# X of Z: MAJOR LEAGUE BASEBALL ATTENDANCE

```X of Z: MAJOR LEAGUE BASEBALL
ATTENDANCE
 Rather than solving for z score first, we may be given
a percentage, then we find the z score, then we find
the specific x value
 Mean MLB attendance in 2009 was 2.43 million with
a standard deviation of .69.
 First, see if percentage is above or below 50%. For
example at least 90% verse at least 44%
X of Z continued…
 Since 90% is above 50%, you know z score must be
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positive and your final x value must be larger than the
mean.
X= &micro;+z(&oacute;)
90% = 1.000-90%= .1000 which is associated with a Z
score of 1.28 using ‘C’ area;
X= 2.43+1.28(.69)= 3.31. For major league team to
draw at least 90% of league attendance they needed to
draw at least 3.31 million in 2009;
What about at least 44% in league attendance?
Since percentage is below 50%, Z must be negative,
Solving for 44%
 –Z(&oacute;) +&micro;
When z score has to be subtracted, write like above.
 44% is associated with a Z score of .15.
 –.15(.69) +2.43= 2.23
 A MLB team would have to draw at least 2.23 million to
be in top 44% of league attendance;
Review of Simple Probability
 The probability of a simple event is a ratio of the
number of favorable outcomes for the event to the
total number of possible outcomes of the event.
 The probability of an event a can be expressed as:
number of favorable outcomes
Pa  
total number of possible outcomes
Find Outcomes of simple events
 For Simple Events – count the outcomes
 Examples:
One Die- 6 outcomes
One coin- 2 outcomes
One deck of cards- 52 outcomes; 4 Aces, 12
face cards, and 36 non-face cards(ie 2-10);
U.S. Roulette Wheel- 38 total slots, 18 red, 18
black, and 2 green;
Basic Probabilities
 What is the probability of getting a red on a
U.S. roulette wheel? U.S. wheel has 18 red
slots, 18 black slots, and 2 green slots.
 Solution: Desired outcome divided by # of
possible outcomes=18/38, .474 or a 47.4%
chance of landing on red;
 What is the probability of drawing a black ace
from a deck of cards; 2/52= .0385 or 3.85%
Compound Events
 Events that cannot occur at the same time
are called mutually exclusive.
 Suppose you want to find the probability of
rolling a 2 or a 4 on a die. P(2 or 4)
 Since a die cannot show both a 2 and a 4 at
the same time, the events are mutually
exclusive.
Compound Mutually Exclusive
or More Events
 P(A or B)= P(A) + P(B) – P( A and B)
 Example: What is probability of drawing a Ace
of Spades or a 2 from a deck of shuffled
cards?
 Solution: 1/52 + 4/52=5/52= .0962=9.62%
 What is the probability of drawing an Ace OR
 Solution: Ace=4/52, spades equal 13/52
minus 1/52= 16/52= .3076 or 30.76%.
1
2
3
4
5
6
1
(1,1)
(1,2)
(1,3)
(1,4)
(1,5)
(1,6)
2
(2,1)
(2,2)
(2,3)
(2,4)
(2,5)
(2,6)
3
(3,1)
(3,2)
(3,3)
(3,4)
(3,5)
(3,6)
4
(4,1)
(4,2)
(4,3)
(4,4)
(4,5)
(4,6)
5
(5,1)
(5,2)
(5,3)
(5,4)
(5,5)
(5,6)
6
(6,1)
(6,2)
(6,3)
(6,4)
(5,6)
(6,6)
Now solve the problem….
Probability of Compound events
P(jack, tails)
4 1
4
( )
 0.04  4%
52 2
104
Compound Event Notations
Compound Events
 When the outcome of one event does not
affect the outcome of a second event, these
are called independent events.
 The probability of two independent events is
found by multiplying the probability of the
first event by the probability of the second
event, minus probability of B given that A has
Multiplication Rule:( ie.
conditional probability]
 What’s the probability of tossing a head and rolling a
6? Is tossing a head independent of rolling a 6? Since
they are independent events we can use
 P (A ∩ B) = P(A) * P(B)|A.
 Prob of tossing a head =1/2 and prob of tossing a 6
=1/6. Solution: Prob of tossing both a head AND a 6
=1/2*1/6= 1/12=.083 =8.3 %.
Birthday continued…
 Event A= friend has B-day on June 15th
 Event B= friend has B-Day on June 15th.
 P (A ∩ B) = P(A) * P(B)|A.
 Are they independent events? Yes.
 Solution: 1/365 * 1/365=.0000075 very small
chance;
Multiplication Rule: Dependent Events
P (A ∩ B) = P(A) * P(B)|A.
 Dependent Events- There are 10 African children and
What’s the probability of choosing an African child
and then another African Child?
 Prob choosing first African child =10/18 but the prob
of choosing second African child drops to 9/17th.
 Then we simply multiply 10/18 by 9/17th= 90/306=
.294 or 29.4% probability of choosing two
consecutive African children without replacement;
Multiplication Rule: Implied AND
 Sometimes AND is implied. Ex: What is the
probability of getting four heads in a row?
 First, ask yourself what is the probability of getting
one head. Which is 50%. multiply this value by the
# of events;
 What is the probability of a woman giving birth to
four consecutive female children?
 First, ask yourself what is the probability of a
woman giving birth to a single female child, we will
assume it is 50%; multiply this value by the # of
events;
A= Jason arrives first
B= Scott arrives last
Are events independent? No.
P (A ∩ B) = P(A) * P(B)|A.
There is only one Jason and one Scott and they
arrive at different times;
You TRY!!
 What is the probability of drawing an Ace and
a King from a deck of cards?
 What is the probability of drawing an Ace OR a
King from a deck of cards?
 What is the probability of drawing a 2 OR a
Heart from a deck of cards?
 Find probability of rolling 3 consecutive 3s with
a fair die?
Fundamental Counting Principle
Fundamental Counting Principle
 The total outcomes of each event are found
by using a tree diagram or by using the
fundamental counting principle.
 Example:
At football games, a student concession
stand sells sandwiches on either wheat or rye
bread. The sandwiches come with salami,
turkey, or ham, and either chips, a brownie,
or fruit. Use a tree diagram to determine the
number of possible sandwich combinations.
Tree diagram with sample space
 Using the fundamental counting principle
2 x
3 x 3 = 18 possibilities
Fundamental Counting Principle
 Professor wants to assign a 3-digit code to each of
his students. How many total codes are possible?
 First step: Decide how many independent events
there are.
 There are three events( ie. one event for each )
digit with no restrictions.
10
10
10
More on the fundamental counting
principle
 Sometimes the number of outcomes changes
after each event depending upon the situation
 Example:
There are 8 students in a student action
group at SJSU. The students want to stand in
a line for their end of the term pictures. How
many different ways could the 8 students
stand for their picture?
Counting principle cont’
The number of ways to arrange the students
can be found by multiplying the number of
choices for each position.
There are eight people from which to choose
for the first position.
After choosing a person for the first position,
there are seven people left from which to
choose for the second position.
Counting Principle
• There are now six choices for the third
position.
• This process continues until there is only one
choice left for the last position.
Let n represent the number of arrangements.
Factorials and Permutations
 Symbol that represents factorial is !
 4! = 4*3*2*1= 24
 7!= 7*6*5*4*3*2*1=5,040
 10!=10*9*8*7*6*5*4*3*2*1=3,628,800
 8! Divided by 4!=
 8*7*6*5*4*3*2*1
4*3*2*1
=8*7*6*5 =1680
Counting Principle Problem
Permutations
 We have four cards each representing one suit. What
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is probability of drawing a heart and a spade without
replacement (ie order matters)? On first draw there is
a one in four chance but in second there is only a one
in three chance. 1/4 * 1/3= 1/12 or 8.33% chance
Permutation formula= n!
= 4!
(n-r)!
(4-2)!
= 4*3*2*1
2*1= 12 different ways(4*3) that two out of four
cards can occur in a specific order. We are only
interested in one of the outcomes(ie a heart then a
spade) divide 1 by 12 = 8.33% chance.
Permutation problem Using
Formula
 We’re only concerned with placing the top
three positions. Since order matters, it is a
permutation problem.
10 possible ways a person can come in first, 9
possible ways a person can come in second,
8 possible ways a person can come in third.
10*9*8=720 possible ways a person can come
in first, second, and third place assuming no
ties
Using the formula: 10!
= 10*9*8=720
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(10-3) !
Combinations: Order Doesn’t Matter!!
Combination Problems
First question we need to ask: Does
order matter?
More combination problems….
First question: Does order
matter?
More difficult Combination problem…..
Does order matter?
Event 1 is picking 2 motors from 8,
and Event 2 is picking 2 switches
from 5;
More Binomial Problems
Binomial Distr Requirements
Binomial Distribution
 P(X) = px q
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Where nCx = n!
n–x
* nCx
=
Combination formula
(n-x)! x!
 p=prob of success in one trial
 q=prob of failure in one trial
 n=number of trials
 x= number of successes in ‘n’ trials
More Binomial Problems
Binomial Distribution Problems
The probability of getting a getting a question correct, a
success, is 1/5 or .20. The probability of not getting
question correct. A failure, is 4/5 or .80
Binomial…
 There are 6 trials, so n=6. We are concerned
with only 2 of them being color blind, so x will
be 2. A ‘success’ =.09 and a failure, or those
who aren’t color blind =1–.09= .91 * 6 men
taken 2 at a time = all different ways men in
study can be ordered;
 p=.09
 q=.91
 n=6
 x=2
One more Binomial cause you love
this stuff!!
p=.20
q= .80
n=18
x=5
Probability Questions
 1. What is the probability of having four female
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children (births) in a row?
2. What is the probability of randomly selecting the
top three finishers of a race, in the exact
order of them finishing, out of a race comprised of six
total racers?
3. What is the probability of getting three heads out of
seven consecutive flips of a coin?
4. What is the probability of having three or fewer
female children out of six consecutive births?
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