AP Statistics Objectives Ch16
Find the probability model for a
discrete random variable.
Find the mean and variance of a
random variable.
Use the proper notation for the
population parameters of mean,
standard deviation, and variance.
AP Statistics Objectives Ch16
Determine the new mean and
standard deviation after adding a
constant, multiplying a constant,
or adding or subtracting two
independent random variables
Interpret expected value and
standard deviation of a random
variable in context.
Vocabulary
Random Variable
Discrete Random Variable
Continuous Random Variable
Probability Model
Expected Value
Chapter 16 Notes
Chapter 16
Assignment
Chapter 15
Answers
Chapter 16 Assignment
Part I: pages 381 – 382 #2&10,
Part II: page 382 #18, 20
Part III: page 383 #24, 26, 27
Part IV: pages 383 – 384 #32, 34&36
Part V: page 384 #38, 40
Chapter 16 Notes
1. Random Variable – a variable that assumes
several different values as a result of some
random event
-Denoted by a capital letter
P(X = 7)
“The Probability
that the”
“equals 7”
“Random
Variable X”
EXAMPLE 1:
For one six-sided die, the event of rolling a five is
P(X = 5) = 1/6
Probability Model
2. Following is the probability model for rolling a six-sided
die:
x
P(X=x)
Probability Model
2. Following is the probability model for rolling a six-sided
die:
x
1
2
3
4
5
6
P(X=x)
Probability Model
2. Following is the probability model for rolling a six-sided
die:
x
1
2
3
4
5
6
1
1
1
1
1
1
P(X=x)
6
6
6
6
6
6
Probability Model
3. What is the probability model for rolling two six-sided
dice?
x
P(X=x)
Probability Model
3. What is the probability model for rolling two six-sided
dice?
x
2
3
4
5
6
7
8
9
10
P(X=x)
1
2
3
1
2
3
4
2
3
4
5
3
4
5
6
4
5
6
7
5
6
7
8
6
7
8
9
4
5
6
7
8
9
10
5
6
6
7
7
8
8
9
9
10
10
11
11
12
11
12
Probability Model
3. What is the probability model for rolling two six-sided
dice?
x
2 3
4
5
6
7
8
9 10 11 12
P(X=x) 𝟏 𝟐
𝟒
𝟒
𝟑
𝟔
𝟑
𝟐
𝟏
𝟓
𝟓
𝟑𝟔 𝟑𝟔 𝟑𝟔 𝟑𝟔 𝟑𝟔 𝟑𝟔 𝟑𝟔 𝟑𝟔 𝟑𝟔 𝟑𝟔 𝟑𝟔
1
2
3
1
2
3
4
2
3
4
5
3
4
5
6
4
5
6
7
5
6
7
8
6
7
8
9
4
5
6
7
8
9
10
5
6
6
7
7
8
8
9
9
10
10
11
11
12
Chapter 16 Notes
4. Discrete random Variable – a variable that can take
on one of a finite number of distinct outcomes
Example: Rolling two dice
5. Continuous random Variable – a variable that can
take on any numeric value within a range of values
Example: Amount of rainfall during a day
Chapter 16 Notes
6. Expected Value
• the theoretical long-run average value of a random
variable
• the center of the probability model (more chp 17)
For discrete random variables:
E(X) = 𝝁𝒙 = 𝒙𝑷(𝒙)
7. What is the expected value of rolling a die?
E(X)
1
1
1
1
1
= 1( ) + 2( ) + 3( ) + 4( ) + 5( )
6
6
6
6
6
1 2 3 4 5
6 21
= + + + + + = = 3.5
6 6 6 6 6
6
6
+
1
6( )
6
Chapter 16 Notes
For discrete random variables:
E(X) = 𝞵 = 𝑥𝑃(𝑥)
8. What is the expected value of rolling 2 dice?
1
2
3
4
5
6
1
2
2
3
3
4
4
5
5
6
6
7
3
4
5
4
5
6
5
6
7
6
7
8
7
8
9
8
9
10
6
7
8
9
10
11
7
8
9
10
11
12
Chapter 16 Notes
For discrete random variables:
E(X) = 𝞵 = 𝑥𝑃(𝑥)
8. What is the expected value of rolling 2 dice?
𝜇𝑥 =
1
2( )
36
+
2
3( )
36
+
3
4( )
36
=
2
6 12 20
+ + +
36 36 36 36
=
252
36
=7
+
30
36
+
4
5( )
36
+
42
36
+
5
6( )
36
40
+
36
+
+
6
7( )
36
36
36
+
+…
30
36
+
1
12( )
36
22
36
+
12
36
Chapter 16 Notes
9. Variance - For discrete random variables:
𝜎 2 = 𝑉𝐴𝑅(𝑋) = (𝒙 − 𝝁)2 𝑃(𝑥)
• What is the variance of rolling 2 dice?
Remember E(X) = 7
𝟏
𝟑𝟔
𝝈𝟐 = (𝟐 − 𝟕)𝟐
= (−𝟓)𝟐
𝟏
𝟑𝟔
=
𝟏
(𝟐𝟓)
𝟑𝟔
=
𝟐𝟓
𝟑𝟔
=
𝟐𝟏𝟎
𝟑𝟔
+
𝟑𝟐
𝟑𝟔
+
+(𝟑 − 𝟕)𝟐
+ (−𝟒)𝟐
+
𝟐𝟕
𝟑𝟔
≈ 𝟓. 𝟖𝟑
𝟐
𝟑𝟔
𝟐
(16)
𝟑𝟔
+
𝟏𝟔
𝟑𝟔
+
𝟐
𝟑𝟔
+ ⋯ + (𝟏𝟏 −
𝟐
𝟐
𝟑𝟔
+ ⋯ + 𝟏𝟔
𝟐
𝟑𝟔
+ … + 𝟒
𝟓
𝟑𝟔
+
𝟎
𝟑𝟔
+
𝟓
𝟑𝟔
+
+
+
𝟏𝟔
𝟑𝟔
+
𝟓
𝟐
𝟏
𝟑𝟔
𝟏
(𝟐𝟓)
𝟑𝟔
𝟐𝟕
𝟑𝟔
+
𝟑𝟐
𝟑𝟔
+
𝟐𝟓
𝟑𝟔
Chapter 16 Notes
10. Standard Deviation -
For discrete random variables:
𝜎 = 𝑆𝐷 𝑋 = 𝑉𝐴𝑅(𝑋)
What is the standard deviation of rolling 2 dice?
𝝈=
𝟐𝟏𝟎
𝟑𝟔
≈ 𝟐. 𝟒𝟐
11. Calculator Steps For Expected Value &
Standard Deviation of a
Discrete Random Variable
1.
2.
3.
4.
5.
6.
Place possible values in L1
Place matching probabilities in L2
STAT key
CALC
1-Var Stats - List: 𝑳𝟏 FreqList: 𝑳𝟐
It’s labeled wrong but 𝝁 (Expected Value) is listed
a the top as 𝒙
7. Standard Deviation is listed as 𝝈
Chapter 16 Notes
12. Use the calculator to check the Expected
Value and standard deviation of rolling 2 dice.
𝝁=𝟕
𝝈=
𝟐𝟏𝟎
𝟑𝟔
≈ 𝟐. 𝟒𝟏𝟓 𝑽𝑨𝑹 𝑿 =
𝟐𝟏𝟎
𝟑𝟔
≈5.83
13. On your own use the calculator to check the
Expected Value and standard deviation of rolling
a single die.
𝝁 = 𝟑. 𝟓
𝝈=
𝟏𝟕.𝟓
≈
𝟔
𝑽𝑨𝑹 𝑿 =
𝟏. 𝟕𝟎𝟖
𝟏𝟕.𝟓
𝟔
≈2.92
Chapter 16 Notes
14. Shifting the outcomes
Remember: Shifting the data (adding or subtracting
a constant) affect measures of center, but NOT
measures of SPREAD.
𝑬 𝑿±𝒄 =𝑬 𝑿 ±𝒄
VAR 𝑿 ± 𝒄 = 𝑽𝑨𝑹 𝑿
S𝐃 𝑿 ± 𝒄 = 𝑺𝑫 𝑿
Chapter 16 Notes
15. If a six-sided die has the values 6, 7, 8, 9, 10, 11
instead of 1 through 6 (i.e. 5 was added to each side),
what is the Expected Value and Standard Deviation
of the die?
𝑬 𝑫𝒊𝒆 + 𝟓 = 𝑬 𝑫𝒊𝒆 + 𝟓
= 𝟑. 𝟓 + 5
= 𝟖. 𝟓
Chapter 16 Notes
15. If a six-sided die has the values 6, 7, 8, 9, 10, 11
instead of 1 through 6 (i.e. 5 was added to each side),
what is the Expected Value and Standard Deviation
of the die?
NOTE: UNCHANGED
VAR 𝑨 𝑫𝒊𝒆 + 𝟓 = 𝑽𝑨𝑹 𝑨 𝑫𝒊𝒆
=
𝟏𝟕.𝟓
𝟔
≈ 𝟐. 𝟗𝟐
NOTE: UNCHANGED
SD(𝑨 𝑫𝒊𝒆 + 𝟓) =
𝟏𝟕.𝟓
𝟔
≈ 𝟏. 𝟕𝟎𝟖
Chapter 16 Notes
16. Rescaling the outcomes
Remember: Rescaling the data (multiplying by a
constant) affect measures of center, AND measures
of spread.
𝑬 𝒂𝑿 = 𝒂𝑬 𝑿
VAR 𝒂𝑿 = 𝒂𝟐 𝑽𝑨𝑹 𝑿
S𝐃 𝒂𝑿 =
𝒂𝟐 𝑽𝑨𝑹(𝑿) = 𝒂S𝐃 𝑿
Chapter 16 Notes
17. Consider a dice game:
No points for rolling a 1, 2, or 3; 5 points for a 4 or 5;
50 points for a 6.
Find the expected value and standard deviation.
For discrete random variables:
E(X) = 𝞵 = 𝑥𝑃(𝑥)
𝐸 𝑋 =0
3
6
=0+
+5
10
6
60
=
= 10
6
+
2
+
6
50
6
50
1
6
Try the calculator.
L1
0
L2
3/6
5
2/6
50
1/6
17. Consider a dice game: no points for rolling a 1, 2, or 3; 5 points for a
4 or 5; 50 points for a 6. Find the expected value and standard
deviation.
From the previous slide: E(X) = 𝜇 = 10
For discrete random variables:
𝜎 2 = 𝑉𝐴𝑅(𝑋) = (𝑥 − 𝜇)2 𝑃(𝑥)
VAR 𝑋 =
=
=
=
2 3
2 2
0 − 10
+ 5 − 10
+ 50
6
6
2 3
2 2
2 1
−10
+ −5
+ 40
6
6
6
3
2
1
100
+ 25
+ 1600
6
6
6
300
50
1600
1950
+
+
=
= 325
6
6
6
6
SD 𝑋 = 𝑉𝐴𝑅(𝑋) =
325 ≈ 𝟏𝟖. 𝟎𝟑
− 10
2 1
6
18.LET’S TRIPLE THE POINTS (Think….Rescale)
no points for rolling a 1, 2, or 3; 15 points for a 4 or 5;
150 points for a 6.
Find the expected value and standard deviation.
For discrete random variables: E(X) = 𝞵 = 𝑥𝑃(𝑥)
𝑬 𝒀 =0
3
6
=0+
+ 15
30
6
+
2
+
6
150
6
180
=
= 30
6
150
1
6
Try the calculator.
L1
0
15
L2
3/6
2/6
150
1/6
OR using #17: E(3X) = 3E(X) = 3(10) = 30
18. LET’S TRIPLE THE POINTS (Think rescale)
No points for rolling a 1, 2, or 3; 15 points for a 4 or 5; 150 points for a
6. Find the expected value and standard deviation.
From the previous slide: E(3X) = 30
For discrete random variables: 𝜎 2 = 𝑉𝐴𝑅(𝑋) = (𝑥 − 𝜇)2 𝑃(𝑥)
VAR 𝒀 = 0 − 30
=
=
=
2 3
6
+ 15 − 30
2 2
6
+ 150 − 30
2 1
6
3
2
1
2
2
2
−30
+ −15
+ 120
6
6
6
3
2
1
900
+ 225
+ 14400
6
6
6
2700
450
14400
17550
+
+
=
= 2925
6
6
6
6
S𝐃 𝒀 = 𝑉𝐴𝑅(𝑋) =
2925 ≈ 54.08
So SD(3X) = 3SD(X) = 3(18.03) = 54.1
Chapter 16 Notes
19. ADDING OR SUBTRACTING
RANDOM VARIABLES
𝑬 𝑿 ± 𝒀 = 𝑬 𝑿 ± 𝑬(𝒀)
If X and Y are independent,
VAR 𝑿 ± 𝒀 = 𝑽𝑨𝑹 𝑿 + 𝑽𝑨𝑹(𝒀)
NOTE: We ALWAYS add the variances.
You cannot take the variance away.
20. Always Add Variance Lab
1. Mark a “fill line” near the top of the large and
small cup.
2. Fill your large cup with water.
3. Record the weight of your full large cup in the
table below.
4. Pour water from the large cup to fill a small cup.
5. Now that you have used the large cup, record
the weight of the used large cup.
(NOTE: It should still have water in it.)
6. Repeat 1-5 five additional times to fill the table
below.
7. Calculate mean and standard deviation of Full
and Used Large Cups.
1.
2.
3.
4.
5.
20. Always Add Variance Lab
Mark a “fill line” near the top of the large and small cup.
Fill your large cup with water.
Record the weight of your full large cup in the table below.
Pour water from the large cup to fill a small cup.
Now that you have used the large cup, record the weight of the used large cup.
(NOTE: It should still have water in it.)
6. Repeat 1-5 five additional times to fill the table below.
Even though the mean weight
of the used large cup is
smaller
________________
than the
mean weight of the full large
cup, the variance of the used
large cup is
larger
___________________
than
the variance of the full large
cup.
Chapter 16 Notes
21. PYTHAGOREAN THEOREM of Statistics
NEVER Add the Standard Deviations!
Always convert to Variance first, add, then
convert back to Standard Deviation.
Standard deviation is like the SIDES of a
right triangle…
a2 + b2 = c2 The squares can be added to
show equality
HOWEVER…
a+b ≠ c
The sides cannot just be b
added
c
a
22. LET’S JUST PLAY THE ORIGINAL GAME THREE
TIMES.
Consider the original dice game: no points for rolling
a 1, 2, or 3; 5 points for a 4 or 5; 50 points for a 6.
Find the total expected value and standard deviation,
if you play three games.
E(𝑋1 + 𝑋2 + 𝑋3 ) = E(𝑋1 ) + 𝐸(𝑋2 ) + 𝐸(𝑋3 )
= 10 + 10 + 10 = 30
Same as E(3X), but what about standard deviation?
22. LET’S JUST PLAY THE ORIGINAL GAME THREE TIMES
Consider a dice game: no points for rolling a 1, 2, or 3; 5 points for a 4
or 5; 50 points for a 6. Find the expected value and standard deviation.
From the early slide: SD(X) = 18.03 VAR(X) = 325
IF INDEPENDENT, then
SD(𝑋1 + 𝑋2 + 𝑋3 ) = 𝑉𝐴𝑅 𝑋1 + 𝑉𝐴𝑅 𝑋2 + 𝑉𝐴𝑅(𝑋3 )
= 𝟑𝟐𝟓 + 𝟑𝟐𝟓 + 𝟑𝟐𝟓 ≈ 31.22
NOT SAME as SD(3X)!!!!
Chapter 16 Notes
𝑬 𝑿 ± 𝒀 = 𝑬 𝑿 ± 𝑬(𝒀)
xx. In the United States, the average adult male is
about 5’10” with a standard deviation of about 3”
and the average adult female is about 5’4” with a
standard deviation of about 3” as well.
a. What is the mean height difference between adult
males and females in the United States?
𝐸 𝑀𝑎𝑙𝑒 − 𝐹𝑒𝑚𝑎𝑙𝑒 = 𝐸 𝑀𝑎𝑙𝑒 − 𝐸(𝐹𝑒𝑚𝑎𝑙𝑒)
= 70 − 64 = 6 𝑖𝑛𝑐ℎ𝑒𝑠
NOTE: 5’10” = 70” and 5’4” = 64”
If X and Y are independent,
VAR 𝑿 ± 𝒀 = 𝑽𝑨𝑹 𝑿 + 𝑽𝑨𝑹(𝒀)
xx. In the United States, the average adult male is about 5’10” with a
standard deviation of about 3” and the average adult female is about
5’4” with a standard deviation of about 3” as well.
b. Assuming that they are independent of each other, what
is the standard deviation of height difference between
adult men and women in the United States?
SD 𝑀 − 𝐹 = 𝑉𝐴𝑅(𝑀 − 𝐹) =
= 9+9=
𝑉𝐴𝑅 𝑀 + 𝑉𝐴𝑅(𝐹)
18 ≈ 4.24 𝑖𝑛𝑐ℎ𝑒𝑠
NOTE: Since SD(M) = 3, then VAR(M) = 9. The same is true for the females.
Chapter 16 - #26a
26. Given independent random variables with means and
standard deviations as shown, find the mean and standard
deviation of each of these variables:
Mean
SD
X
80
12
Y
12
3
a) 2Y + 20
𝐸 2𝑌 + 20 = 2𝐸 𝑌 + 20 = 2 12 + 20 = 44
SD 2𝑌 + 𝟐𝟎 = 2𝑆𝐷 𝑌 = 2 3 = 6
BEACAUSE: S𝐡𝐢𝐟𝐭𝐢𝐧𝐠 𝐛𝐲 𝟐𝟎 𝐝𝐨𝐞𝐬 𝐧𝐨𝐭 𝐚𝐟𝐟𝐞𝐜𝐭 𝐬𝐩𝐫𝐞𝐚𝐝
Chapter 16 - #26b
Given independent random variables with means and
standard deviations as shown, find the mean and standard
deviation of each of these variables:
Mean
SD
X
80
12
Y
12
3
b) 3X
𝐸 3𝑋 = 3𝐸 𝑋 = 3 80 = 240
𝑆𝐷 3𝑋 = 3𝑆𝐷 𝑋 = 3 12 = 36
Chapter 16 - #26e
Given independent random variables with means and
standard deviations as shown, find the mean and standard
deviation of each of these variables:
Mean
SD
X
80
12 VAR(X) = 122 = 144
Y
12
3
c) 𝑋1 + 𝑋2 + 𝑋3
𝐸 𝑋1 + 𝑋2 + 𝑋3 = 𝐸 𝑋1 + 𝐸 𝑋2 + 𝐸 𝑋3
= 80 + 80 + 80 = 240 NO CHANGE FROM E(3X)
𝑆𝐷 𝑋1 + 𝑋2 + 𝑋3 = 𝑉𝐴𝑅 𝑋1 + 𝑉𝐴𝑅 𝑋2 + 𝑉𝐴𝑅(𝑋3 ) =
= 144 + 144 + 144 = 20.78
Chapter
16
#26c
Given independent random variables with means and standard
deviations as shown, find the mean and standard deviation of each of
these variables:
Mean
80
12
X
Y
SD
VAR(X) = 122
12 = 144
VAR(X) = 32 =
3 9
d) 0.25X + Y
𝐸
1
𝑋
4
+𝑌 =
1
𝐸
4
𝑋 +𝐸 𝑌
1
𝑆𝐷 𝑋 + 𝑌 =
4
=
1
(144)
16
+9=
1
=4
80 + 12 = 32
1
𝑉𝐴𝑅 𝑋 + 𝑉𝐴𝑅 𝑌 =
16
9 + 9 = 18 ≈ 4.24
NOTE:
1 2
4
=
1
16
Chapter 16 - #26d
Given independent random variables with means and standard
deviations as shown, find the mean and standard deviation of each of
these variables:
Mean
SD
X
Y
VAR(X) = 122
= 144
80
12
12
VAR(X) = 32 =
3 9
e) X – 5Y
𝐸 𝑋 − 5𝑌 = 𝐸 𝑋 − 5𝐸 𝑌 = 80 − 5 12 = 80 − 60 = 20
𝑆𝐷 𝑋 − 5𝑌 = 𝑉𝐴𝑅 𝑋 + 25𝑉𝐴𝑅 𝑌 =
= 144 + 25(9) = 369 ≈ 19.21
NOTE: 5
2
= 25
Chapter 16
25. Explain the difference between 𝑥 and 𝜇𝑋 .
𝒙 represents the mean of raw data, 𝝁𝑿
represents the expected value (aka mean) of a
random variable X. You need to remember
that 𝝁 is used for models. Here a probability
model.
Chapter 16
26. Suppose 𝜇𝑋 = 5 and 𝜇𝑌 = 10. According to the rules for
means, what is 𝜇𝑋+𝑌 ?
𝐸 𝑋 ± 𝑌 = 𝑬 𝑿 ± 𝑬(𝒀)
𝝁𝑿+𝒀 = 𝝁𝑿 + 𝝁𝒀
= 𝟓 + 𝟏𝟎
= 𝟏𝟓
Chapter 16
27. Suppose 𝜇𝑋 = 2 . According to the rules for means, what is
𝜇3+4𝑋 ?
Same as: 𝑬 𝟑 + 𝟒𝑿
𝐸 𝑋±𝑐 =𝑬 𝑿 ±𝒄
𝝁𝟑+𝟒𝑿 = 𝟑 + 𝝁𝟒𝑿
= 𝟑 + 𝟒𝝁𝑿
=𝟑+𝟒 𝟐
= 𝟏𝟏
𝐸 𝑎𝑋 = 𝒂𝑬 𝑿
Chapter 16
28. Suppose 𝜎𝑥 = 4 and 𝜎𝑌 = 9 and X and Y are independent
random variables, what is 𝜎𝑋+𝑌 ?
𝑉𝐴𝑅 𝑋 ± 𝑌 = 𝑽𝑨𝑹 𝑿 + 𝑽𝑨𝑹(𝒀)
𝑅𝐸𝑀𝐸𝑀𝐵𝐸𝑅:
𝜎 = 𝑺𝑫 𝑿 𝑎𝑛𝑑 𝜎 2 = 𝑽𝑨𝑹 𝑿
𝝈𝑿+𝒀 = 𝑽𝑨𝑹(𝑿 + 𝒀)
=
𝑉𝐴𝑅 𝑋 + 𝑉𝐴𝑅(𝑌)
= 𝟏𝟔 + 𝟖𝟏
= 𝟗𝟕 ≈ 𝟗. 𝟖𝟓
29. Suppose 𝜎𝑥 = 3 and X is an independent random variable,
what is 𝜎3+2𝑋 ?
𝑅𝐸𝑀𝐸𝑀𝐵𝐸𝑅:
′
𝑠ℎ𝑖𝑓𝑡𝑖𝑛𝑔 𝑑𝑜𝑒𝑠𝑛 𝑡 𝑎𝑓𝑓𝑒𝑐𝑡 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑠
𝑜𝑓 𝒔𝒑𝒓𝒆𝒂𝒅
𝑆𝐷 𝑋 ± 𝑐 = 𝑺𝑫 𝑿
𝑆𝐷 𝑎𝑋 = aSD(X)
𝜎3+2𝑋 =
𝝈𝟐𝑿= 𝟐𝑺𝑫(𝑿)
= 𝟐(𝟑)
=𝟔
BVD Chapter 16 # 33
30. The amount of cereal that can be poured into a
small bowl varies with a mean of 1.5 ounces and a
standard deviation of 0.3 ounces. A large bowl holds a
mean of 2.5 ounces with a standard deviation of 0.4
ounces. You open a new box of cereal and pour one
large and one small bowl.
a) How much more cereal do you expect to be in the large
bowl? E(large bowl – small bowl)
= E(large bowl) – E(small bowl)
= 2.5 – 1.5
= 1 ounce
BVD Chapter 16 # 33
30. The amount of cereal that can be poured into a small bowl
varies with a mean of 1.5 ounces and a standard deviation of 0.3
ounces. A large bowl holds a mean of 2.5 ounces with a
standard deviation of 0.4 ounces. You open a new box of cereal
and pour one large and one small bowl.
b) What’s the standard deviation of this difference?
SD(large bowl – small bowl)
= 𝑽𝒂𝒓 𝑳𝒂𝒓𝒈𝒆 𝒃𝒐𝒘𝒍 + 𝑽𝒂𝒓(𝑺𝒎𝒂𝒍𝒍 𝒃𝒐𝒘𝒍)
= 𝟎. 𝟒𝟐 + 𝟎. 𝟑𝟐
= 𝟎. 𝟏𝟔 + 𝟎. 𝟎𝟗
= 𝟎. 𝟐𝟓
= 0.5 ounces
BVD Chapter 16 # 33
30. The amount of cereal that can be poured into a small bowl varies with a mean of
1.5 ounces and a standard deviation of 0.3 ounces. A large bowl holds a mean of 2.5
ounces with a standard deviation of 0.4 ounces. You open a new box of cereal and
pour one large and one small bowl.
c) If the difference follows a Normal model, what’s the probability the
small bowl contains more cereal than the large one?
𝑥−𝜇
z=
𝜎
=
No
difference
0 −1
0.5
= -2
By 68-95-99.7 Rule, what percent of the graph is less
than 2 standard deviations below the mean?
BVD Chapter 16 # 33
30. The amount of cereal that can be poured into a small bowl varies with a mean of
1.5 ounces and a standard deviation of 0.3 ounces. A large bowl holds a mean of 2.5
ounces with a standard deviation of 0.4 ounces. You open a new box of cereal and
pour one large and one small bowl.
c) If the difference follows a Normal model, what’s the probability the
small bowl contains more cereal than the large one?
68-95-99.7 Rule
BVD Chapter 16 # 33
30. The amount of cereal that can be poured into a small bowl varies with a mean of
1.5 ounces and a standard deviation of 0.3 ounces. A large bowl holds a mean of 2.5
ounces with a standard deviation of 0.4 ounces. You open a new box of cereal and
pour one large and one small bowl.
c) If the difference follows a Normal model, what’s the probability the
small bowl contains more cereal than the large one?
𝑥−𝜇
z=
𝜎
=
No
difference
0 −1
0.5
= -2
By 68-95-99.7 Rule, what percent of the graph is less
than 2 standard deviations below the mean?
100 – 95 = 5
𝟓
= 𝟐. 𝟓%
𝟐
BVD Chapter 16 # 33
30. The amount of cereal that can be poured into a small bowl varies with a mean of 1.5 ounces and a
standard deviation of 0.3 ounces. A large bowl holds a mean of 2.5 ounces with a standard deviation of 0.4
ounces. You open a new box of cereal and pour one large and one small bowl.
c) If the difference follows a Normal model, what’s the probability the small bowl contains more cereal than the large one?
z=
=
𝟐. 𝟓%
No
difference
𝑥−𝜇
𝜎
0 −1
0.5
= -2
Using the 68-95-99.7 Rule, the probability the
small bowl contains more cereal than the
large one is about 2.5%.
BVD Chapter 16 # 33
30. The amount of cereal that can be poured into a small bowl varies with a mean of 1.5 ounces and a
standard deviation of 0.3 ounces. A large bowl holds a mean of 2.5 ounces with a standard deviation of 0.4
ounces. You open a new box of cereal and pour one large and one small bowl.
c) If the difference follows a Normal model, what’s the probability the small bowl contains more cereal than the large one?
z=
=
𝟐. 𝟑%
𝑥−𝜇
𝜎
0 −1
0.5
= -2
More accurately use the TI-84 table lookup:
nd VARS
1.
2
No
difference 2. 2: normalcdf(
3. Lower: -99 (lowest value)
P(z<-2) = 2.3%
4. Upper: -2 (our z-score)
5. ALWAYS leave 𝝁: 𝟎
6. ALWAYS leave 𝝈 ∶ 𝟏
BVD Chapter 16 # 33
31. The amount of cereal that can be poured into a small bowl
varies with a mean of 1.5 ounces and a standard deviation of 0.3
ounces. A large bowl holds a mean of 2.5 ounces with a
standard deviation of 0.4 ounces. You open a new box of cereal
and pour one large and one small bowl.
a) What are the mean and standard deviation of the total
amount of cereal in the two bowls?
SD(large bowl – small bowl)
E(large + small)
= E(large) + E(small)
= 𝑽𝒂𝒓 𝑳𝒂𝒓𝒈𝒆 + 𝑽𝒂𝒓(𝑺𝒎𝒂𝒍𝒍 )
= 2.5 + 1.5
= 4 ounces
= 𝟎. 𝟒𝟐 + 𝟎. 𝟑𝟐
= 𝟎. 𝟏𝟔 + 𝟎. 𝟎𝟗
= 𝟎. 𝟐𝟓
= 0.5 ounces
SD is the same as
difference
BVD Chapter 16 # 33
31. The amount of cereal that can be poured into a small bowl varies with a mean of
1.5 ounces and a standard deviation of 0.3 ounces. A large bowl holds a mean of 2.5
ounces with a standard deviation of 0.4 ounces. You open a new box of cereal and
pour one large and one small bowl.
b) If the total follows a Normal model, what’s the probability you
poured out more than 4.5 ounces of cereal in the two bowls together?
𝑥−𝜇
z=
𝜎
=
4.5 − 4
0.5
=1
Requested value
By 68-95-99.7 Rule, what percent of the graph is more
than 1 standard deviations above the mean?
100 – 68 = 32
𝟑𝟐
= 𝟏𝟔%
𝟐
BVD Chapter 16 # 33
31. The amount of cereal that can be poured into a small bowl varies with a mean of 1.5 ounces and a
standard deviation of 0.3 ounces. A large bowl holds a mean of 2.5 ounces with a standard deviation of 0.4
ounces. You open a new box of cereal and pour one large and one small bowl.
b) If the total follows a Normal model, what’s the probability you poured out more than 4.5 ounces of cereal in the two bowls together?
z=
𝟏𝟔%
𝑥−𝜇
𝜎
=
4.5 − 4
0.5
=1
Requested value
Using the 68-95-99.7 Rule, the probability you poured
out more than 4.5 ounces of cereal in the two bowls
together is about 16%.
BVD Chapter 16 # 33
31. The amount of cereal that can be poured into a small bowl varies with a mean of 1.5 ounces and a
standard deviation of 0.3 ounces. A large bowl holds a mean of 2.5 ounces with a standard deviation of 0.4
ounces. You open a new box of cereal and pour one large and one small bowl.
b) If the total follows a Normal model, what’s the probability you poured out more than 4.5 ounces of cereal in the two bowls together?
z=
𝟏𝟓. 𝟗%
𝑥−𝜇
𝜎
=
4.5 − 4
0.5
=1
Requested value
More accurately use the TI-84 table lookup:
1. 2nd VARS
2. 2: normalcdf(
3. Lower: 1 (our z-score)
4. Upper: 99 (highest value)
P(z>1) = 15.9%
5. ALWAYS leave 𝝁: 𝟎
6. ALWAYS leave 𝝈 ∶ 𝟏
BVD Chapter 16 # 33
31. The amount of cereal that can be poured into a small bowl
varies with a mean of 1.5 ounces and a standard deviation of 0.3
ounces. A large bowl holds a mean of 2.5 ounces with a
standard deviation of 0.4 ounces. You open a new box of cereal
and pour one large and one small bowl.
c) The amount of cereal the manufacturer puts in the boxes
is a random variable with a mean of 16.3 ounces and a
standard deviation of 0.2 ounces. Find the expected amount
of cereal left in the box, and the standard deviation.
BVD Chapter 16 # 33
31. The amount of cereal that can be poured into a small bowl varies with a mean of
1.5 ounces and a standard deviation of 0.3 ounces. A large bowl holds a mean of 2.5
ounces with a standard deviation of 0.4 ounces. You open a new box of cereal and
pour one large and one small bowl.
c) The amount of cereal the manufacturer puts in the boxes is a random variable
with a mean of 16.3 ounces and a standard deviation of 0.2 ounces. Find the
expected amount of cereal left in the box, and the standard deviation.
E(Box – L Bwl – S Bwl) = E(Box) – E(L Bwl) + E(s Bwl)
= 16.3 – 2.5 – 1.5 = 12.3 ounces
SD(Box – L Bwl – s Bwl)
=
𝑽𝒂𝒓 𝑩𝒐𝒙 + 𝑽𝒂𝒓 𝑳 𝑩𝒘𝒍 + 𝑽𝒂𝒓(𝒔 𝑩𝒘𝒍 )
= 𝟎. 𝟐𝟐 +𝟎. 𝟒𝟐 +𝟎. 𝟑𝟐 = 𝟎. 𝟎𝟒 + 𝟎. 𝟏𝟔 + 𝟎. 𝟎𝟗
= 𝟎. 𝟐𝟗 = 0.54 ounces
BVD Chapter 16 # 33
31. The amount of cereal that can be poured into a small bowl varies with a mean of
1.5 ounces and a standard deviation of 0.3 ounces. A large bowl holds a mean of 2.5
ounces with a standard deviation of 0.4 ounces. You open a new box of cereal and
pour one large and one small bowl.
c) The amount of cereal the manufacturer puts in the boxes is a random variable
with a mean of 16.3 ounces and a standard deviation of 0.2 ounces. Find the
expected amount of cereal left in the box, and the standard deviation.
E(Box – L Bwl – S Bwl) = E(Box) – E(L Bwl) + E(s Bwl)
= 16.3 – 2.5 – 1.5 = 12.3 ounces
SD(Box – L Bwl – s Bwl)
=
𝑽𝒂𝒓 𝑩𝒐𝒙 + 𝑽𝒂𝒓 𝑳 𝑩𝒘𝒍 + 𝑽𝒂𝒓(𝒔 𝑩𝒘𝒍 )
= 𝟎. 𝟐𝟐 +𝟎. 𝟒𝟐 +𝟎. 𝟑𝟐 = 𝟎. 𝟎𝟒 + 𝟎. 𝟏𝟔 + 𝟎. 𝟎𝟗
= 𝟎. 𝟐𝟗 = 0.54 ounces
Greedy Pig
Example 2:
Round
Bob’s
Roll
Bob’s
Score
Ed’s
Roll
Ed’s
Score
1
2
2
4
4
2
2
4
1
5
-
Sits
4
3
4
Sitting
Sitting
4
4
6
5
11
0
Greedy Pig - Let’s Play
Start: Stand and roll die once. What you roll is your
current score. For example: You roll a six , so you
have six points.
Next you may remain standing and roll again or sit
and keep your score. If you roll again, you add to
your previous score your new roll. UNLESS you roll
a FIVE. If you roll a five, you loss all points and you
must sit down. Game continues until everyone is
sitting. High score at that time wins.
Greedy Pig
No homework. Before you leave you must write a
statement on what you think the best winning
strategy is for playing GREEDY PIG.
Play several games keeping in mind that you are
looking for the best strategy.
START AGAIN.
Chapter 15 Questions
8&14. A survey of students in a large
Introductory Statistics class asked about their
birth order (first or only child, second, etc.) and
which college of the university they were
enrolled in. Here are t data:
College
Birth Order
Arts & Science
Agriculture
Human Ecology
Other
First or
Only
34
52
15
12
Total
113
Second
or Later
23
41
28
18
110
Total
57
93
43
30
223
Chapter 15 Questions
College
8&14. A survey of students in a large Introductory Statistics
class asked about their birth order (first or only child, second,
etc.) and which college of the university they were
enrolled in. Here are t data: Birth Order
Arts & Science
Agriculture
Human Ecology
Other
First or
Only
34
52
15
12
Total
113
Second
or Later
23
41
28
18
110
Total
57
93
43
30
223
a. Suppose we select a student at random from this class.
i. What is the probability we select a Human Ecology student?
Chapter 15 Questions
College
8&14. A survey of students in a large Introductory Statistics
class asked about their birth order (first or only child, second,
etc.) and which college of the university they were
enrolled in. Here are t data: Birth Order
Arts & Science
Agriculture
Human Ecology
Other
First or
Only
34
52
15
12
Total
113
Second
or Later
23
41
28
18
110
Total
57
93
43
30
223
a. Suppose we select a student at random from this class.
ii. What is the probability that we select a first-born student?
Chapter 15 Questions
College
8&14. A survey of students in a large Introductory Statistics
class asked about their birth order (first or only child, second,
etc.) and which college of the university they were
enrolled in. Here are t data:
Arts & Science
Agriculture
Human Ecology
Other
Birth Order
First or Second
Only
or Later
34
23
52
41
15
28
12
18
Total
113
110
Total
57
93
43
30
223
a. Suppose we select a student at random from this class.
iii. What is the probability that the person is first-born and a Human
Ecology student?
Chapter 15 Questions
8&14. A survey of students in a large Introductory Statistics class asked about their birth order
(first or only child, second, etc.) and which college of the university they were
enrolled in. Here are t data:
Birth Order
College
First or Second Total
Only
or Later
Arts & Science
34
23
57
Agriculture
52
41
93
Human Ecology
15
28
43
Other
12
18
30
Total
113
110
223
a. Suppose we select a student at random from this class.
iv. What is the probability that the person is first-born or a Human
Ecology student?
Chapter 15 Questions
College
8&14. A survey of students in a large Introductory Statistics
class asked about their birth order (first or only child, second,
etc.) and which college of the university they were
enrolled in. Here are t data: Birth Order
Arts & Science
Agriculture
Human Ecology
Other
First or
Only
34
52
15
12
Total
113
Second
or Later
23
41
28
18
110
Total
57
93
43
30
223
b. If we select a student at random, what is the probability the person is
an Arts and Sciences student who is a second child (or more)?
Chapter 15 Questions
College
8&14. A survey of students in a large Introductory Statistics
class asked about their birth order (first or only child, second,
etc.) and which college of the university they were
enrolled in. Here are t data: Birth Order
Arts & Science
Agriculture
Human Ecology
Other
First or
Only
34
52
15
12
Total
113
Second
or Later
23
41
28
18
110
Total
57
93
43
30
223
c. Among the Arts and Sciences students, what’s the probability a
student was a second child (or more)?
Chapter 15 Questions
College
8&14. A survey of students in a large Introductory Statistics
class asked about their birth order (first or only child, second,
etc.) and which college of the university they were
enrolled in. Here are t data: Birth Order
Arts & Science
Agriculture
Human Ecology
Other
First or
Only
34
52
15
12
Total
113
Second
or Later
23
41
28
18
110
Total
57
93
43
30
223
d. Among second children (or more), what’s the probability the student
is enrolled in Arts and Sciences?
Chapter 15 Questions
College
8&14. A survey of students in a large Introductory Statistics
class asked about their birth order (first or only child, second,
etc.) and which college of the university they were
enrolled in. Here are t data: Birth Order
Arts & Science
Agriculture
Human Ecology
Other
First or
Only
34
52
15
12
Total
113
Second
or Later
23
41
28
18
110
Total
57
93
43
30
223
e. What’s the probability that a first or only child is enrolled in the
Agriculture College?
Chapter 15 Questions
College
8&14. A survey of students in a large Introductory Statistics
class asked about their birth order (first or only child, second,
etc.) and which college of the university they were
enrolled in. Here are t data: Birth Order
Arts & Science
Agriculture
Human Ecology
Other
First or
Only
34
52
15
12
Total
113
Second
or Later
23
41
28
18
110
Total
57
93
43
30
223
f. What is the probability that an Agriculture student is a first or only
child?
Chapter 15 Questions
22. Fifty-six percent of all American workers have a workplace
retirement plan, 68% have health insurance, and 49% have
both benefits. We select a worker at random.
a. What’s the probability he has neither employer-sponsored
health insurance nor a retirement plan?
Chapter 15 Questions
22. Fifty-six percent of all American workers have a workplace
retirement plan, 68% have health insurance, and 49% have
both benefits. We select a worker at random.
b. What’s the probability he has health insurance if he has a
retirement plan?
Chapter 15 Questions
22. Fifty-six percent of all American workers have a workplace
retirement plan, 68% have health insurance, and 49% have
both benefits. We select a worker at random.
c. Are having health insurance and a retirement plan independent
events? Explain.
Chapter 15 Questions
22. Fifty-six percent of all American workers have a workplace
retirement plan, 68% have health insurance, and 49% have
both benefits. We select a worker at random.
d. Are having these two benefits mutually exclusive? Explain.
Chapter 15 Questions
28. In Exercises 8 and 14 (see first page of this worksheet) we
looked at the birth orders and college choices of some Intro
Stats students.
a. Are enrolling in Agriculture and Human Ecology disjoint? Explain.
Chapter 15 Questions
28. In Exercises 8 and 14 (see first page of this worksheet) we
looked at the birth orders and college choices of some Intro
Stats students.
b. Are enrolling in Agriculture and Human Ecology independent?
Explain.
Chapter 15 Questions
28. In Exercises 8 and 14 (see first page of this worksheet) we
looked at the birth orders and college choices of some Intro
Stats students.
c. Are being first-born and enrolling in Human Ecology disjoint?
Explain.
Chapter 15 Questions
28. In Exercises 8 and 14 (see first page of this worksheet) we
looked at the birth orders and college choices of some Intro
Stats students.
d. Are being first-born and enrolling in Human Ecology
independent? Explain.
Chapter 15 Questions
36&38. A private college report contains these statistics:
70% of incoming freshmen attended public schools.
75% of public school students who enroll as freshmen
eventually graduate.
90% of other freshmen eventually graduate.
a. Is there any evidence that a freshman’s chances to graduate may
depend upon what kind of high school the student attended? Explain.
Chapter 15 Questions
36&38. A private college report contains these statistics:
70% of incoming freshmen attended public schools.
75% of public school students who enroll as freshmen
eventually graduate.
90% of other freshmen eventually graduate.
a. Is there any evidence that a freshman’s chances to graduate may
depend upon what kind of high school the student attended? Explain.
Chapter 15 Questions
36&38. A private college report contains these statistics:
70% of incoming freshmen attended public schools.
75% of public school students who enroll as freshmen
eventually graduate.
90% of other freshmen eventually graduate.
a. Is there any evidence that a freshman’s chances to graduate may
depend upon what kind of high school the student attended? Explain.
P(Grad) = .525 + .27 = .795
P(Grad|Private) = .90
P(Grad) ≠ P(Grad|Private)
So the two are not independent
and there appears to be an
association between graduating
and the type of high school
attended.
Chapter 15 Questions
b. What percent of freshmen eventually graduate?
Chapter 15 Questions
C. Among college graduates, what percent of students attended a
public high school? (Hint: It isn’t 75%. Use a tree diagram.)
𝑃(𝑃𝑢𝑏 ∩ 𝐺𝑟𝑎𝑑)
𝑃 𝑃𝑢𝑏 𝐺𝑟𝑎𝑑 =
𝑃(𝐺𝑟𝑎𝑑)
=
0.525
0.795
≈ 66%
Chapter 15 Questions
43. Police often set up sobriety checkpoints --- roadblocks where drivers
are asked a few brief questions to allow the officer to judge whether or
not the person may have been drinking. If the officer does not suspect a
problem, drivers are released to go on their way. Otherwise, drivers are
detained for a Breathalyzer test that will determine whether or not they
are arrested. The police say that based on the brief initial stop, trained
officers can make the right decision 80% of the time. Suppose the police
operate a sobriety checkpoint after 9 p.m. on a Saturday night, a time
when national traffic safety experts suspect that about 12% of drivers
have been drinking.
Chapter 15 Questions
43. Police often set up sobriety checkpoints --- roadblocks where drivers are asked a
few brief questions to allow the officer to judge whether or not the person may have
been drinking. If the officer does not suspect a problem, drivers are released to go on
their way. Otherwise, drivers are detained for a Breathalyzer test that will determine
whether or not they are arrested. The police say that based on the brief initial stop,
trained officers can make the right decision 80% of the time. Suppose the police
operate a sobriety checkpoint after 9 p.m. on a Saturday night, a time when national
traffic safety experts suspect that about 12% of drivers have been drinking.
Chapter 15 Questions
43. Police often set up sobriety checkpoints --- roadblocks where drivers are asked a
few brief questions to allow the officer to judge whether or not the person may have
been drinking. If the officer does not suspect a problem, drivers are released to go on
their way. Otherwise, drivers are detained for a Breathalyzer test that will determine
whether or not they are arrested. The police say that based on the brief initial stop,
trained officers can make the right decision 80% of the time. Suppose the police
operate a sobriety checkpoint after 9 p.m. on a Saturday night, a time when national
traffic safety experts suspect that about 12% of drivers have been drinking.
a. You are stopped at the checkpoint
and, of course, have not been drinking.
What’s the probability that you are
detained for further testing?
Chapter 15 Questions
43. Police often set up sobriety checkpoints --- roadblocks where drivers are asked a
few brief questions to allow the officer to judge whether or not the person may have
been drinking. If the officer does not suspect a problem, drivers are released to go on
their way. Otherwise, drivers are detained for a Breathalyzer test that will determine
whether or not they are arrested. The police say that based on the brief initial stop,
trained officers can make the right decision 80% of the time. Suppose the police
operate a sobriety checkpoint after 9 p.m. on a Saturday night, a time when national
traffic safety experts suspect that about 12% of drivers have been drinking.
b. What’s the probability that any given driver will be detained?
Chapter 15 Questions
43. Police often set up sobriety checkpoints --- roadblocks where drivers are asked a
few brief questions to allow the officer to judge whether or not the person may have
been drinking. If the officer does not suspect a problem, drivers are released to go on
their way. Otherwise, drivers are detained for a Breathalyzer test that will determine
whether or not they are arrested. The police say that based on the brief initial stop,
trained officers can make the right decision 80% of the time. Suppose the police
operate a sobriety checkpoint after 9 p.m. on a Saturday night, a time when national
traffic safety experts suspect that about 12% of drivers have been drinking.
c. What’s the probability that a driver who is detained has actually
been drinking?
Chapter 15 Questions
43. Police often set up sobriety checkpoints --- roadblocks where drivers are asked a
few brief questions to allow the officer to judge whether or not the person may have
been drinking. If the officer does not suspect a problem, drivers are released to go on
their way. Otherwise, drivers are detained for a Breathalyzer test that will determine
whether or not they are arrested. The police say that based on the brief initial stop,
trained officers can make the right decision 80% of the time. Suppose the police
operate a sobriety checkpoint after 9 p.m. on a Saturday night, a time when national
traffic safety experts suspect that about 12% of drivers have been drinking.
d. What’s the probability that a driver who was released had actually
been drinking?
Chapter 15 Questions
43. Police often set up sobriety checkpoints --- roadblocks where drivers are asked a
few brief questions to allow the officer to judge whether or not the person may have
been drinking. If the officer does not suspect a problem, drivers are released to go on
their way. Otherwise, drivers are detained for a Breathalyzer test that will determine
whether or not they are arrested. The police say that based on the brief initial stop,
trained officers can make the right decision 80% of the time. Suppose the police
operate a sobriety checkpoint after 9 p.m. on a Saturday night, a time when national
traffic safety experts suspect that about 12% of drivers have been drinking.
d. What’s the probability that a driver who was released had actually
been drinking?