current call, how many callers do you expect to be
waiting by that time? What is the probability that none will be waiting?
d. If no calls are currently being processed, what is the probability that
the agent can take 3 minutes for personal time without being
interrupted by a call?

Poisson – problem 46 – solution

Hypergeometric Probability Distribution
• The hypergeometric distribution is closely related to
the binomial distribution.
• However, for the hypergeometric distribution:
• the trials are not independent
• the probability of success changes from trial to trial.

Continuous Probability Distributions
• Uniform Probability Distribution
f (x)
x
Uniform
x
f (x)
Normal
x
f (x) Exponential
 Normal Probability Distribution
 Exponential Probability Distribution
 Normal Approximation of Binomial Probabilities

Normal Distribution
• Applied to single variable continuous data e.g.
• heights of plants, weights of lambs, lengths of time
• Used to calculate the probability of occurrences less than, more than,
between given values e.g.
• “the probability that the plants will be less than 70mm”,
• “the probability that the lambs will be heavier than 70kg”,
• “the probability that the time taken will be between 10 and 12 minutes”

Normal distribution illustrating compatibility
of z values and standard deviation
• Aprox 68% of all values in normal
distributed population lie within + or –
1 SD from the mean
• Aprox 95% of all values in normal
2 SD from the mean
• Aprox 99.7 % of all values in normal
3 SD from the mean

Area under normal curve and z values
P(X<=40)
X= 40
Mean = μ =50
SD = σ = 25
-25 0 25 50 75 100
125
-3 -2 -1 0 1 2 3
Z= x- μ / σ
Z= number of standard
deviations from x to the mean of
this distribution
X= value of random variable
which is being tested
μ = mean of distribution of
random variable
σ = SD of given distribution
https://www.youtube.com/watch?v=mtbJbDwqWLE
Z= (40 - 50) / 25

Standard Normal Probability Distribution
z
x
=
 

We can think of z as a measure of the number of standard deviations x is from 𝜇.
To find the associated probability, find the table value of Z from normal table
To compute x value from a given probability Eg 20%
1 Convert the % to proportion – 0.20
2 Find the corresponding value of 0.20 from normal
table -0.8
3 Use formula 𝑥 = 𝜇 + 𝑧𝜎 to find x

Normal distribution calculation
Find table value of 1 in normal distribution table =0.8413
(P <= 6 ) = 0.8413
(P> 6 ) = 1- 0.8413 = 0.1587
X= 2 ; 𝑍 = −
2
2
= −1 ; 𝑃 𝑋 ≤ 2 = 0.1587
𝑃 2 ≤ 𝑥 ≤ 6 = |0.1587 − 0.8413| = 0.6826

Question 1
•Wool fibre breaking strengths are normally
distributed with mean m = 23.56 Newtons and
standard deviation, σ = 4.55. What proportion of
fibres would have a breaking strength of 14.45 or
less?
•X= 14.45 ; Mu = 23.56 ; SD = 4.55

• From a new batch If 30.15% of wool fibre is found to be breaking
what could be the breaking strength in newtons
• 𝑥 = 𝜇 + 𝑧𝜎 ---------------------------------------------- (1)
• Where z is table value of 30% or 0.3015 = -0.5
• Sub in (1)
• 𝑥 = 23.56 + −0.5) ∗4.55 = 21.285 newtons is the breaking strength
in new batch

Q12 Pg 384
• Given that Z is a standard normal random variable, compute the
following probabilities
a) 𝑃 0 ≤ 𝑧 ≤ 0.83
b) 𝑃 −1.57 ≤ 𝑧 ≤ 0
c) 𝑃 𝑍 > 0.44
d) 𝑃 𝑧 ≥ −0.23
e) 𝑃 𝑧 < 1.20
f) 𝑃 𝑧 ≤ −0.71)

Q 12 solution
a. P(0 ≤ z ≤ .83) = .7967 - .5000 = .2967
b. P(-1.57 ≤ z ≤ 0) = .5000 - .0582 = .4418
c. P(z > .44) = 1 - .6700 = .3300
d. P(z ≥ -.23) = 1 - .4090 = .5910
e. P(z < 1.20) = .8849
f. P(z ≤ -.71) = .2389

Q19 Pg 385
• In an article about the cost of healthcare, money magazine reported that a
visit to a hospital emergency room for something as simple as a sore throat
has a mean cost of $328. Assume that the cost for this type of hospital
emergency room visit is normally distributed with standard deviation of
$92.Answer the following questions about the cost of hospital emergency
room visit for this medical service.
a. Probability that cost will be more than $500?
b. Probability that cost will be less than $250?
c. Probability that cost will be between $300 and $400?
d. If the cost to the patient is in lower 8% of charges for this medical service ,
what was the cost of this patents emergency room visit?

Probability-1.pptx

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