Introduction to Business Statistics QM 120 Chapter …Chapter 5: Random Variables Random variables ¾Let x denote the number of PCs owned by a family. Then x can take any of the four

DEPARTMENT OF QUANTITATIVE METHODS & INFORMATION SYSTEMS

Introduction to Business StatisticsQM 120

Chapter 5

Discrete random variables and their probability distribution

Dr. Mohammad ZainalSpring 2008

Chapter 5: Random Variables

Random variables

Let x denote the number of PCs owned by a family. Then xcan take any of the four possible values (0, 1, 2, and 3).

A random variable (RV) is a variable whose value isdetermined by the outcome of a random experiment.

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# of PC’s owned Frequency Relative Frequency

0 120 .12

1 180 .18

2 470 .47

3 230 .23

N = 1000 Sum = 1.000

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Discrete random variableA random variable that assumes countable values is called adiscrete random variable.

Examples of discrete RVs:

Number of cars sold at a dealership during a week

Number of houses in a certain block

Number of fish caught on a fishing trip

Number of costumers in a bank at any given day

Continuous random variableA random variable that can assume any value contained inone or more intervals is called a continuous random variable.

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Examples of continuous RVs:Height of a personTime taken to complete a testWeight of a fishPrice of a car

Example: Classify each of the following RVs as discrete or continuous.

The number of new accounts opened at a bank during a weekThe time taken to run a marathonThe price of a meal in fast food restaurantThe score of a football gameThe weight of a parcel

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Chapter 5: Probability Distribution of a Discrete RV

The probability distribution of a discrete RV lists all thepossible value that the RV can assume and theircorresponding probabilities.

Example: Write the probability distribution of the number ofPCs owned by a family.

# of PC’s owned

Frequency Relative Frequency

0 120 .12

1 180 .18

2 470 .47

3 230 .23

N = 1000 Sum = 1.000

X

0

1

2

3

P(x)

.12

.18

.47

.23

ΣP(x) = 1.000

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The following two characteristics must hold for any discrete probability distribution:

The probability assigned to each value of a RV x lies in the range 0to 1; that is 0 ≤ P(x) ≤ 1 for each x.The sum of the probabilities assigned to all possible values of x isequal to 1.0; that is ΣP(x) = 1.

Example: Each of the following tables lists certain values of xand their probabilities. Determine whether or not each tablerepresents a valid probability distribution.


x P(x)

0 .08

1 .11

2 .39

3 .27

x P(x)

0 .25

1 .34

2 .28

3 .13

x P(x)

4 .2

5 .3

6 .6

8 ‐.1

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Example: The following table lists the probability distribution of a discrete RV x.

a) P(x = 3) b) P(x ≤ 2) c) P(x ≥ 4) d) P(1 ≤ x ≤ 4)e) Probability that x assumes a value less than 4f) Probability that x assumes a value greater than 2g) Probability that x assumes a value in the interval 2 to 5

6543210x

.06.09.12.15.28.19.11P(x)


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Example: For the following table

a) Construct a probability distribution table. Draw a graph of the probability distribution.

b) Find the following probabilitiesi. P(x = 3) ii. P(x < 4) iii. P(x ≥ 3) iv. P(2 ≤ x ≤ 4)P(x = 3) = .3 P(x < 4) = .1 + .25 + .3 = .65P(x ≥ 3) = .3 + .2 + .15 = .65 P(2 ≤ x ≤ 4) = .25 + .3 + .2 = .75


0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

1 2 3 4 5

54321x

121624208f

x f

1 8

2 20

3 24

4 16

5 12

P(x)

.1

.25

.3

.2

.15

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Mean of a discrete RV

The mean μ ‐or expected value E(x)‐ of a discrete RV is thevalue that you would expect to observe on average if theexperiment is repeated again and again

It is denoted by

Illustration: Let us toss two fair coins, and let x denote thenumber of heads observed. We should have the followingprobability distribution table

Suppose we repeat the experiment a large number of times,say n =4,000. We should expect to have approximately

( ) ( )E x xp x=∑

210x

1/41/21/4P(x)

Chapter 5: Mean of a discrete RV

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1 thousand zeros, 2 thousand ones, and 1 thousand twos. Then the average value of x would equal

Similarly, if we use the , we would have

Sum of measurements 1,000

(0) 2,000(1) 1000(2)4,000

1 1 1(0) (1) (2)4 4

2

n+ +

=

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

( ) ( )E x xp x=∑( ) 0 (0) 1 (1) 2 (2)

0(1/ 4) 1(1/ 2) 2(1/ 4)1

E x P P P= + += + +=


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Example: Recall “ number of PC’s owned by a family” example. Find the mean number of PCs owned by a family.

Thus, on average, we expect to see 1.81 PC owned by a family!


x P(x)

0 .12

1 .18

2 .47

3 .23

We need to find x.p(x)for each value of x and

then add them up together

x P(x)

0 .12

1 .18

2 .47

3 .23

x.P(x)

0(.12) = 0.00

1(.18) = 0.18

2(.47) = 0.94

3(.23) = 0.69

ΣxP(x) = 0.00 + 0.18 + 0.94

0.69 =1.81

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Example: In a lottery conducted to benefit the local firecompany, 8000 tickets are to be sold at $5 each. The prize is a$12,000. If you purchase two tickets, what is your expectedgain?

Your gain is:o Either the $12,000 but don’t forget the ticket price youalready paid

oor you buy the ticket for a $10 but get nothing

Expected gain = E(x) = ‐ $7, that is to lose 7 dollars


x P(x)

12000 ‐ 10 2/8000

‐10 7998/8000

Sum

x.P(x)

2.9975

‐9.9975

‐7


Example: Determine the annual premium for a $1000insurance policy covering an event that over a long period oftime, has occurred at the rate of 2 times in 100. Letx : the yearly financial gain to the insurance companyresulting from the sale of the policyC : unknown premiumCalculate the value of C such that the expected gain E(x) willequal to zero so that the company can add the administrativecosts and profit.

E(x) = 0 =.98C + .02C – 20→ C =$20, the insurance company shouldask a minimum of $20 and add the administrative costs and profitto that amount.

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x P(x)

C 98/100

C‐1000 2/100

x.P(x)

0.98C

0.02C‐20


Example: You can insure a $50,000 diamond for its total valueby paying a premium of D dollars. If the probability of theft ina given year is estimated to be .01, what premium should theinsurance company charge if it wants the expected gain toequal $1000.

E(x) = 1000 =.99D + .01D – 500 → D =$1500, the insurance companyshould ask for that amount as a premium

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x P(x)

D .99

D‐50,000 .01

x.P(x)

0.99D

0.01D‐500

Chapter 5: Standard Deviation of a Discrete RV

The standard deviation of a discrete RV x, denoted by σ,measures the spread of its probability distribution.

A higher value of σ indicates that x can assume values overa larger range about the mean. While, a smaller valueindicates that most of the values that can x assume areclustered closely around the mean.

The standard deviation σ can be found using the followingformula:

Hence, the variance σ2 can be obtained by squaring itsstandard deviation σ.

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2 2( )x p xσ μ= −∑


Example: Recall “ number of PC’s owned by a family”example. Find the standard deviation of PCs owned by afamily.

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x P(x)

0 .12

1 .18

2 .47

3 .23

x P(x)

0 .12

1 .18

2 .47

3 .23

x.P(x)

0(.12) = 0.00

1(.18) = 0.18

2(.47) = 0.94

3(.23) = 0.69

Here we need to find two columns x.p(x) and

x2.p(x)

Σx2P(x) = 0.00 + 0.18 + 1.88+ 2.07 =4.13

x2.P(x)

0(.00) = 0.00

1(.18) = 0.18

2(.94) = 1.88

3(.69) =2.07

2 24.13 1.81 .8539 .8539 .924σ σ= − = ⇒ = =


Example: An electronic store sells a particular model of a computernotebook. There are only four notebooks in stocks, and the managerwonders what today’s demand for this particular model will be. Shelearns from marketing department that the probability distributionfor x, the daily demand for the laptop, is as shown in the table.

Find the mean, variance, and the standard deviation of x. Is it likelythat five or more costumers will want to buy a laptop?Solution:

μ= 1.9σ= 1.34P(X ≥ 5) = 0.05It means it is unlikely the 5or more costumers willwant to buy laptops at anygiven day

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543210x

.05.10.15.20.40.10P(x)

x P(x)

0 .10

1 .40

2 .20

3 .15

4 .10

5 .05

x.P(x)

0.00

0.40

0.40

0.45

0.40

0.25

x2.P(x)

0.00

0.40

0.80

1.35

1.60

1.25


Example: A farmer will earn a profit of $30,000 in case ofheavy rain next year, $60,000 in case of moderate rain, and$15,000 in case of little rain. A meteorologist forecasts that theprobability is .35 for heavy rain, .40 for moderate rain, and .25for little rain next year. Let x be the RV that represents nextyear’s profits in thousands of dollars for this farmer. Write theprobability distribution of x. Find the mean, variance, and thestandard deviation of x.

μ= 38.250→ $ 38,250

σ= 18.66→ $ 18,660

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x P(x) xP(x) x2P(x)30 0.35 10.50 315.0060 0.40 24.00 1440.0015 0.25 3.75 56.25

Sum 38.25 1811.25


Example: An instant lottery ticket costs $2. Out of a total of10,000 tickets printed for this lottery, 1000 tickets contain aprize of $5 each. 100 tickets contain a prize of $10 each, 5tickets contain a prize of $1000 each, and 1 ticket has the prizeof $5000. Let x be the RV that denotes the net amount playerwins by playing this lottery. Write the probability distributionof x. Determine the mean and the standard deviation of x.

μ= ‐$0.4→ Lose $ 0.4

Thus, on the average, the players who play the game are expectedto lose $ 0.4 per ticket.

σ= $ 54.78

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x P(x) xP(x) x2P(x)5-2 0.1 0.3 0.910-2 0.01 0.08 0.64

1000-2 0.0005 0.499 498.0025000-2 0.0001 0.4998 2498

-2 0.8894 -1.7788 3.5576Sum -0.4 3001.1


Example: Based in its analysis of future demand for itsproducts, the financial department a company has determinedthat there is a .17 probability that the company will lose $1.2million during the next year, a .21 probability that it will lose$.7 million, a .37 probability that it will make a profit of $0.9million, and a .25 probability that it will make a profit of $2.3million.a) Let x be a RV that denotes the profit earned by thiscompany during the next year. Write the probabilitydistribution of x.

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x P(x) xP(x) x2P(x)

-1.2 0.17 -0.204 0.2448

-0.7 0.21 -0.147 0.1029

0.9 0.37 0.333 0.2997

2.3 0.25 0.575 1.3225

Sum 0.557 1.9699


b) Find the mean and standard deviation of the probability ofpart a.

μ = $ 557,000

σ= $ 1.288 million

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Chapter 5: Factorials & Combinations

FactorialsThe symbol n!, reads as “n factorial,” represents theproduct of all integers from n to 1. In other words,

n! = n(n ‐ 1)(n ‐ 2)(n ‐ 3)…3.2.1

Example: Evaluate 7!7! = 7.6.5.4.3.2.1 = 5040Example: Evaluate (12‐4)!(12‐4)! = (8)! = 8.7.6.5.4.3.2.1 =40,320

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Example: Evaluate (8‐8)!(8‐8)! = (0)! = 1

Example: Find the value of 8!By using factorial Table or your calculator.Locate 8 in the column labeled n.Then read the value for n! nextto 8.

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CombinationsCombinations give the number of ways x elements can beselected from n elements. The notation used to denote thetotal number of combinations is

It can be found using the following formula

Remember: n is always greater than or at least equal to x.

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)!(!!

xnxnCC xn

nx −

==

)(nx

nxxn CC ==


Example: Three members of a jury will be randomly selectedfrom five people. How many different combinations arepossible?

We have total of n = 5 and we are to select r = 3 of them. Hence,

5C3 = 5!/[3! (5‐3)!] = 5!/(3! . 2!) = 120/(6 . 2) = 10

If we assume their names are A, B, C, D, and E. Then the 10combinations of the jury are

ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE

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Documents

Introduction to Business Statistics QM 120 Chapter …Chapter 5: Random Variables Random variables ¾Let x denote the number of PCs owned by a family. Then x can take any of the four