Download MATH 110 Test Three Outline of Test Material EXPECTED VALUE

MATH 110 Test Three Outline of Test Material
EXPECTED VALUE (8.5)
 Super easy ones
(when the PDF is already given to you as a table and all you need to
do is multiply down the columns and add across)
Example: Find the expected value of the random variable X.
X
2
4
6
7
P(X)
0.3
0.2
0.1
0.4
Solution:
a) Add a row for the product X·P(X).
X
2
4
P(X)
0.3
0.2
X·P(X)
2(0.3)
4(0.2)
6
0.1
6(0.1)
7
0.4
7(0.4)
b) Carry out the multiplication.
X
2
4
P(X)
0.3
0.2
X·P(X)
0.6
0.8
6
0.1
0.6
7
0.4
2.4
c) Add across the bottom row of products.
E(X) = 0.6 + 0.8 + 0.6 + 2.4 = 4.4
EXPECTED VALUE (8.5)
 Still easy but the probabilities are given graphically
Example:
Find the expected value of the
random variable X.
Solution:
Read the probabilities from the graph, creating the usual PDF
table and proceed as usual.
X
P(X)
X·P(X)
1
0.15
0.15
2
0.2
0.4
3
0.25
0.75
4
0.05
0.20
5
0.35
1.75
E(X) = 0.15 + 0.4 + 0.75 + 020 + 1.75 = 3.25
Note: When reading probabilities from a graph, you should
always add them up and make sure that you get 1. Otherwise,
you made a mistake.
Here: 0.15 + 0.2 + 0.25 + 0.05 + 0.35 = 1.00
EXPECTED VALUE (8.5)
 Still easy but the probabilities are given in a ‘word problem’.
Example: A wedding photographer has a big event that will yield a
profit of $2000 with a probability of 0.8 or a loss (due to
unforeseen circumstances) of $500 with a probability of 0.2. What
is the photographer’s expected profit?
Solution: Extract the X values (profit/loss amount) and the
associated probabilities from the problem, put them in a PDF table
and proceed as usual. (Remember that a profit is POSITIVE and a
loss is NEGATIVE.)
X
P(X)
X·P(X)
$2000
0.8
$1600
-$500
0.2
-$100
E(X) = $1600 + (-$100) = $1600 - $100 = $1500
EXPECTED VALUE (8.5)
 Lotteries, Raffles, etc.
(Practice makes perfect. Please do a lot of these. I posted lots of
problems like these…including some YouTube videos.)
Example: Suppose you buy 1 ticket for $2 in a lottery with 1000 tickets.
The prize for the one winning ticket is $300. What are your expected
winnings?
Solution:
a) Extract the possibilities (the X values) and their probabilities.
Here there are only 2 things that can happen. You win $300 (minus the
cost of the ticket) or you lose the cost of your
ticket. The probability of winning is 1 out of 1000 (0.001) and the
probability of losing is 999 out of 1000 (0.999).
b) Create the PDF table. Also, I find it easier to solve the problem by first
calculating the expected value as if the ticket
didn’t cost anything (free ticket) and then adjusting at the end by
subtracting off the actual cost of the ticket. (You are
free to do it differently…by incorporating the cost of the ticket from the
beginning. You get the same answer either way.)
(PDF: If you assume the ticket is free, you lose $0 if you lose and win
$300 if you win.)
X
$300
-$0
P(X)
0.001
0.999
X·P(X)
$0.30
$0.00
So, E(X) = $0.30 + $0.00 = $0.30.
But the ticket is not free so we must subtract the real cost ($2.00)
Expected winnings = $0.30 - $2.00 = -$1.70 (Remember, a negative
here represents a loss.)
EXPECTED VALUE (8.5)
 Lotteries, Raffles, etc.
Example: Find the expected payback for a game in which you bet
$4 on any number from 0 to 199 if you get $400 if your number
comes up.
Solution: As I did earlier, I will first assume that it costs nothing to
bet and then subtract off the $4 bet at the end.
You have 1 out of 200 (0 to 199 = 200 numbers) of winning $400 &
199 out of 200 chances of losing $0 (assuming free bet).
X
P(X)
X·P(X)
$400
1/200 =
0.005
$2.00
-$0
199/200 =
0.995
$0.00
E(X) = $2.00 - $0.00 = $2.00
But actually thebet is not free so now we have to subtract off the
cost of the bet ($4.00)
Expected winnings = $2.00 - $4.00 = -$2.00
negative here represents a loss.)
(Remember, a
EXPECTED VALUE (8.5)
 Lotteries, Raffles, etc.
Example: In roulette, there are 18 red compartments, 18 black
compartments & 2 compartments that are not red or black.
If you bet $2 on red and the ball lands on red, you get to keep the
$2 you paid to play and you win another $2. Otherwise, you lose
your $2 bet. What is your expected payback if you bet $2 on red?
Solution: There are18 red and 20 non-red compartments.
So the probability of red is 18/38 and the probability of non-red is
20/38.
(Note: Here the arithmetic is so easy here it is probably not worth
the trouble of first assuming that the bet is free. Also, if you do try
that technique here, you have to be very careful because, unlike
things like buying a lottery ticket, in this situation, you only pay the
$2 if you lose.)
X
P(X)
X·P(X)
$2
18/38 =
$36/38
( )
(
E(X) = -$0.11
)
-$2
20/38 =
-$40/38
BASIC STATISTICS (9.1 & 9.2)
 Easy…mode, median, mean, range and standard deviation
Example: For the following set of numbers, find the mode, the
median, the mean (round to nearest tenth), the range and the
standard deviation (round to the nearest hundredth):
41 60 56 35 40 36
Solution:
MODE: The most common number. (There can be no mode, 1
mode or more than one mode.)
Here there is NO mode because no number occurs more than once.
MEDIAN: The middle number when the numbers are put in order.
(If there’s no single middle number, average the 2 middle numbers.)
The numbers in order: 35 36 40 41 56 60
So the median is (40+41)/2 = 81/2 = 40.5
MEAN: ̅
= 44.7
(Note: Especially because you will also calculate the standard
deviation here, it would probably be better to use the Stats Mode
of your calculator to calculate the mean.)
RANGE: Largest – Smallest = 60 – 35 = 25
STANDARD DEVIATION: Use the calculator Stats Mode to get: 10.65
[Note: When using the calculator to get the standard deviation, IN
THIS COURSE, always use
(or its equivalent if your calculator
uses a different symbol for sample standard deviation). In this
course, never use (or its equivalent for the population standard
deviation) which we do not cover.]
BASIC STATISTICS (9.1 & 9.2)
 Mean and standard deviation from a Frequency table
Example: Find the mean (round to the nearest tenth) and standard
deviation (round to the nearest hundredth) of the placement
scores in the table below.
Value
4
7
8
3
Frequency
3
2
1
4
Solution:
From the table, we have 4 three times, 7 two times, 8 once, and 3
four times so the actual dataset is: 4 4 4 7 7 8 3 3 3 3.
MEAN: ̅
= 4.6
(or use the Stats Mode on your calculator)
STANDARD DEVIATION: Use the calculator Stats Mode to get: 1.96
[Note: When using the calculator to get the standard deviation, IN
THIS COURSE, always use
(or its equivalent if your calculator
uses a different symbol for sample standard deviation). In this
course, never use (or its equivalent for the population standard
deviation) which we do not cover.]
BASIC STATISTICS (9.1 & 9.2)
 Grouped means and standard deviations
Example: Find the mean (round to the nearest tenth) and standard
deviation (round to the nearest hundredth) of the data below:
Interval
1-4
5-8
9-12
13-16
Frequency
3
2
1
4
Solution: Replace each interval with its midpoint & proceed as usual.
1 – 4 midpoint = (1+4)/2 = 5/2 = 2.5
5 – 8 midpoint = (5+8)/2 = 13/2 = 6.5
9 – 12 midpoint = (9+12)/2 = 21/2 = 10.5
13 – 16 midpoint = (13+16)/2 = 29/2 = 14.5
Interval Interval midpoint Frequency
1-4
2.5
3
5-8
6.5
2
9-12
10.5
1
13-16
14.5
4
So the dataset is: 2.5 2.5 2.5 6.5 6.5 10.5 14.5 14.5 14.5 14.5
Now use your calculator’s Stats Mode to get:
Mean: 8.9
AND Standard Deviation: 5.40
BASIC STATISTICS (9.1 & 9.2)
 Chebyshev’s Theorem
Example: Find the fraction of all the numbers of a data set that
must lie within 3 standard deviations from the mean.
Solution:
Chebyshev’s Theorem: For any set of numbers, a fraction of at least
of them will be within k standard deviations from the mean.
Here k=3, and…
So, 8/9 of the data must lie within 3 standard deviations from the
mean.
BASIC STATISTICS (9.1 & 9.2)
 Chebyshev’s Theorem
Example: In a certain distribution, the mean is 50 with a standard
deviation of 6. Use Chebyshev’s Theorem to find the probability
that a number lies between 26 and 74. Write your final answer
rounded to the nearest thousandth.
Solution:
Chebyshev’s Theorem: For any set of numbers, a fraction of at least
of them will be within k standard deviations from the mean.
First, find k.
Looking at the number line above, we see that k = 4, so…
We were asked to write the final answer rounded to the nearest
thousandth, so…
The probability that a number lies between 26 and 74 is
15/16 = .938