Download Random Variable Formulas

Random Variables - Terms and Formulas
Random Variable – a variable whose value is a numerical outcome of a random phenomenon.
Example 1: Put all the letters of the alphabet in a hat. If you choose a consonant, I pay you $1. If you choose a
vowel, I pay you $5. X is the random variable representing the outcome of the experiment. Its possible values
are 1 and 5.
Example 2: In most college courses, you get as a grade, either an A, B, C, D, or F. For credit purposes, A’s are
given 4 points, B’s are given 3 points, C’s are given 2 points, Ds are given 1 point, and F’s are no points. Let X
be the random variable representing the points a student gets. The possible values of X are 4, 3, 2, 1, and 0.
Discrete Random Variables have a countable number of possible values. The probability distribution of X
lists all possible values of X and their probabilities:
Value of X
Probability of X
x1
p1
x2
p2
…
…
x3
p3
The probabilities must satisfy two requirements:
!
!
!
!
!
Every probability p is a number !
between 0 and 1.
i
xn
pn
!
!
p1 + p2 + p3 + ...+ pn = 1
Example 3: Put all the letters of the alphabet in a hat. If you choose a consonant, I pay you $1. If you choose a
vowel, I pay you!$5. X is the random variable representing the outcome of the experiment. Create the
! distribution of X.
Example 4: A college instructor teaching a large class traditionally gives 10% A’s, 20% B’s, 45% C’s, 15%
D’s, and 10% F’s. If a student is chosen at random from the class, the student’s grade on a 4-point scale (A = 4)
is a random variable X. Create the distribution of X.
What is the probability that a student has a grade point of 3 or better in this class?
What is the probability that a student has a grade point of 2 or worse in this class?
Draw a probability histogram to picture the probability distribution of the random variable X.
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Continuous Random Variables take on all possible values in an interval of numbers. The probability
distribution of X is described by a density curve. The probability of any event is the area under the density
curve and above the values of X that makes up that area.
Example 5: A random number is chosen from 0 to 1. Answer the following questions:
a. Find P(X > .5)
b. Find P(X ≥ .5)
c. Find P(X ≤ .8)
d. Find P(X ≥ .15)
e. Find P(.2 <X < .4)
f. Find P(.2 <X ≤ .4)
g. Find P(X < 1)
h. Find P(X = .5)
Mean of a Random Variable – if an experiment with random variable X is done over a long period of time, we
can calculate the mean (average) value of that random variable. Another term for mean of a random variable X
is the expected value of X.
If X is a discrete random variable whose distribution is given by
Value of X
Probability of X
x1
p1
x2
p2
…
…
x3
p3
xn
pn
To find the mean (or expected value) of X, multiply every possible value of X by its probability
!
!
!
!
Then add the results. The symbol for the mean of X is µX .
!
!
!
!
µX = x1 p1 + x 2 p2 + x 3 p3 + ... + x n pn = " x i pi
!
Example 6: A fair coin is flipped 3 times. Find the mean of the discrete random variable X that counts the
number of heads.
!
Example 7: The daily lottery costs $1 to play. You pick a 3 digit number. If you win, you win $500. Find the
expected value of the lottery.
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Law of Large Numbers: Draw independent observations from any population with finite mean µ . As the
number of observations increases, the mean x of the observed values eventually approaches the mean µ of that
population.
! will approach 50%.
Example 8: If I flip a fair coin, over a long period of time, the percentage of heads
!
That does not mean that the number of heads will approach the number of tails. Here!is a typical
experiment in tossing coins.
Heads
7
Tails
3
Percentage of heads 70%
42
58
42%
463
537
46.3%
4,875
5,125
48.75%
49,660
50,340
49.66%
Example 9: If you play blackjack perfectly at a casino, the expected value for every dollar played is
98.5 cents. That means on the average of hand you play, you lose 1.5 cents. Can an individual player
win playing blackjack at a casino. The “law of small numbers” says that over a short run, a player can
certainly win. But over the long run, the law of large numbers says that a player will eventually lose.
And given the huge number of players who play daily, the law of large numbers states that the casino
has to win. There is a reason that casinos look the way they do – it is guaranteed money. Still, players
can win over the short haul. The question that crops up is what is considered “large?” There is no
answer to that. What you have to know is that the larger n (the number of trials) gets, the closer the
average win (or loss) per play approaches the true average of losing 1.5 cents a play.
Rules for Means
Rule 1: If X and Y are random variables, then µX +Y = µX + µY
Rule 2: If X is a random variable and a and b are constants, then µa +bX = a + bµX
!
Example 10: You play two casino games. Game 1 has an expectation of losing $1 a play and game 2 has an
expectation of losing $2 a play. If you play both games, what is your expectation for both games.
!
Example 11: You go to a casino and play the same slot machine which averages losing 15 cents a play. You
play the machine 100 times and then leave, paying $5 for parking. Find your expectation for your casino visit.
Example 12: Depending on the attendance of a minor league baseball team, the number of hot dogs and sodas
sold at a game is given by the following table.
Hot dogs sold
Probability
100
.05
200
.1
500
.3
1,000 2,500
.35
.2
Sodas sold
Probability
100 200
.1
.15
400
.25
800
.3
1,500
.2
If the profit on a hot dog is $1.50 and the profit on a soda is $0.75, find the average profit on hot dogs and sodas
per game.
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Stu Schwartz
Variance and standard deviation of a Discrete Random Variable
If X is a discrete random variable whose distribution is given by
Value of X
Probability of X
x1
p1
x2
p2
…
…
x3
p3
xn
pn
We know that the mean (expected value) of X (µX ) is given by
!
!
!
!
!
!µX = x1 p!1 + x 2 p2 + x 3 p3 + ...! + x n pn = " x i pi
To find the variance of X (" 2 X ) , we !
employ this formula:
2
2
2
2
"!2 X = ( x1 # µX ) p1 + ( x 2 # µX ) p2 + ( x 3 # µX ) p3 + ...+ ( x n # µX ) pn
To find the standard
! deviation of X (" X ) , we know that
(" X ) = " 2 X
!
Example 13: A fair coin is flipped
! 3 times. Find the standard deviation of the discrete random variable X that
counts the number of heads.
!
Example 14: The daily lottery costs $1 to play. You pick a 3 digit number. If you win, you win $500. Find the
standard deviation of the lottery.
Example 15: Choose an American household at random and let the random variable X be the number of
persons living in the household. Find the standard deviation of the average American household.
X
Probability
1
.25
2
.32
3
.17
4
.15
5
.07
6
.03
7 or more
.01
When a problem has many values of X like this one, this can be done using your list feature of your calculator:
1. Input the data into Lists L1 and L2.
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2. To get µX , use L3 to multiply L1 and L2.
3. To find µX , find the sum of L3.
!
!
4. We no longer need L3. In L3, we will put our formula for variance:
5. To find the variance " 2 X , find the sum of L3. The standard deviation is the square root.
Rules for Variances of Random Variables:
!
Rule 1: If X is a random variable and a and b are constants, then
Variance
" 2 a +bX = b 2" 2 X
Standard Deviation
" a +bX = b" X
Rule 2: If X and Y are independent random variables, then
!
!
!
!
Variance
" 2 X +Y = " 2 X + " 2Y
Standard Deviation
" X +Y = " 2 X + " 2Y (Note that you don’t add standard deviations)
" 2 X #Y = " 2 X + " 2Y
" X #Y = " 2 X + " 2Y
(Note that even though you are finding the variance of the difference between X and Y, you add the
variances).
!
!
Example 16: Suppose two pro bowlers Adam (A) and Bart (B) have the following distribution of scores.
µA = 209
" A = 14
µB = 221
" A = 22
Find the following.
!
a) µB +A
b) µB "A
c) 3" A
d) " B +A
e) " B #A
Example 17: Depending on the attendance of a minor league baseball team, the number of hot dogs and sodas
! at a game is given!by the following table.!
!
!
sold
Hot dogs sold
Probability
100
.05
200
.1
500
.3
1,000 2,500
.35
.2
Sodas sold
Probability
100 200
.1
.15
400
.25
800
.3
1,500
.2
Find the standard deviation of the number of hot dogs plus the number of sodas sold.
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Discrete Random Variables
1. Choose an American household at random and let the random variable X be the number of persons living in
the household. The probability of X is:
X
Probability
a. Find P(X > 4)
1
.25
2
.32
3
.17
4
.15
b. Find P(X ≥ 4)
5
.07
6
.03
c. Find P(2 <X ≤ 4)
7 or more
.01
d. Find P(X ≠ 2)
2. A fair coin is flipped 4 times. Find the probability distribution of the discrete random variable X that counts
the number of heads. Then draw the probability histogram for X.
a. Find P(X > 2)
b. Find P(X ≥ 2)
c. Find P(X ≥ 1)
d. Find P(X ≥ 0)
3. A coin that is rigged is flipped 3 times. The probability of heads is twice the probability of tails. Find the
probability distribution of the discrete random variable X that counts the number of heads. Then draw the
probability histogram for X.
a. Find P(X > 2)
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b. Find P(X ≥ 2)
c. Find P(X ≤ 3)
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d. Find P(X ≠ 2)
Stu Schwartz
4. The numbers 1, 2, 3, 4, 5 are placed in a hat. A number is chosen at random, replaced, and another number is
chosen at random. Let X be the number of odd numbers that are chosen. Find the probability distribution of
the random variable X and draw a probability histogram.
a. Find P(X > 1)
b. Find P(X ≥ 1)
c. Find P(X ≤ 1)
d. Find P(X ≠ 2)
5. An SRS of 3 students are chosen to be on a committee that advises a university on its computer policies.
70% of the student population are PC users and 30% are Mac users. Let X be the number of PC users on the
student committee. Find the probability distribution of the random variable X and draw a probability
histogram.
a. Find P(X > 2)
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b. Find P(X ≥ 2)
c. Find P(X ≤ 1)
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d. Find P(X ≠ 3)
Stu Schwartz
6. A boss of a company chooses 4 workers at random to act as advisers to him. The company is 75% white and
25% African American. Let X be the number of African-Americans on the committee. Find the probability
distribution of the random variable X and draw a probability histogram.
a. Find P(X > 2)
b. Find P(X < 2)
c. Find P(1 ≤ X ≤ 3)
d. Find P(X ≠ 0)
7. A density curve is shown on the figure on the right. Find the
following probabilities.
a. Find P(X > 1)
b. Find P(X < 1.5)
c. Find P(.5 ≤ X ≤ 1.5)
d. Find P(X <2)
8. Two random numbers from 0 to 1 are chosen and added. Let X be
their sum. X is thus a continuous random variable between 0 and
2. The density curve is shown to the right. Find the following
probabilities:
a. Find P(X > 1)
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b. Find P(X > 1.5)
c. Find P(X < .2)
-8-
d. Find P(.5 ≤ X ≤ 1.5)
Stu Schwartz
Mean of a Random Variable (Expected Value) Problems
1) Choose an American suburban household at random and let the random variable X be the number of cars
that people in the house own. Find the number of cars owned by the average suburban American household.
X
Probability
0
.08
1
.29
2
.32
3
.17
4
.08
5
.04
6 or more
.02
2) A huge cookie jar has 60% chocolate cookies and 40% vanilla cookies. Sam chooses 3 cookies blindfolded.
Let X be the number of chocolate cookies he chooses. Construct the probability distribution of X and the
average number of chocolate cookies he chooses.
7
3) A contractor is bidding on a road construction job that promises a profit of $20,000 with a probability of 10
and a loss, due to strikes, weather conditions, late arrival of building materials, and so on, of $40,000 with a
3
probability of 10 . What is the contractor’s mean expectation?
4 ) Pete selects one card from a deck of 52 cards. If it is an ace, he wins $5. If it is a club, he wins only $1.
However, if it is the ace of clubs, then he wins an extra $10. What is his mathematical expectation? Suppose
it costs him $1 to play the game. What could he expect to win or lose per game? Suppose he plays 100
games. How much would he be expected to win or lose?
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5) A box contains 8 green marbles, 7 yellow marbles, and 5 black marbles. One marble is to be selected from
the box. You get $10 if the marble selected is black, but lose $3 if the marble is green and lose $5 if the
marble is yellow. What is your mean expectation?
6) A coin is tossed three times. If heads appears on all 3 tosses, Mary will win $16. If heads appears on 2 of the
tosses, she will win $2. The game costs $5 to play. What is her mean expectation?
7) A carnival has set up the following game. You are blindfolded and have to pick a coin from 9 pennies, 8
nickels, 12 dimes, 16 quarters, 11 half-dollars, and 4 dollar pieces. It costs a quarter to play the game. What
could you be expected to win or lose every time you play the game?
8) A pair of dice is rolled. Depending on the sum of the dots on both dice, Jake can win or lose money as
shown in the following chart. What is Jake’s mean expectation?
Sum of Dots
6 or 8 or 9
2 or 4 or 5
10
3 or 12
7 or 11
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Outcome
win $6
lose $4
win $7
win $5
lose $1
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Stu Schwartz
9) An oil company will only invest in an oil well if the can expect to make at least $1 million in profit. They
find a possible area in which to drill in Canada. It will cost the company $3 million to make the attempt. If
they find oil, they can expect to clear $7 million in profits. Geologists have estimated that the probability of
striking oil is 0.35. Should the company make the attempt and why or why not?
10) A wheel of fortune below costs $2 to play. If the spinner stops on black or
blue after one spin, the prize is $5. If it stops on green, the prize is $3. If it
stops on yellow, the prize is $0.50. If it stops on brown or red, there is no
payoff. What is your mean expectation every time you spin the wheel.
Should you play the game?
Black Green
Yellow
Blue
Red
Brown
11) In a raffle costing $1 a ticket, 225 tickets are sold. First place gets $50. 2 second place tickets gets $30 and 3
third place tickets gets $20. What is the mean value of your return on a ticket?
12) An ecologist collected the data shown in the table below on the life span of a certain species of deer. Based
on this sample, what is the expected lifespan of this species?
Age at death (years)
Number
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1
2
2
30
3
86
4
132
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5
173
6
77
7
40
8
10
Stu Schwartz
13) A drug is administered to sets of three patients. Over a period of time, it is determined that the probability of
3 cures, 2 cures, 1 cure, and no cures are .70, .20, .09, and .01 respectively. What is the mean number of
cures that can be expected in a group of three?
14) A man on his 64th birthday obtains a $1,000 one-year insurance policy at a cost of $50. Based on mortality
tables, there is a .963 probability that the man will live for at least one more year. How much can the
insurance company expect to earn on this policy?
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Mean and Standard Deviation of a Random Variable
Randy and Dan are businessmen who spend time on the road weekly (Mon-Fri). They always eat dinner out.
Randy spends money for dinner according to the following probability chart.
X
$15
$20
$25
$30
P(X)
.2
.4
.25
.15
Dan spends money for dinner according to the following probability chart.
X
$20
$25
$30
$35
P(X)
.35
.25
.3
.1
1. Find the average amount of money Randy spends.
2. Find the average amount of money Dan spends.
3. Find the standard deviation of the amount of money Randy spends.
4. Find the standard deviation of the amount of money Dan spends.
5. Find the average amount of money they spend for dinner together.
6. Find the standard deviation of the total money they spend for dinner together.
7. Find the average difference in money they spend for dinner.
8. Find the standard deviation of the difference they spend for dinner.
9. If Randy eats out 4 times weekly, find the average amount of money he spends on dinner weekly as well
as the standard deviation.
10. If Dan eats out 5 times weekly and on Friday treats himself to wine worth $10, what is the average
amount and standard deviation that he spends on dinner weekly.
12. If Randy eats out 5 times weekly and Dan 4 times weekly(no wine), find the average amount of money
they spend on dinner weekly together. How about the standard deviation?
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9. Standard Deviation of a random variable:
Go back to pages 9 - 12 on the mean of a random variable and use that figure to calculate the standard deviation
of the random variable.
Standard Deviation Rules
Go back to page 13 and calculate the questions dealing with standard deviation.
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Random Variables - Terms and Formulas
Random Variable – a variable whose value is a numerical outcome of a random phenomenon.
Example 1: Put all the letters of the alphabet in a hat. If you choose a consonant, I pay you $1. If you choose a
vowel, I pay you $5. X is the random variable representing the outcome of the experiment. Its possible values
are 1 and 5.
Example 2: In most college courses, you get as a grade, either an A, B, C, D, or F. For credit purposes, A’s are
given 4 points, B’s are given 3 points, C’s are given 2 points, Ds are given 1 point, and F’s are no points. Let X
be the random variable representing the points a student gets. The possible values of X are 4, 3, 2, 1, and 0.
Discrete Random Variables have a countable number of possible values. The probability distribution of X
lists all possible values of X and their probabilities:
Value of X
Probability of X
x1
p1
x2
p2
…
…
x3
p3
xn
pn
The probabilities must satisfy two requirements: Every probability pi is a number between 0 and 1.
!
!
!
!
p1 + p2 + p3 + ...+ pn = 1
!
!
!
!
Example 3: Put all the letters of the alphabet in a hat. If you choose a consonant, I pay you $1. If you choose a
!
vowel, I pay you $5. X is the random variable representing the outcome
of the experiment. Create the
! distribution of X.
Value of X
P(X)
1
21
26
5
5
26
Example 4: A college instructor teaching a large class traditionally gives 10% A’s, 20% B’s, 45% C’s, 15%
D’s, and 10% F’s. If a student is chosen at random from the class, the student’s grade on a 4-point scale (A = 4)
! distribution of X.
is a random variable X. !
Create the
Value of X
P(X)
4
.1
3
.2
2
.45
1
.15
0
.1
What is the probability that a student has a grade point of 3 or better in this class? .3
What is the probability that a student has a grade point of 2 or worse in this class? .7
Draw a probability histogram to picture the probability distribution of the random variable X.
!
Histogram
Collection 1
!
50%
40%
30%
20%
10%
0
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1
2
3
grades
- 15 -
4
5
Stu Schwartz
Continuous Random Variables take on all possible values in an interval of numbers. The probability
distribution of X is described by a density curve. The probability of any event is the area under the density
curve and above the values of X that makes up that area.
Example 5: A random number is chosen from 0 to 1. Answer the following questions:
a. Find P(X > .5)
b. Find P(X ≥ .5)
.5
.5
e. Find P(.2 <X < .4)
!
c. Find P(X ≤ .8)
.8
f. Find P(.2 <X ≤ .4)
!
.2
d. Find P(X ≥ .15)
g. Find P(X < 1)
!
.2
.85
h. Find P(X = .5)
!
.1
0
Mean of a Random Variable – if an experiment with random variable X is done over a long period of time, we
can calculate the mean (average) value of that random variable. Another term for mean of a random variable X
! is the expected value of X. !
!
!
If X is a discrete random variable whose distribution is given by
Value of X
Probability of X
x1
p1
x2
p2
…
…
x3
p3
xn
pn
To find the mean (or expected value) of X, multiply every possible value of X by its probability
!
!
!
!
Then add the results. The symbol for the mean of X is µX .
!
!
!
!
µX = x1 p1 + x 2 p2 + x 3 p3 + ... + x n pn = " x i pi
!
Example 6: A fair coin is flipped 3 times. Find the mean of the discrete random variable X that counts the
number of heads.
!
X
P(X)
3
1
8
2
3
8
1
3
8
0
1
8
" 1 % " 3 % " 3% " 1 %
µX = 3$ ' + 2$ ' + 1$ ' + 0$ ' = 1.5
# 8!
& # 8 & ! # 8 & !# 8 &
!
Example 7: The daily lottery costs $1 to play. You pick a 3 digit number. If you win, you win $500. Find the
expected value of the lottery.
!
X
500
0
1
999
P(X)
1000
1000
" 1 % " 999 %
µX = 500$
' + 0$
' = .50 (1 = (.50 (You have to subtract the $1 cost)
# 1000 & #1000 &
!
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!
!
- 16 -
Stu Schwartz
!
Law of Large Numbers: Draw independent observations from any population with finite mean µ . As the
number of observations increases, the mean x of the observed values eventually approaches the mean µ of that
population.
! will approach 50%.
Example 8: If I flip a fair coin, over a long period of time, the percentage of heads
!
That does not mean that the number of heads will approach the number of tails. Here!is a typical
experiment in tossing coins.
Heads
7
Tails
3
Percentage of heads 70%
42
58
42%
463
537
46.3%
4,875
5,125
48.75%
49,660
50,340
49.66%
Example 9: If you play blackjack perfectly at a casino, the expected value for every dollar played is
98.5 cents. That means on the average of hand you play, you lose 1.5 cents. Can an individual player
win playing blackjack at a casino. The “law of small numbers” says that over a short run, a player can
certainly win. But over the long run, the law of large numbers says that a player will eventually lose.
And given the huge number of players who play daily, the law of large numbers states that the casino
has to win. There is a reason that casinos look the way they do – it is guaranteed money. Still, players
can win over the short haul. The question that crops up is what is considered “large?” There is no
answer to that. What you have to know is that the larger n (the number of trials) gets, the closer the
average win (or loss) per play approaches the true average of losing 1.5 cents a play.
Rules for Means
Rule 1: If X and Y are random variables, then µX +Y = µX + µY
Rule 2: If X is a random variable and a and b are constants, then µa +bX = a + bµX
!
Example 10: You play two casino games. Game 1 has an expectation of losing $1 a play and game 2 has an
expectation of losing $2 a play. If you play both games, what is your expectation for both games.
µX +Y = !
"1" 2 = "3
Example 11: You go to a casino and play the same slot machine which averages losing 15 cents a play. You
play the machine 100 times and then leave, paying $5 for parking. Find your expectation for your casino visit.
!
µ100X "5 = 100(".15) " 5 = "$20
Example 12: Depending on the attendance of a minor league baseball team, the number of hot dogs and sodas
sold at a game is given by the following table.
!
Hot dogs sold
Probability
100
.05
200
.1
500
.3
1,000 2,500
.35
.2
Sodas sold
Probability
100 200
.1
.15
400
.25
800
.3
1,500
.2
If the profit on a hot dog is $1.50 and the profit on a soda is $0.75, find the average profit on hot dogs and sodas
per game.
µhotdogs = 100(.05) + 200(.1) + 500(.3) + 1000(.35) + 2500(.2) = $1,025
µsoda = 100(.1) + 200(.15) + 400(.25) + 800(.3) + 1500(.2) = $680
µhotdogs+sodas = $1,705
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Stu Schwartz
Variance and standard deviation of a Discrete Random Variable
If X is a discrete random variable whose distribution is given by
Value of X
Probability of X
x1
p1
x2
p2
…
…
x3
p3
xn
pn
We know that the mean (expected value) of X (µX ) is given by
!
!
!
!
!
!µX = x1 p!1 + x 2 p2 + x 3 p3 + ...! + x n pn = " x i pi
To find the variance of X (" 2 X ) , we !
employ this formula:
2
2
2
2
"!2 X = ( x1 # µX ) p1 + ( x 2 # µX ) p2 + ( x 3 # µX ) p3 + ...+ ( x n # µX ) pn
To find the standard
! deviation of X (" X ) , we know that
(" X ) = " 2 X
!
Example 13: A fair coin is flipped
! 3 times. Find the standard deviation of the discrete random variable X that
counts the number of heads.
!
2$ 1 '
2$ 3'
2$ 3'
2$ 1 '
" 2 X = ( 3 #1.5) & ) + (2 #1.5) & ) + (1#1.5) & ) + (0 #1.5) & ) = .75
" X = .866
% 8(
%8(
% 8(
%8(
Example 14: The daily lottery costs $1 to play. You pick a 3 digit number. If you win, you win $500. Find the
standard deviation of the lottery.
!
2$ 1 '
2 $ 999 '
" 2 X = (500 # .5) &
) + (0 # .5) &
) = 249.75
% 1000 (
%1000 (
" X = 15.803
Example 15: Choose an American household at random and let the random variable X be the number of
persons living in the household. Find the standard deviation of the average American household.
!
X
Probability
1
.25
2
.32
3
.17
4
.15
5
.07
6
.03
7 or more
.01
µX = .25(1) + .32(2) + .17( 3) + .15( 4 ) + .07(5) + .03(6) + .01( 7) = 2.6
2
2
2
2
2
2
2
" 2 X = .25(1# 2.6) + .32(2 # 2.6) + .17( 3 # 2.6) + .15( 4 # 2.6) + .07(5 # 2.6) + .03(6 # 2.6) + .01( 7 # 2.6) = 2.02
" X = 1.421
When a problem has many values of X like this one, this can be done using your list feature of your calculator:
1. Input the data into Lists L1 and L2.
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- 18 -
Stu Schwartz
2. To get µX , use L3 to multiply L1 and L2.
3. To find µX , find the sum of L3.
!
!
4. We no longer need L3. In L3, we will put our formula for variance:
5. To find the variance " 2 X , find the sum of L3. The standard deviation is the square root.
Rules for Variances of Random Variables:
!
Rule 1: If X is a random variable and a and b are constants, then
Variance
" 2 a +bX = b 2" 2 X
Standard Deviation
" a +bX = b" X
Rule 2: If X and Y are independent random variables, then
!
Variance
" 2 X +Y = " 2 X + " 2Y
Standard Deviation
" X +Y = " 2 X + " 2Y (Note that you don’t add standard deviations)
" 2 X #Y = " 2 X + " 2Y
" X #Y = " 2 X + " 2Y
(Note that even though you are finding the variance of the difference between X and Y, you add the
variances).
!
!
!
!
!
Example 16: Suppose two pro bowlers Adam (A) and Bart (B) have the following distribution of scores.
µA = 209
" A = 14
µB = 221
Find the following.
a) µB +A
b) µB "A
!
221+ 209 = 430
221" 209 = 12
" A = 22
c) 3" A
3(209) = 627
d) " B +A
2
22 + 14
26.077
e) " B #A
2
22 2 + 14 2
26.077
!
!
!
!
!
Example 17: Depending on the attendance of a minor league baseball team, the number of hot dogs and sodas
!
! by the following table.
sold at a game is given
!
!
!
Hot dogs sold
Probability
100
.05
200
.1
500
.3
1,000 2,500
.35
.2
Sodas sold
Probability
100 200
.1
.15
400
.25
800
.3
1,500
.2
Find the standard deviation of the number of hot dogs plus the number of sodas sold.
" 2 hotdogs = 628875
" 2 sodas = 226600
" hotdogs + sodas = 628875 + 22660 = 924.919
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!
- 19 -
Stu Schwartz
Discrete Random Variables
1. Choose an American household at random and let the random variable X be the number of persons living in
the household. The probability of X is:
X
Probability
1
.25
a. Find P(X > 4)
2
.32
3
.17
4
.15
b. Find P(X ≥ 4)
.1
5
.07
6
.03
7 or more
.01
c. Find P(2 <X ≤ 4)
.25
d. Find P(X ≠ 2)
.32
.68
2. A fair coin is flipped 4 times. Find the probability distribution of the discrete random variable X that counts
the number of heads. Then draw the probability histogram for X.
!
!
!
!
X
Probability
Collection 1
40%
4
3
2
1
1
4
3
8
1
4
1
16
0
1
16
Histogram
!
!
!
!
!
30%
20%
10%
0
1
a. Find P(X > 2)
!
3
4
b. Find P(X ≥ 2)
c. Find P(X ≥ 1)
d. Find P(X ≥ 0)
1
5 16
11 16
15 16
3. A coin that is rigged is flipped 3 times. The probability of heads is twice the probability of tails. Find the
probability distribution of the discrete random variable X that counts the number of heads. Then draw the
probability histogram for X.
!
!
!
X
Probability
Collection 1
60%
50%
40%
30%
20%
10%
3
2
8
27
1
12
27
0
6
27
1
27
Histogram
!
0
!
1
a. Find P(X > 2)
2
Heads
!
3
!
4
b. Find P(X ≥ 2)
1
8
27
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!
2
Heads
c. Find P(X ≤ 3)
1
d. Find P(X ≠ 2)
5
9
- 20 -
!
!
Stu Schwartz
!
4. The numbers 1, 2, 3, 4, 5 are placed in a hat. A number is chosen at random, replaced, and another
number is chosen at random. Let X be the number of odd numbers that are chosen. Find the probability
distribution of the random variable X and draw a probability histogram.
X
Probability
2
Collection 1 !
60%
50%
40%
30%
20%
10%
!
1
9
25
0.0
a. Find P(X > 1)
0
12
25
4
25
Histogram
!
1.0
Heads
2.0
b. Find P(X ≥ 1)
9
25
c. Find P(X ≤ 1)
21
25
d. Find P(X ≠ 2)
16
25
16
25
5. An SRS of 3 students are chosen to be on a committee that advises a university on its computer policies.
70% of the student population are PC users and 30% are Mac users. Let X be the number of PC users on the
!
! a probability
student committee. Find!the probability distribution of!the random variable X and draw
histogram.
X
Probability
3
.343
Collection 1
60%
50%
40%
30%
20%
10%
1
.189
0
.027
Histogram
0
1
a. Find P(X > 2)
2
PC
3
4
b. Find P(X ≥ 2)
.343
.784
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!
2
.441
c. Find P(X ≤ 1)
d. Find P(X ≠ 3)
.216
.657
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!
!
Stu Schwartz
!
6. A boss of a company chooses 4 workers at random to act as advisers to him. The company is 75% white and
25% African American. Let X be the number of African-Americans on the committee. Find the probability
distribution of the random variable X and draw a probability histogram.
X
Probability
4
.004
Collection 1
3
.047
2
.211
1
.422
0
.316
Histogram
40%
30%
20%
10%
0
1
2
PC
3
4
a. Find P(X > 2)
b. Find P(X < 2)
.051
.738
c. Find P(1 ≤ X ≤ 3)
.680
.684
7. A density curve is shown on the figure on the right. Find the
following probabilities.
!
!
!
a. Find P(X > 1)
d. Find P(X ≠ 0)
b. Find P(X < 1.5)
!
c. Find P(.5 ≤ X ≤ 1.5)
d. Find P(X <2)
15
3
.25
1
16
16
8. Two random numbers from 0 to 1 are chosen and added. Let X be their sum. X is thus a continuous random
variable between 0 and 2. The density curve is shown to the right. Find the following probabilities:
!
!
a. Find P(X > 1)
b. Find P(X > 1.5)
1
2
!
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c. Find P(X < .2)
7
8
!
!
!
d. Find P(.5 ≤ X ≤ 1.5)
1
50
!
- 22 -
3
4
!
Stu Schwartz
Mean of a Random Variable (Expected Value) Problems
1) Choose an American suburban household at random and let the random variable X be the number of cars
that people in the house own. Find the number of cars owned by the average suburban American household.
X
Probability
0
.08
1
.29
2
.32
3
.17
4
.08
5
.04
6 or more
.02
2
µX = # X " P ( X ) = 2.08
$ = P ( X )( X % µX ) = 1.332
2) A huge cookie jar has 60% chocolate cookies and 40% vanilla cookies. Sam chooses 3 cookies blindfolded.
Let X be the!
number of chocolate cookies he chooses. Construct the probability distribution of X and the
average number of chocolate cookies he chooses.
X
Probability
3
.216
2
.432
1
.288
0
.064
2
µX = # X " P ( X ) = 1.80
$ = P ( X )( X % µX ) = 0.849
7
3) A contractor!is bidding on a road construction job that promises a profit of $20,000 with a probability of 10
and a loss, due to strikes, weather conditions, late arrival of building materials, and so on, of $40,000 with a
3
probability of 10 . What is the contractor’s mean expectation?
X
Probability
20000
.7
-40000
.3
2
µX = # X " P ( X ) = $2,000
$ = P ( X )( X % µX ) = $27,495
4 ) Pete selects one card from a deck of 52 cards. If it is an ace, he wins $5. If it is a club, he wins only $1.
However, if it is the ace of clubs, then he wins an extra $10. What is his mathematical expectation? Suppose
! $1 to play the game. What could he expect to win or lose per game. Suppose he plays 100
it costs him
games. How much would he be expected to win or lose?
X
Probability
5
3 52
1
12 52
µX = # X " P ( X ) = .827 $1 = $.173
!
!
!
16
1 52
0
36 52
2
% = P ( X )( X $ µX ) = $2.43 100 games - loss of $17.30
!
!
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Stu Schwartz
5) A box contains 8 green marbles, 7 yellow marbles, and 5 black marbles. One marble is to be selected from
the box. You get $10 if the marble selected is black, but lose $3 if the marble is green and lose $5 if the
marble is yellow. What is your mean expectation?
X
Probability
10
.25
-3
.4
-5
.35
2
µX = # X " P ( X ) = $.45
% = P ( X )( X $ µX ) = 6.095
6) A coin is tossed three times. If heads appears on all 3 tosses, Mary will win $16. If heads appears on 2 of the
tosses, she !
will win $2. The game costs $5 to play. What is her mean expectation?
X
Probability
16
.125
2
.375
0
.5
µX = # X " P ( X ) = 2.75 $ 5 = $2.25
2
% = P ( X )( X $ µX ) = 5.09
7) A carnival has set up the following game. You are blindfolded and have to pick a coin from 9 pennies, 8
! 12 dimes, 16 quarters, 11 half-dollars, and 4 dollar pieces. It costs a quarter to play the game. What
nickels,
could you be expected to win or lose every time you play the game?
X
Probability
.01
9 60
.05
8 60
.10
12 60
.25
16 60
.50
11 60
1
4 60
2
µX = # X " P ( X ) = .253 $ .25 = .003
% = P ( X )( X $ µX ) = .260
!
!
!
!
!
!
8) A pair of dice is rolled. Depending on the sum of the dots on both dice, Jake can win or lose money as
shown in the following chart. What is Jake’s mean expectation?
!
Sum of Dots
6 or 8 or 9
2 or 4 or 5
10
3 or 12
7 or 11
X
Probability
Outcome
win $6
lose $4
win $7
win $5
lose $1
6
14 36
-4
8 36
7
3 36
-1
8 36
2
µX = # X " P ( X ) = 2.222
!
!
!
!
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!
5
3 36
$ = P ( X )( X % µX ) = 4.360
!
- 24 -
Stu Schwartz
9) An oil company will only invest in an oil well if the can expect to make at least $1 million in profit. They
find a possible area in which to drill in Canada. It will cost the company $3 million to make the attempt. If
they find oil, they can expect to clear $7 million in profits. Geologists have estimated that the probability of
striking oil is 0.35. Should the company make the attempt and why or why not?
X
Probability
7
.35
0
.65
2
µX = # X " P ( X ) = 2.45 $ 3 = $$550,000 (no - don't drill)
!
% = P ( X )( X $ µX ) = 3.339 million
10) A wheel of fortune below costs $2 to play. If the spinner stops on black or
blue after one spin, the prize is $5. If it stops on green, the prize is $3. If it
stops on yellow, the prize is $0.50. If it stops on brown or red, there is no
payoff. What is your mean expectation every time you spin the wheel?
Should you play the game?
X
Probability
5
13
3
16
.50
16
Black Green
Yellow
Red
Blue
0
13
Brown
2
µX = # X " P ( X ) = 2.25 $ 2 = .25 (play the game)
!
!
!
!
% = P ( X )( X $ µX ) = 2.194
11) In a raffle costing $1 a ticket, 225 tickets are sold. First place gets $50. 2 second place tickets gets $30 and 3
!third place tickets gets $20. What is the mean value of your return on a ticket?
X
Probability
50
1 225
30
2 225
20
3 225
0
219 225
2
µX = # X " P ( X ) = .75 $1 = $.25
!
!
!
!
% = P ( X )( X $ µX ) = 4.886
12) An ecologist collected the data shown in the table below on the life span of a certain species of deer. Based
on this !
sample, what is the expected lifespan of this species?
Age at death (years)
Number
1
2
2
30
µX = # X " P ( X ) = 4.609
3
86
4
132
5
173
6
77
7
40
8
10
2
$ = P ( X )( X % µX ) = 1.362
!
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Stu Schwartz
13) A drug is administered to sets of three patients. Over a period of time, it is determined that the probability of
3 cures, 2 cures, 1 cure, and no cures are .70, .20, .09, and .01 respectively. What is the mean number of
cures that can be expected in a group of three?
X
Probability
3
.70
2
.20
1
.09
0
.01
2
µX = # X " P ( X ) = 2.590
$ = P ( X )( X % µX ) = .694
14) A man on his 64th birthday obtains a $10,000 one-year insurance policy at a cost of $500. Based on
!
mortality tables,
there is a .963 probability that the man will live for at least one more year. How much can
the insurance company expect to earn on this policy?
X
Probability
1000
.037
µX = # X " P ( X ) = 370 $ 500 = $$130
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0
.963
Insurance company makes $130
- 26 -
2
% = P ( X )( X $ µX ) = 1887.618
Stu Schwartz
Mean and Standard Deviation of a Random Variable
Randy and Dan are businessmen who spend time on the road weekly (Mon-Fri). They always eat dinner out.
Randy spends money for dinner according to the following probability chart.
X
$15
$20
$25
$30
P(X)
.2
.4
.25
.15
Dan spends money for dinner according to the following probability chart.
X
$20
$25
$30
$35
P(X)
.35
.25
.3
.1
1. Find the average amount of money Randy spends. .2(15) + .4 (20) + .25(25) + .15( 30) = $21.75
2. Find the average amount of money Dan spends. .35(20) + .25(25) + .3( 30) + .1( 35) = $25.75
!
3. Find the standard deviation of the amount of money Randy spends.
" 2 = .2(15 - 21.75) + .4 (20 # 21.75) + .25(25 # 21.75) + .15( 30 # 21.75) = 23.19
!
4. Find the standard deviation of the amount of money Dan spends.
" 2 = .35(20 - 25.75) + .25(25 # 25.75) + .3( 30 # 25.75) + .1( 35 # 25.75) = 25.69
!
" = $4.82
" = $5.07
5. Find the average amount of money they spend for dinner together. 25.75 + 21.75 = $47.50
!
6. Find the standard deviation of the total money they spend for dinner together. 4.82 2 + 5.07 2 = $7.00
!
7. Find the average difference in money they spend for dinner. 25.75 " 21.75 = $4.00
!
8. Find the standard deviation of the difference they spend
for dinner.
!
4.82 2 + 5.07 2 = $7.00
9. If Randy eats out 4 times weekly, find the average amount of money he spends on dinner weekly as well
" 4 x!= 4 ( 4.82) = $19.28
as the standard deviation. µ4 x = 4 (21.75) = $87.00
10. If Dan eats out 5 times weekly and on Friday treats himself to wine worth $10, what is the average
!
amount and standard
deviation that he spends on dinner weekly.
µ10+5x = 10 + 5(25.75) = $138.75 "10+5x = 5(5.07) = $23.25
!
11. If Randy eats out 5 times weekly and Dan 4 times weekly(no wine), find the average amount of money
they spend on dinner weekly together. How about the standard deviation?
µ5x +4 y = 5(21.75) + 4 (25.75) = $211.75
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!
2
2
" 5x +4 y = 25( 4.82) + 16(5.07) = $31.50
- 27 -
Stu Schwartz
Standard Deviation of a random variable:
Go back to pages 9 – 12 on the mean of a random variable and use that figure to calculate the standard
deviation of the random variable.
Standard Deviation Rules
Go back to page 13 and calculate the questions dealing with standard deviation.
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Stu Schwartz