Title: Random Variable
1 Random Variable
2Random variable
- A random variable ? is a function (rule) that
assigns a number to each outcome of a chance
experiment. - A function ? acts on the elements of its domain
(the sample space) associating each element with
a unique real number. The set of all values
assigned by the random variable is the range of
this function. If that set is finite, then the
random variable is said to be finite discrete. If
that set is infinite but can be written as a
sequence, then the random variable is said to be
infinite discrete. If that set is an interval
then the random variable is said to be
continuous.
3Example (1)
- A coin is tossed 3 times. Let the random variable
? denotes the number of heads that occur in 3
tosses. - 1. List the outcomes of the experiment (Find the
domain of the function ?) - Answer
- The outcomes of the experiments are those of the
sample space - S HHH, HHT, HTH, THH, HTT, THT, TTH, TTT
- the domain of the function (random variable) ?
- 2. Find the value assigned to each outcome by ?
- See the table on the following slide.
4Value of ? Outcome
3 HHH
2 HHT
2 HTH
2 THH
1 HTT
1 THT
1 TTH
0 TTT
5- 3. Let E be the event containing the outcomes to
which a value of 2 has been assigned by ?. Find
E! - Answer
- E HHT, HTH, THH
- 4. Is ? finite discrete, infinite discrete or
continuous? - Answer
- The set of the values assigned by ? is
- 0, 1, 2, 3, which is a finite set, hence the
random variable is ? finite discrete
6Example (2)
- A coin is tossed repeatedly until a head occurs.
Let the random variable ? denotes the number coin
tosses in the experiment - 1. List the outcomes of the experiment (Find the
domain of the function ?) - Answer
- The outcomes of the experiments are those of the
sample space - S H, TH, TTH, TTTH, TTTTH, TTTTTH, TTTTTTH,
. - the domain of the function (random variable) ?
- 2. Find the value assigned to each outcome by ?
- See the table on the following slide which shows
some of these values.
7Value of ? Outcome
1 H
2 TH
3 TTH
4 TTTH
5 TTTTH
6 TTTTTH
7 TTTTTTH
.. .Etc etc
8- 3. Is ? finite discrete, infinite discrete or
continuous? - Answer
- The set of the values assigned by ? is
- 1, 2, 3, 4, 5, 6, 7, 8, 9, .., which is a
infinite set, that can written as a sequence,
hence the random variable is ? is infinite
discrete
9Example (3)
- A flashlight is turned on until the battery runs
out. Let the Radom variable ? denote the length
(time) of the life of the battery. What values
may ? assume ? Is ? finite discrete, infinite
discrete or continuous? - Answer
- The set of possible values is the interval 0,8),
and hence the random variable ? is continuous. -
10 Probability Distribution of Random variable
11Probability Distribution of Random variable
- A table showing the probability distribution
associated with the random variable ( which is
associated with the experiment) rather than with
the outcomes (which are related to the random
variable)
12Example (4)
- A coin is tossed 3 times. Let the random variable
? denotes the number of heads that occur in 3
tosses. - Show the probability distribution of the random
variable associated with the experiment.
13Value of ? Outcome
3 HHH
2 HHT
2 HTH
2 THH
1 HTT
1 THT
1 TTH
0 TTT
14probability distribution of the random variable X
P(Xx) Value of ?
1/8 0
3/8 1
3/8 2
1/8 3
15Example (5)
- Two dice are rolled. Let the random variable ?
denotes the sum of the numbers on the faces that
fall uppermost. - Show the probability distribution of the random
variable X. - Answer
- The values assumed by the random variable X are
2, 3, 4, 5, 6,.,12, corresponding to the events
E2, E3, E4,.,E12. The probabilities associated
with the random variable X corresponding to 2, 3,
4, ,12 are the probabilities p(E2), p(E3),,
p(E4), ., p(E12)..
16Sum ofuppermost numbers Event Ek
(1 ,1) 2
(1,2) , (2 ,1) 3
(1,3) , (2 , 2) , (3 ,1) 4
(1, 4) , (2 , 3) , (3 , 2) , (4 ,1) 5
(1 , 5 ) , (2 , 4) , (3 , 3) , (4 , 2) , (5 ,1) 6
(1, 6) , (2 , 5) , (3,4) , (4 , 3) , (5 , 2) , (6 ,1) 7
(2 , 6) , (3 , 5) , (4 , 4) , (5 ,1) , (6 , 2) 8
(3 , 6) , (4 , 5) , (5 , 4) , (6 , 3) 9
(4 , 6) , (5 , 5) , (6 , 4) 10
(5 , 6) , (6 , 5) 11
(6,6) 12
17Probability Distribution of the Random Variable X
p(Xx) x
1/36 2
2/36 3
3/36 4
4/36 5
5/36 6
6/36 7
5/36 8
4/36 9
3/36 10
2/36 11
1/36 12
18Example (6)
- The following table shows the number of cars
observed waiting in line at the beginning of 2
minutes interval from 10.00 am to 12.00 noon at
the drive-in ATM of QNB branch in Ghrafa and the
corresponding frequency of occurrence. Show the
probability distribution table of the random
variable x denoting the number of cars observed
waiting in line.
Frequency of Occurrence Cars
2 0
9 1
16 2
12 3
8 4
6 5
4 6
2 7
1 8
19- Dividing the frequency by 60 (the sum of these
numbers- the ones indicating frequency), we get
the respective probability associated with random
variable X. - Thus
- P(X0) 2/60 1/30 0.03
- P(X1) 9/60 3/20 0.15
- P(X2) 16/60 1/10 0.1
- .etc
- See the opposite table
Frequency of Occurrence p(X x) Cars x
2/60 0.03 0
9/60 0.15 1
16/60 0.27 2
12/60 0.20 3
8/60 0.13 4
6/60 0.10 5
4/60 0.07 6
2/60 0.03 7
1/60 0.02 8
20Bar Charts ( Histograms )
- A bar chart or histogram is a graphical means of
exhibiting probability distribution of a random
variable. - For a given probability distribution, a histogram
is constructed as follows - 1. Locate the values of the random variable on
the number line (x-axis) - 2. Above each number (value) erect a rectangle of
width 1 and height equal to the probability
asociated with that number (value).
21Remarks
- 1. The area of rectangle associated with the
value of the random variable x - The probability associated with the value x
(notice tat the width of the rectangle is 1) - 2. The probability associated with more than one
value of the random variable x is given by the
sum of the areas of the rectangles associated
with those values.
22Example
- Back to the example of three tosses of the coin
23Value of ? Outcome
3 HHH
2 HHT
2 HTH
2 THH
1 HTT
1 THT
1 TTH
0 TTT
24Probability Distribution
x p(Xx)
0 1/8
1 3/8
2 3/8
3 1/8
Value of ? Outcome
3 HHH
2 HHT
2 HTH
2 THH
1 HTT
1 THT
1 TTH
0 TTT
253
1
2
26- The event F of obtaining at least two heads
- F HHH, HHT, HTH, THH
- P(F) 4/8 ½
- The same result can be obtained from the
following reasoning - F is the event (X2) or (X3)
- The probability of this event is equal to
- p(X2) p(X3)
- the sum of the areas of the rectangles
associated with the values 2 and 3 of the random
variable - 1(3/8) 1(1/8)
- 4/8 ½
- on the following slide, these areas are
identified (the shaded areas)
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28 Expected Value of a Random Variable
29Mean (Average)
- The average (mean) of the numbers a1, a2,
- a3,. an is denoted by a and equal to
- (a1a2a3.an ) / n
- Example
- Find the average of the following numbers
- 5, 7, 9, 11, 13
- Answer The average of the given numbers ie equal
to ( 5791113) / 5 45 / 5 9
30Expected Value of a Random Variable( the average
or the mean of the random variable)
- Let x be a random variable that assumes the
values x1, x2, x3,. xn with associated
probabilities p1, p2, p3,. pn respectively.
Then the expected value of x denoted by E(x) is
equal to - x1 p1 x2 p2 x3 p3 xn pn
31- The the expected value of x ( the average or the
mean of X) is a measure of central tendency of
the probability distribution associated with X. - As the number of the trails of an experiment gets
larger and larger the values of x gets closer and
closer to the expected value of x. - Geometric Interpretation
- Consider the histogram of the probability
distribution associated with the random variable
X. If a laminate (thin board or sheet) is made of
this histogram, then the expected value of X
corresponds to the point on the base of the
laminate at which the laminate will balance
perfectly when the point is directly over the a
fulcrum (balancing object).
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33Example (7)
- Let X be the random variable giving the sum of
the dots on the faces that fall uppermost in the
two dice rolling experiment. Find the expected
value E(X) of X.
34Recall the probability distribution of the this
random variable
P(Xx) x
1/36 2
2/36 3
3/36 4
4/36 5
5/36 6
6/36 7
5/36 8
4/36 9
3/36 10
2/36 11
1/36 12
35- Solution
- E(X)
- 2(1/36) 3(2/36) 4(3/36) 5(4/36) 6(5/36)
7(6/36) 8(5/36) 9(4/36) 10(3/36)
11(2/36) 12(1/36) - (26122030424036302212) / 36 252 / 36
- 7
- Inspecting the histogram on the next slide, we
notice that the symmetry of the histogram with
respect to the vertical line x 7, which is the
expected value of the random variable X.
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37Example (8)
- The occupancy rates with corresponding
probability of hotels A (Which has 52 rooms) B
(which has 60 rooms) during the tourist season
are given by the tables on the next slide. The
average profit per day for each occupied room is
QR 200 and QR 180 for hotels A B respectively.
Find - 1. The average number of rooms occupied per day
in each hotel. - 2. Which hotel generates the higher daily profit.
-
38Hotel B
Occupancy Rate Probability
0.75 0.35
0.80 0.21
0.85 0.18
0.90 0.15
0.95 0.09
1.00 0.02
Hotel A
Occupancy Rate Probability
0.80 0.19
0.85 0.22
0.90 0.31
0.95 0.23
1.00 0.05
39Steps
- I. For each hotel
- 1. Find the expected value of the random
variable defined to be the occupancy rate in the
hotel. - 2. Multiply that by the number of rooms of the
hotel to find the average number of rooms
occupied per day. - 3. Multiply that by the profit made on each room
per day to find the hotel daily profit. - II. Compare the result of step 3., for hotels A
B.
40Solution
- 1. Let the occupancy rate in Hotels A B be the
random numbers X Y respectively. Then the
average daily occupancy rate is given by the
expected values E(X) and E(Y) of X and Y
respectively. Thus - E(X) (0.80)(0.19) (0.85)(0.22) (0.90)(0.31)
(0.95)(0.23) (1.00)(0.05) 0.8865 - The average number of rooms occupied per day in
Hotel A - 0.8865 (52) 46.1 rooms
- E(Y) (0.75)(0.35) (0.80)().21) (0.85)(0.18)
(0.90)(0.15) (0.95)(0.09) (1.00)(0.02)
0.8240 - The average number of rooms occupied per day in
Hotel A - 0.8240 (60) 49.4 rooms
41- 2. The expected daily profit at hotel A
- (46.1 )(200) 9220
- The expected daily profit at hotel B
- (49.4 )(180) 8890
- ? hotel A generates a higher profit