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Chapter 17. The Expected Value and Standard Error
Using a Box Model we want to answer the questions:
1. What should be the EXPECTED SUM of the DRAWS? This is the
EXPECTED VALUE for the SUM.
2. When the game is actually played, how much will my ACTUAL sum of
draws differ from the EXPECTED sum of draws? This is the STANDARD
ERROR for the Sum.
Key Words
Sum of the Draws – Is found by drawing tickets at random from a box and then adding
up the numbers on the tickets.
Expected Value – the value expected for the sum of draws made at random with
replacement from a box.
Standard Error – The amount the numbers drawn at random differ from the expected
value.
EXPECTED VALUE
The expected value from the sum of draws made at random with replacement from a box
equals
(number of draws)  (average of box)
Example 3
Find the expected value for the sum of 100 draws with replacement from the box:
-2, 4, 4, 7, 7
Solution
The average of the box is
24477
5
= 4
There are 100 draws.
replacement is
Therefore, the expected value from the sum of draws with
(number of draws)  (average of box)
(100)  (4) = 400
STANDARD ERROR
The likely size of the chance error in a sum of draws process is given by the standard
error or SE. For instance, the sum of the draws from a box will be around the expected
value, give or take a standard error or so.
There is a formula to use in computing the SE for a sum of draws made at random with
replacement from a box. It is called the square root law.
Standard Error for sum = SquareRoot(Number of Draws)  SDofBOX

Example 4, Exercise B, Page 293, #1
Solution
(a)
Expected value = (number of draws)  (average of box)
Number of draws = 100
1 2  3  4  5  6  7
7
Average of box =
= 4.
Expected value = (100)  (4) = 400
To find the standard error, apply the following equation:.
Standard error =
number _ of _ draws  (SD _ of _ box)
The deviations from the average are (-3), (-2), (-1), 0, 1, 2, 3. The SD of the box is
(3) 2  (2) 2  ( 1) 2  (0) 2  (1) 2  ( 2) 2  (3) 2
7
=2
The standard error is 100  2 = 20.
(b)
The sum of the draws will be around 400, give or take 20 or so.
(c)
You should guess that the sum is 400 and that the sum would be off by 20 or so.
USING THE NORMAL CURVE
The purpose of this section is to find out what is the chance that the sum of the
draws will be in a given range when the number of draws from a box is large and is made
with replacement. Provided the number of draws is sufficiently large, the normal curve
can be used to figure chances from the sum of the draws. To find the area underneath the
normal curve, find the expected value and standard error and then covert to standard
units.
If we take the sum of 25 draws made at random with replacement from a box,
what is the chance that the sum will fall between 50 and 100? The numbers in the box
are: 0, 2, 3, 4 and 6. The sum would fall within the range of 0 and 150 because (25
draws)  0 = 0 and (25 draws)  6 = 150. To find the chance, convert to standard units and
use the normal curve. Standard units say how many SE’s a number is away from the
expected value.
Expected value = (number of draws)  (average of box)
Number of draws = 25
Average of box =
0 23 46
=3
5
Expected value = (25)  (3) = 75
Standard Error =
number _ of _ draws  (SD of box)
SD of box = 2
Standard Error =
25  (2) = 10
The expected value for the sum is 75 and the SE is 10, so 100 is 2.5 SE’s above
the expected value
100  75
= +2.5
10
and 50 is -2.5 SE’s below the expected value
50  75
= -2.5.
10
The interval from 50 to 100 is the interval within 2.5 SE’s of the expected value, so the
area underneath the curve according to the z-table is 99%.
WHAT IS THE DIFFERENCE BETWEEN THE SE AND THE SD?
The SD measures the spread between the numbers on the tickets in the box. The SE
measures the spread between the SUMs of the numbers on the tickets in the box.
SUM of DRAWS FOLLOWS the NORMAL CURVE!!!!
The SUMs of the DRAWS will follow the NORMAL CURVE as the histogram EVEN
when the numbers on the tickets in the box do NOT. That is what the central limit theorm
in probability theory says. Look at the spreadsheet Central Limit Theorem Spreadsheet to
see that this is true.
In the spreadsheet we draw 50 times from a box with 18 1’s and 20 0’s, and we take the
sum of the 50 draws. We do this 500 times and draw the histogram for the sum of draws.
You can see in the picture that it looks like the normal curve.
A SHORT-CUT For THE SD OF A BOX
There is a short-cut to finding the SD for boxes with only two kinds of tickets.
This shortcut applies ONLY to boxes with only two different numbers on the tickets.
There can be many tickets in the box but they can only have some tickets with one of the
numbers and the rest of the tickets have the other number. The numbers do NOT have to
be ZERO and ONE. When the tickets in the box show only two different numbers there
is a larger number (“big”) and a smaller number (“small”), the SD of the box equals
big _ number  small _ number 
( fraction _ with _ big _ number)  ( fraction _ with _ small _ number)
For example, take the box 1, 1, 1, 5. Since there are only two different numbers in the
box, use the short-cut.
5  1 
1 3
  1.73
4 4
Example 5, Page 304, # 1
Solution
(a)
The draws can be as small as 100  1=100 or as large as 100  10=1,000.
(b)
Expected value = (number of draws)  (average of box)
Number of draws = 100
1  6  7  9  9  10
=7
6
Expected value = (100)  (7) = 700
Average of box =
Standard error =
SD of box =
number _ of _ draws  (SD _ of _ box)
(1  7) 2  (6  7) 2  (7  7) 2  (9  7) 2  (9  7) 2  (10  7) 2
=3
6
Standard error = 100  3 =30
Converting to Standard Units:
750  700 5
=  +1.67
30
3
650  700
5
= -  -1.67
30
3
Referring to the z-table, the area underneath the curve when the z-values are –1.67 and
1.67 is approximately 90%.
Example 6, Page 304, # 3
Solution
Using the short-cut equation:
big _ number  small _ number 
( fraction _ with _ big _ number)  ( fraction _ with _ small _ number)
(a)
(iii) because
[1-(-2)] 
1 2
 = 3
3 3
(b)
(i) because
(16-15) 
1 2
 =
3 3
(c)
(v) because
[1-(-1)] 
1 3
 = 2
4 4
(d)
(iv) because
(1-0) 
1 3
 =
4 4
1 2

3 3
1 2

3 3
1 3

4 4
1 3

4 4
(e)
(ii) because
(2-0) 
1 2
 =2 
3 3
1 2

3 3
CLASSIFYING AND COUNTING
Some chance processes involve counting. The box model needs to be setup
correctly and then the square root law can be used. In a counting problem we are
interested in COUNTING the number of times something of interest happens. We
COUNT that as drawing a ONE out of the box, and when it doesn’t happen, that is like
drawing a ZERO out of the box. So the box in such problems is a ZERO-ONE box. The
steps for this are
STEP 1) CLASSIFY the event of interest as a ONE. When it doesn’t happen that
is a ZERO.
STEP 2) The number of ONEs in the box should be set up so that the chances of
drawing a ONE should be the same as the chance the event of interest occurs.
The following example shows how this is done.
Example 7, Pg. 303, #2
Solution
The expected value for the number of fives is
1
 100  20.
5
The box model will be applied to figure out the deviation from the expected value
of 20 to 8 and 32 respectively.
The box model for this problem
1 2 3 4 5
becomes
0 0 0 0 1
because we are concerned with what is the chance of getting 5’s. Notice that on each
draw, the number of 5’s either goes up by 1 or stays the same:


1 is added to the count if the draw is 5
0 is added to the count if the draw is 1, 2, 3, or 4
The short-cut method for calculating the SD of the box involves two numbers, in this
case the two numbers are 0 and 1. Using the short-cut method, the SD is
SD = (1  0) 
1 4
 
5 5
4
2

25 5
2
 4.
5
To find out the chance of getting between 8 and 32 tickets marked “5”, use the
following logic. The expected number of 5s is 20. So we need the area under the
normal curve centered at 20 from 8 to 32.
The SE is 100 
FIRST Convert 8 and 32 to standard units to get two z values.
z1=(8-20)/4=-3,
z2=(32-20)/4=3.
NOW go to the z-table and find the area under the Standard normal curve from –3 to
+3. This is about 99%, and that is the chances of getting between 8 and 32 5s.
.