Introduction to Probability Distributions

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Introduction to Probability Distributions



Random Variable



A variable that can assume more than one value



Represents a possible numerical value from a random event



Example: What grade do you anticipate to earn in UGBS 301?

Random Variables

Discrete Random Variable

Continuous Random Variable

Possible outcomes: Grade A – F.

Let A = 1, B+ = 2, B=3, C+ =4, C=5, D+ = 6 and F = 7. Then a random variable x representing possible grades are:

x = {1,2,3,4,5,6,7}.

Note that the original letter grades were changed to numbers.

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Discrete Random Variables



Can only assume a countable number of values

Examples:



Roll a die twice

Let x be the number of times 4 comes up (then x could be 0, 1, or 2 times)



Toss a coin 5 times.

Let x be the number of heads

(then x = 0, 1, 2, 3, 4, or 5)

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Experiment: Toss 2 Coins. Let x = # heads.

T T

Discrete Probability Distribution

4 possible outcomes

T T

H H

Probability Distribution

0 1 2 x x Value Probability

0 1/4 = .25 1 2/4 = .50 2 1/4 = .25

.50 .25

Probability

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

A list of all possible [ x

_i

, P(x

_i

) ] pairs

x

_i

= Value of Random Variable (Outcome) P(x

_i

) = Probability Associated with Value



x

_i

’s are mutually exclusive (no overlap)



x

_i

’s are collectively exhaustive (nothing left out)



0  P(x

_i

)  1 for each x

_i



S P(x

_i

) = 1

Discrete Probability Distribution

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Discrete Random Variable Summary Measures



Expected Value of a discrete distribution

(Weighted Average)

E(x) = Sx _i P(x _i )



Example: Toss 2 coins, x = # of heads,

compute expected value of x:

E(x) = (0 x .25) + (1 x .50) + (2 x .25) = 1.0

x P(x) 0 .25 1 .50 2 .25

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

Standard Deviation of a discrete distribution

where:

E(x) = Expected value of the random variable x = Values of the random variable

P(x) = Probability of the random variable having the value of x

Discrete Random Variable Summary Measures

P(x) E(x)}

σ

_x

  {x 

²

(continued)

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

Example: Toss 2 coins, x = # heads,

compute standard deviation (recall E(x) = 1)

Discrete Random Variable Summary Measures

P(x) E(x)}

σ

_x

  {x 

²

.707 .50

(.25) 1)

(2 (.50)

1) (1

(.25) 1)

σ

_x

 (0 

²

 

²

 

²

 

(continued)

Possible number of heads

= 0, 1, or 2

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

A probability distribution assigns probability to each of the possible outcomes of a random variable.

Example,



Suppose it would rain exactly once next week. If you’ve scheduled your wedding for next Saturday, what is the probability (chance) it would rain on Saturday?

Probability Distributions

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

Number of days in a week = 7



Prob(random day is Saturday) = 1/7 or 14.3% chance



Same probability or chance for other days.



This probability distribution is uniform: the chances that the random day would fall on any particular day are the same.



The graph of the above probability distribution would be a straight line.

Probability Distributions

Day Sun Mon Tue Wed Thu Fri Sat

Probability 1/7 1/7 1/7 1/7 1/7 1/7 1/7

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Discrete and Continuous

Probability Distributions

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Section Goals

After completing this section, you should be able to:



Apply the binomial distribution to applied problems



Compute probabilities for the Poisson and hypergeometric distributions



Find probabilities using a normal distribution table and apply the normal distribution to business problems



Recognize when to apply the uniform and exponential

distributions

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Probability Distributions

Continuous Probability Distributions Binomial

Hypergeometric Poisson

Probability Distributions Discrete

Probability Distributions

Normal

Uniform

Exponential

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

A discrete random variable is a variable that can assume only a countable number of values

Many possible outcomes:



number of complaints per day



number of TV’s in a household



number of cars passing between 7am and 9am



number of rings before the phone is answered Only two possible outcomes:



gender: male or female



defective: yes or no



Game: Win or loose

Discrete Probability Distributions

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Continuous Probability Distributions



A continuous random variable is a variable that can assume any value on a continuum (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 measure

accurately.

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The Binomial Distribution

Binomial

Hypergeometric Poisson

Probability Distributions Discrete

Probability

Distributions

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The Binomial Distribution



The binomial distribution is used to describe real world phenomena where:



An event results in only two possible outcomes.



The same event is repeated multiple times



It describes the distribution of "success" in a series of trials, i.e., out of

N tries, what is the probability that X of them succeed?

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Binomial Distribution Settings

Examples:



A manufacturing plant labels items as either defective or acceptable



A firm bidding for a contract will either get the 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

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The Binomial Distribution



Characteristics of the Binomial Distribution:

 A trial event has only two possible outcomes – “success” or “failure”, “yes or no”

 There is a fixed number, n, of identical trials

 The trials of the experiment are independent of each other

 The probability of a success, p, remains constant from trial to trial

 If p represents the probability of a success, then (1-p) = q is the probability of a failure

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Counting Rule for Combinations



A combination is the number of ways of picking x unordered outcomes from n possibilities

)!

x n

(

! x

! C

ⁿ_x

n

 

where

:

n! =n(n - 1)(n - 2) . . . (2)(1) x! = x(x - 1)(x - 2) . . . (2)(1) 0! = 1 (by definition)

𝐶

_𝑥^𝑛

is also called the Binomial Coefficient

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Counting Rule for Combinations

p

q p

q

p q

Suppose a coin is tossed four times.

Let probability of obtaining a head be p and a tail be q.

What is the probability of obtaining 4 heads?

𝑃𝑟𝑜𝑏 𝐻 = 4 = 𝑝

⁴

𝑞

⁰

What is the probability of obtaining 3 heads?

𝑃𝑟𝑜𝑏 𝐻 = 3 = 𝑝

³

𝑞

¹

What about 3 tails?

𝑃𝑟𝑜𝑏 𝑇 = 3 = 𝑝

¹

∗ 𝑞

³

In general, if n tosses, probability of x heads is 𝑃𝑟𝑜𝑏 𝐻 = 𝑥 = 𝑝

^𝑥

𝑞

^𝑛−𝑥

Probability tree showing possible outcomes

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P(x) = probability of x successes in n trials,

with probability of success p on each trial x = number of ‘successes’ in sample,

(x = 0, 1, 2, ..., n)

p = probability of “success” per trial q = probability of “failure” = (1 – p) n = number of trials (sample size)

P(x) n

x ! n ! x p

^x

q

ⁿ ^x

( )!

 



Example: Flip a coin four times, let x = # heads:

n = 4 p = 0.5 q = (1 - .5) = .5 x = 0, 1, 2, 3, 4

Binomial Distribution Formula

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n = 5 p = 0.1

n = 5 p = 0.5 Mean

0 .2 .4 .6

0 1 2 3 4 5

X P(X)

.2 .4 .6

0 1 2 3 4 5

X P(X)

0

Binomial Distribution



The shape of the binomial distribution depends on the values of p and n

 Here, n = 5 and p = .1

 Here, n = 5 and p = .5

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Mean

Binomial Distribution

If on the basis of past experience, the store manager of ShopRite at Accra mall estimates that the probability of a customer making a purchase when

he/she enters the store is 0.30, what is the probability that two of the next three customers will make a purchase?

Random variable x = {0, 1, 2, 3}

P(x=2) = ?

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Mean

Binomial Distribution

1. The experiment can be described as a sequence of three identical trials, one trial for each of the three customers who will enter the store.

2. Two outcomes—the customer makes a purchase (success) or the customer does not make a purchase (failure)—are possible for each trial.

3. The probability that the customer will make a purchase (.30) or will not make a purchase (.70) is assumed to be the same for all customers.

4. The purchase decision of each customer is independent of the decisions of the other customers.

𝑃 𝑥 = 2 = 𝐶

₂³

∗ 0.3

²

∗ 0.7

¹

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Mean

Binomial Distribution

At UGBS, 30% of students are majoring in finance.

a. In a sample of 10 UGBS students, what is the probability that exactly three of them are majoring in finance?

b. In a sample of 10 students, what is the probability that at least three students are majoring in finance?

a. 𝑃 𝑥 = 3 = 𝐶₃¹⁰ ∗ 0.3³ ∗ 0.7⁷ = 120 ∗ 0.0022 = 0.2668 b. 1 − 𝑃 𝑥 ≤ 2 = 1 − 𝑃 𝑥 = 0 − 𝑃 𝑥 = 1 − 𝑃 𝑥 = 2

𝑃 𝑥 = 3 = 1 − 𝐶₀¹⁰ ∗ (0.3⁰ ∗ 0.7¹⁰) − 𝐶₁¹⁰ ∗ (0.3¹ ∗ 0.7⁹) − 𝐶₂¹⁰ ∗ (0.3² ∗ 0.7³) = 0.617

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Binomial Distribution Characteristics



Mean



Variance and Standard Deviation

np μ  E(x) 

σ

²

 npq σ  npq

Where n = sample size

p = probability of success

q = (1 – p) = probability of failure

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n = 5 p = 0.1

n = 5 p = 0.5 Mean

0 .2 .4 .6

0 1 2 3 4 5

X P(X)

.2 .4 .6

0 1 2 3 4 5

X P(X)

0

0.5 (5)(.1)

μ  np  

0.6708

.1) (5)(.1)(1

σ npq







2.5 (5)(.5)

μ  np  

1.118 .5) (5)(.5)(1

σ npq







Binomial Characteristics

Examples

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Using Binomial Tables

n = 10

x p=.15 p=.20 p=.25 p=.30 p=.35 p=.40 p=.45 p=.50

0 1 2 3 4 5 6 7 8 9 10

0.1969 0.3474 0.2759 0.1298 0.0401 0.0085 0.0012 0.0001 0.0000 0.0000 0.0000

0.1074 0.2684 0.3020 0.2013 0.0881 0.0264 0.0055 0.0008 0.0001 0.0000 0.0000

0.0563 0.1877 0.2816 0.2503 0.1460 0.0584 0.0162 0.0031 0.0004 0.0000 0.0000

0.0282 0.1211 0.2335 0.2668 0.2001 0.1029 0.0368 0.0090 0.0014 0.0001 0.0000

0.0135 0.0725 0.1757 0.2522 0.2377 0.1536 0.0689 0.0212 0.0043 0.0005 0.0000

0.0060 0.0403 0.1209 0.2150 0.2508 0.2007 0.1115 0.0425 0.0106 0.0016 0.0001

0.0025 0.0207 0.0763 0.1665 0.2384 0.2340 0.1596 0.0746 0.0229 0.0042 0.0003

0.0010 0.0098 0.0439 0.1172 0.2051 0.2461 0.2051 0.1172 0.0439 0.0098 0.0010

10 9 8 7 6 5 4 3 2 1 0

p=.85 p=.80 p=.75 p=.70 p=.65 p=.60 p=.55 p=.50 x

Examples:

n = 10, p = .35, x = 3: P(x = 3|n =10, p = .35) = .2522 n = 10, p = .75, x = 2: P(x = 2|n =10, p = .75) = .0004

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Additional Information about Binomial Distribution

 If p, the probability of a success, is 0.5, the binomial distribution is symmetrical and bell-shaped, regardless of the sample size n.

 When the value of p differs from 0.5 in either direction, the binomial distribution is skewed. The skewness is more pronounced when n is small and p

approaches 0 or 1.

 The binomial distribution however becomes more bell-shaped as n increases.

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The Poisson Distribution

Binomial

Hypergeometric Poisson

Probability Distributions Discrete

Probability

Distributions

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The Poisson Distribution



When the total number of possible outcomes cannot be determined, the binomial distribution cannot be applied



For example, we can count the number of potholes per mile, but we cannot count the number of non- potholes



In such situations, we need a different kind of

probability distribution to describe such natural

phenomenon.

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The Poisson Distribution



Characteristics of the Poisson Distribution:



The outcomes of interest are rare relative to the possible outcomes



The average number of outcomes of interest per unit time or space interval is 



Mean number of occurrences in an interval t is   t



The number of outcomes of interest are random, and the occurrence of one outcome does not influence the chances of another outcome of interest



The probability that an outcome of interest occurs in a given

segment is the same for all segments

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Poisson Distribution Formula

where:

t = size of the segment of interest (say 2 hours) x = number of successes in segment of interest

 = expected number of successes in a segment of unit size ( e.g. 1 hour)

 = mean number of occurrences in an interval

e = base of the natural logarithm system (2.71828...)

t = 𝜇

Example: If 𝜇 = 20 per every 2 hours, then t =2 hours, and  = 10 per hour

!

! ) ) (

( x

e x

e x t

P

x t

x   

 ^ _ ^



(34)

Poisson Distribution Formula

A study conducted at Whole Foods Grocery shows that the average number of arrivals at the checkout section of the store per hour is 16.

Further the distribution for the number of arrivals is considered to be Poisson distributed. What is the probability of x = 12 number of

customers arriving at the checkout in one hour?

𝜇 = 𝜆𝑡 = 16 ∗ 1 = 16 𝑃 𝑥 = 12 = 𝜇

^𝑥

𝑒

^−𝜇

𝑥! = 16

¹²

𝑒

⁻¹⁶

12! = 0.0661

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Example: Mercy Hospital



Using the Poisson Probability Function

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?

𝜇 = λ𝑡 = 6 ∗ 0.5 = 3 𝑝𝑒𝑟 30 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 𝑥 = 4

𝑓 4 = 3

⁴

2.71828

⁻³

4! = 0.1680

(36)

Poisson Distribution Characteristics



Mean



Variance and Standard Deviation

where  = number of successes in a segment of unit size t = the size of the segment of interest

𝜇 = λ𝑡

𝜎 ² = λ𝑡

𝜎 = λ𝑡

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

x 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 0 .1225 .1108 .1003 .0907 .0821 .0743 .0672 .0608 .0550 .0498 1 .2572 .2438 .2306 .2177 .2052 .1931 .1815 .1703 .1596 .1494 2 .2700 .2681 .2652 .2613 .2565 .2510 .2450 .2384 .2314 .2240 3 .1890 .1966 .2033 .2090 .2138 .2176 .2205 .2225 .2237 .2240 4 .0992 .1082 .1169 .1254 .1336 .1414 .1488 .1557 .1622 .1680 5 .0417 .0476 .0538 .0602 ..0668 .0735 .0804 .0872 .0940 .1008 6 .0146 .0174 .0206 .0241 .0278 .0319 .0362 .0407 .0455 .0504 7 .0044 .0055 .0068 .0083 .0099 .0118 .0139 .0163 .0188 .0216 8 .0011 .0015 .0019 .0025 .0031 .0038 .0047 .0057 .0068 .0081

Example: Mercy Hospital



Using the Tables of Poisson Probabilities

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Using Poisson Tables

X

t

0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90

0 1 2 3 4 5 6 7

0.9048 0.0905 0.0045 0.0002 0.0000 0.0000 0.0000 0.0000

0.8187 0.1637 0.0164 0.0011 0.0001 0.0000 0.0000 0.0000

0.7408 0.2222 0.0333 0.0033 0.0003 0.0000 0.0000 0.0000

0.6703 0.2681 0.0536 0.0072 0.0007 0.0001 0.0000 0.0000

0.6065 0.3033 0.0758 0.0126 0.0016 0.0002 0.0000 0.0000

0.5488 0.3293 0.0988 0.0198 0.0030 0.0004 0.0000 0.0000

0.4966 0.3476 0.1217 0.0284 0.0050 0.0007 0.0001 0.0000

0.4493 0.3595 0.1438 0.0383 0.0077 0.0012 0.0002 0.0000

0.4066 0.3659 0.1647 0.0494 0.0111 0.0020 0.0003 0.0000

Example: Find P(x = 2) if  = .05 and t = 100

.0758 2!

e (0.50)

! x

e ) t ) (

2 x

( P

0.50 2

t

x

 

 



^^ ^

(39)

Graph of Poisson Probabilities

0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70

0 1 2 3 4 5 6 7

x

P(x)

X

t = 0.50 0

1 2 3 4 5 6 7

0.6065 0.3033 0.0758 0.0126 0.0016 0.0002 0.0000

0.0000

P(x = 2) = .0758

Graphically:

 = .05 and t = 10

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Poisson Distribution Shape



The shape of the Poisson Distribution depends on the parameters  and t:

0.00 0.05 0.10 0.15 0.20 0.25

1 2 3 4 5 6 7 8 9 10 11 12

x

P(x)

0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70

0 1 2 3 4 5 6 7

x

P(x)

t = 0.50 t = 3.0

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The Hypergeometric Distribution

Binomial Poisson

Probability Distributions Discrete

Probability Distributions

Hypergeometric

(42)

The Hypergeometric Distribution



“n” trials in a sample taken from a finite population of size N



Sample taken without replacement



Trials are dependent



Concerned with finding the probability of “x”

successes in the sample where there are “X”

successes in the population

(43)

Hypergeometric Distribution Formula

N n

X x X

N x n

C

C ) C

x ( P



  ^.

Where

N = Population size

X = number of successes in the population n = sample size

x = number of successes in the sample n – x = number of failures in the sample (Two possible outcomes per trial)

(44)

Hypergeometric Distribution Formula

120 0.3 (6)(6) C

C C

C

C 2) C

P(x

₁₀

3 4 2 6

1 N

n

X x X

N x

n

  



^^

■ Example: 3 Light bulbs were selected from 10. Of the 10 there were 4 defective. What is the probability that 2 of the 3 selected are defective?

N = 10 n = 3

X = 4 x = 2

(45)

The Normal Distribution

Continuous Probability Distributions Probability

Distributions

Normal

Uniform

Exponential

(46)

The Normal Distribution



‘Bell Shaped’



Symmetrical



Mean, Median and Mode are Equal

Location is determined by the mean, μ

Spread is determined by the standard deviation, σ

The random variable has an infinite theoretical range:

+  to  

Mean

= Median

= Mode

x f(x)

μ

σ

(47)

The Normal Distribution Shape

x f(x)

μ σ

Changing μ shifts the distribution left or right.

Changing σ increases or decreases the

spread.

(48)

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

Many Normal Distributions

(49)

Finding Normal Probabilities

Probability is the area under the

curve!

a b ^x

f(x) P ( a  x  b )

Probability is measured by the area

under the curve

(50)

Probability as

Area Under the Curve

The total area under the curve is 1.0, and the curve is symmetric, so half is above the mean, half is below

f(x)

μ x

0.5 0.5

1.0 )

x

P(     

0.5 )

P(μ  x    μ) 0.5

x

P(    

(51)

Empirical Rules

μ ± 1 σ

encloses about 68% of x’s

p(x)

μ μ1σ x μ1σ

What can we say about the distribution of values around the mean? There are some general rules:

σ σ

68.26%

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The Empirical Rule

μ ± 2σ covers about 95% of x’s μ ± 3σ covers about 99.7% of x’s

μ x

2σ 2σ

μ x

3σ 3σ

95.44% 99.72%

(continued)

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Importance of the Rule



If a value is about 2 or more standard

deviations away from the mean in a normal distribution, then it is far from the mean



The chance that a value that far or farther

away from the mean is highly unlikely , given

that particular mean and standard deviation

(54)

Normal Distribution

P(x) =

¹

𝜎 2𝜋

𝑒

^{− 𝑥−𝑢} ²^/2𝜎²

Where u = Mean

𝜎 = standard deviation 𝜋 = 3.14159

𝑒 = 2.71828

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The Standard Normal Distribution



Also known as the “z” distribution



Mean is defined to be 0



Standard Deviation is 1



Possess the same probability distribution as the normal

z f(z)

0 1

Values above the mean have positive z-values,

values below the mean have negative z-values

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Normal to Standard Normal



Any normal distribution (with any mean and

standard deviation combination) can be transformed into the standard normal

distribution (z)



Need to transform x units into z units



Translate from x to the standard normal (the “z” distribution) by subtracting the mean of x and

dividing by its

standard deviation:

σ

μ z x 



(57)

Example



If x is distributed normally with mean of 100 and standard deviation of 50, the z value for x = 250 is



This says that x = 250 is three standard deviations (3 increments of 50 units) above the mean of 100.

50 3.0

100 250

σ μ

z x  

 



(58)

Comparing x and z units

z 100

3.0 0

250 x

Note that the distribution is the same, only the

scale has changed. We can express the problem in original units (x) or in standardized units (z)

μ = 100

σ = 50

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The Standard Normal Distribution

P(z) =

¹

2𝜋

𝑒

^{− 𝑧} ²^/2

Where z = standard normal variable 𝜋 = 3.14159

𝑒 = 2.71828

(60)

The Standard Normal Table

(61)

The Standard Normal Table

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0 2.00 z

.4772

Always,

𝑃 𝑎 ≤ 𝑧 ≤ 𝑏 = 𝑃 𝑧 ≤ 𝑏 − 𝑃(𝑧 ≤ 𝑎)

Example:

𝑃 0 ≤ 𝑧 ≤ 2 = 𝑃 𝑧 ≤ 2 − 𝑃 𝑧 ≤ 0 = 0.9772 − 0.5000 = 0.4772

𝑎 𝑏 z

𝑧

𝑃 𝑎 ≤ 𝑧 ≤ 𝑏

The Standard Normal Table

(63)

General Procedure for Finding Probabilities



Draw the normal curve for the problem in terms of x



Translate x-values to z-values



Use the Standard Normal Table

To find P(a < x < b) when x is distributed

normally:

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Z Table example



Suppose x is normal with mean 8.0 and standard deviation 5.0. Find P(8 < x < 8.6)

P(8 < x < 8.6)

= P(0 < z < 0.12)

Z 0.12

0

x 8.6

8

5 0 8 8

σ μ

z x  

 



5 0.12 8 8.6

σ μ

z x  

 



Calculate z-values:

(65)

Finding Normal Probabilities



Suppose x is normal with mean 8.0 and standard deviation 5.0.



Now Find P(x < 8.6)

Z

8.6 8.0

(66)

0 2.00 z

.4772

Example:

P(0 < z < 2.00)

=P(z<2.00) - P(z<0)

= 0.9772 - 0.5 = .4772

Z-table

(67)



Suppose x is normal with mean 8.0 and standard deviation 5.0.



Now Find P(x < 8.6)

(continued)

Z

0.12 0.00

.5478

P(x < 8.6) = P(z < 0.12) = 0.5478

Z-table

(68)

Upper Tail Probabilities



Suppose x is normal with mean 8.0 and standard deviation 5.0.



Now Find P(x > 8.6)

Z

8.6 8.0

(69)



Now Find P(x > 8.6)…

(continued)

Z

0.12 0

0.5478

1- 0.5478 = 0.4522 P(x > 8.6) = P(z > 0.12) = 1 - P(z < 0.12)

= 1- 0.5478 = 0.4522

Upper Tail Probabilities

(70)

Lower Tail Probabilities



Suppose x is normal with mean 8.0 and standard deviation 5.0.



Now Find P(7.4 < x < 8) = P(-0.12 < z < 0)



P(z < 0) – P(z < -0.12) = 0.5 – 0.4522 = 0.0478

z -0.12

0 x

7.4 8.0

(71)

The Uniform Distribution

Continuous Probability Distributions Probability

Distributions

Normal

Uniform

Exponential

(72)

The Uniform Distribution



The uniform distribution is a

probability distribution that has

equal probabilities for all possible

outcomes of the random variable

(73)

The Continuous Uniform Distribution:

otherwise

0 b x

a a if

b

1  



where

f(x) = value of the density function at any x value a = lower limit of the interval

b = upper limit of the interval

The Uniform Distribution

(continued)

f(x) =

(74)

Uniform Distribution

Example: Uniform Probability Distribution Over the range 2 ≤ x ≤ 6:

2 6

.25

f(x) = = .25 for 2 ≤ x ≤ 6 6 - 2 1

x

f(x)

(75)

The Exponential Distribution

Continuous Probability Distributions Probability

Distributions

Normal

Uniform

Exponential

(76)

The Exponential Distribution



Used to measure the time that elapses

between two occurrences of an event (the time between arrivals)



Examples:



Time between trucks arriving at an unloading dock



Time between transactions at an ATM Machine



Time between phone calls to the main operator

(77)

The Exponential Distribution

λ a

e 1

a) x

P(0     ^



The probability that an arrival time is equal to or less than some specified time a is

where 1/ is the mean time between events

Note that if the number of occurrences per time period is Poisson with mean , then the time between occurrences is exponential with mean time 1/ 

(78)

Exponential Distribution



Shape of the exponential distribution

(continued)

f(x)

x

 = 1.0

(mean = 1.0)

= 0.5

(mean = 2.0)

 = 3.0

(mean = .333)

(79)

Example

Example: Customers arrive at the claims counter at the rate of 15 per hour (Poisson distributed). What is the probability that the arrival time between

consecutive customers is less than five minutes?

 Time between arrivals is exponentially distributed with mean time between arrivals of 4 minutes (15 per 60 minutes, on average)

 1/ = 4.0, so  = .25

 P(x < 5) = 1 - e

^-a

= 1 – e

^-(.25)(5)

= .7135

(80)

Section Summary



Reviewed key discrete distributions



binomial, poisson, hypergeometric



Reviewed key continuous distributions



normal, uniform, exponential



Found probabilities using formulas and tables



Recognized when to apply different distributions



Introduction to Probability Distributions