Transcript Probability Distributions

Chapter 5
Discrete Probability
Distributions
Overview
Introduction
O 5-1 Probability Distributions
O 5-2 Mean, Variance, Standard Deviation , and
Expectation
O 5-3 The Binomial Distribution
Probability Distributions
O A random variable (x) is a variable whose values
are
discrete
determined by chance.
continuous
O A discrete probability distribution consists of the
values a random variable can assume and the
corresponding probabilities of the values.
Important remarks
The sum of the probabilities of all events in
a sample space add up to 1. ∑ p(x) = 1
Each probability is between 0 and 1,
inclusively. 0 ≤ P(x) ≤ 1
For example:
tossed a coin S={T,H}
O X= NUMBER OF HEADS
P(x=0)
P(x=1)
𝟏
= P(T) =
𝟐
𝟏
=P(H) =
𝟐
X
P(X)
0
𝟏
𝟐
1
𝟏
𝟐
Example 5-1: Rolling a Die
Construct a probability distribution for rolling a
single die.
Solution:
For example:
tossed 2 coins S={TT,HT,TH,HH}
O X= NUMBER OF HEADS
P(x=0) = P(TT)
𝟏𝟏 𝟏
= =𝟒
𝟐𝟐
P(x=1) =P(HT)+P(TH)
𝟏 𝟏 𝟏 𝟏 𝟐
= . + . =
𝟐 𝟐 𝟐 𝟐 𝟒
𝟏𝟏 𝟏
P(x=2)=P(HH)= =
𝟐𝟐 𝟒
X
P(X)
0
𝟏
𝟒
1
𝟐
𝟒
2
𝟏
𝟒
Example 5-2: Tossing Coins
Represent graphically the probability distribution for the
sample space for tossing three coins.
X=number of head.
.
Solution:
•A probability distribution can be graph by using:
a) Ogive
b) Polygon
c) Bar chart.
------------------------------------------------------------•
A box contains 6 balls. One is numbered 2, three are
numbered 3 and two are numbered 4. Construct a
probability distribution for the numbers on the balls.
Mean, Variance, Standard
Deviation, and Expectation
O Mean:
𝜇 = ∑𝑋. 𝑃(𝑋)
O Variance:
𝝈𝟐 =∑ 𝑋 2 . 𝑃
𝑋
− 𝝁𝟐
Example (5-5) + (5-9) : Rolling a Die
Find the mean, variance and standard deviation of
the number of spots that appear when a die is tossed.
.
Solution:
mean:
   X  P X 
 1 16  2  16  3  16  4  16  5  16  6  16

21
6
 3.5
Variance and s.d:
 2    X 2  P  X     2
  1   2   3   4  16
2
2
2
1
6
2
1
6
 5   6    3.5 
2
  2.9
2
2
1
6
,
2
1
6
1
6
  1.7
2
Example 5-8: Trips of 5 Nights or More
The probability distribution shown represents the
number of trips of five nights or more that
American adults take per year. (That is, 6% do
not take any trips lasting five nights or more,
70% take one trip lasting five nights or more per
year, etc.) Find the mean.
.
Solution :
   X  P X 
 0  0.06   1 0.70   2  0.20 
 3  0.03  4  0.01
 1.2
16
Example 5-11: On Hold for Talk Radio
A talk radio station has four telephone lines. If the
host is unable to talk (i.e., during a commercial) or is
talking to a person, the other callers are placed on
hold. When all lines are in use, others who are trying
to call in get a busy signal. The probability that 0, 1,
2, 3, or 4 people will get through is shown in the
distribution. Find the variance and standard deviation
for the distribution.
Solution:
  0  0.18   1 0.34   2  0.23
 3  0.21  4  0.04   1.6
  0  0.18  1  0.34   2  0.23
2
2
2
2
 3  0.21  4  0.04   1.6 
2
  1.2
2
2
,
  1.1
2
18
Expectation
O The expected value, or expectation, of a discrete
random variable of a probability distribution is the
theoretical average of the variable.
O The expected value is, by definition, the mean of the
probability distribution.
EX    X PX
One thousand tickets are sold at $1 each for a color
television valued at $350. What is the expected
value of the gain (‫ )مكسب‬if you purchase one ticket?
Solution :
Solution :
Gain(X)
Win
Lose
$349
-$1
Probability P(X)
An alternate solution :
Note: This PowerPoint is only a summary and your main source should be the book.
Example 5-13: Winning Tickets
One thousand tickets are sold at $1 each for four
prizes of $100, $50, $25, and $10. After each
prize drawing, the winning ticket is then returned
to the pool of tickets. What is the expected value
if you purchase two tickets?
Solution :
$98
Gain X
Probability
P(X)
$48
$23
$8
-$2
2
2
2 992
2
1000 1000 1000 1000 1000
2
2
2
E  X   $98  1000
 $48  1000
 $23  1000
992
2
 $8  1000
  $2   1000
 $1.63
An alternate solution :
The Binomial Distribution
5-3 The Binomial Distribution
•
Many types of probability problems have
only two possible outcomes or they can be
reduced to two outcomes.
•
Examples include: when a coin is tossed it
can land on heads or tails, when a baby is
born it is either a boy or girl, etc.
The Binomial Distribution
The binomial experiment is a probability
experiment that satisfies these requirements:
1. Each trial can have only two possible
outcomes—success or failure.
2. There must be a fixed number of trials.
3. The outcomes of each trial must be
independent of each other.
4. The probability of success must remain the
same for each trial.
•The number of outcome for each trail a binomial experiment
is:
a) 2
b) 1
c) 0
-------------------------------------------------------------------------•Which of the following is a binomial experiment:
a) Asking 200 people what kind of exercise they play.
b) Asking 300 people if they playing exercise.
c) Asking 200 people about their favorite drink.
** Which of the following is not a binomial experiment?
a) Rolling a die 40 times to see how many even number occur.
b) Observe the gender of the babies born at a local hospital.
c) Asking 100 people if they smoke.
d) Tossing a coin 50 times to see how many heads occur.
Notation for the Binomial Distribution
P(S)
The symbol for the probability of success
P(F)
The symbol for the probability of failure
p
The numerical probability of success
q
The numerical probability of failure
P(S) = p and P(F) = 1 – p = q
n
The number of trials
X
The number of successes
Note that X = 0, 1, 2, 3,...,n
The Binomial Distribution
In a binomial experiment, the probability
of exactly X successes in n trials is
n!
X
n X
P X  
 p q
 n - X ! X !
or
P X  
n
Cx
number of possible
desired outcomes
 p q
X
n X
probability of a
desired outcome
Mean, Variance and Standard
Deviation for binomial
The mean , variance and SD of a variable
that the binomial distribution can be
found by using the following formulas:
Mean:
𝝁 = 𝒏𝒑
Variance:
𝝈𝟐 = 𝒏𝒑𝒒
SD:
𝝈 = 𝒏𝒑𝒒
A coin is tossed 3 times. Find the probability of getting exactly
heads.
Solution :
Note: This PowerPoint is only a summary and your main source should be the book.
Example 5-16: Survey on Doctor Visits
A survey found that one out of five Americans say
he or she has visited a doctor in any given month.
If 10 people are selected at random, find the
probability that exactly 3 will have visited a doctor
last month.
Solution:
P X  
n!
 p X  q n X
 n - X ! X !
n  10,"one out of five"  p  15 , X  3
10!  1 
P  3 
 
7!3!  5 
3
7
4
    0.201
5
Example 5-17: Survey on Employment
A survey from Teenage Research Unlimited
(Northbrook, Illinois) found that 30% of
teenage consumers receive their spending
money (‫ )يستلم مال إنفاقه‬from part-time jobs. If
5 teenagers are selected at random, find the
probability that at least 3 of them will have
part-time jobs.
Solution:
n  5, p  0.30,"at least 3"  X  3, 4,5
5!
3
2
P  3 
  0.30    0.70 
2!3!
 0.132
5!
4
1
P  4 
  0.30    0.70   0.028
1!4!
P X
5!
5
0
P  5 
  0.30    0.70   0.002
0!5!
 3  0.132
0.028
0.002
 0.162
Example 5-21: tossing a coin
A coin is tossed 4 times
Find the mean, variance and standard deviation of
number of heads that will be obtained.
Example 5-22: Rolling a die
A die is rolled 360 times , find the mean ,
variance and standerd deviation of the
number of 4s that will be rolled .
Solution:
n= 360,
𝟏
p=
𝟔
,
𝟓
q=
𝟔
1
𝜇 = 𝑛. 𝑝 = 360
= 60
6
1 5
2
𝜎 = 𝑛. 𝑝. 𝑞 = 360
= 50
6 6
𝜎 = 50 = 7.07
** How many times a die is rolled when the
mean of the numbers greater than 4 that will
be rolled = 20?
a) 60
1.A student takes a 3 questions
multiple choices quiz with 4
choices for each question. If the
student guesses (‫ )تخمين‬at
random on each question,
What is probability that
student gets exactly 2 question
is wrong?
Revision
• At a local university 62.3% of incoming first-year students
have computers. If 3 students are selected at random, find the
probability at least one has a computer.
a) 0.946
--------------------------------------------------------------------------• Using the probability distribution to answer a questions:
X
1
2
3
4
P(X)
0.2
0.3
K
0.2
** What is the value of K ?
a) 0.2
b) 0.3
** What is the probability value of x=2?
a) 0.2
b) 0.3
• A coin tossed 56 times.
Answer the following questions:
*1* the mean for the number of heads that will be tossed:
a) 14
b) 28
--------------------------------------------------------------------------*2* the variance for the number of tails that will be tossed:
a) 14
b) 28
--------------------------------------------------------------------------*3* the standard deviation for the number of tails that will be tossed:
a) 14
b) 28
• What is the value K would be needed to complete the
following probability distribution:
X
1
2
3
4
P(X)
0.15
K
0.33
0.20
a) 0.32
b) 0.40
--------------------------------------------------------------------------• If a family has 3 children, find the probability distribution for
the number of boys in a family:
a)
X
0
1
2
3
b)
P(X)
2/8
3/8
3/8
1/8
X
0
1
2
3
P(X)
1/8
3/8
3/8
1/8
• If we have the following probability distribution:
P( X ) 5 C x ( 14 ) x ( 34 ) 5 x ; x  0,1,2,3,4,5
Find the mean of the distribution:
a) 5/4
b) 1/4
Find the variance of the distribution:
a) 15/16
b) 16/18
•Home work
• In the past year, 80% of businesses have eliminate jobs. If 6
businesses are selected at random, find the probability that at
least 5 have eliminated jobs during the last year.
a) 0.655
b) 0.233
c) 0.455
• If 60% of all women are employed outside the home, find the
probability that in a sample of 20 women, exactly 15 are not
employed outside the home.
a) 0.075
b) 0.925
c) 0.999
--------------------------------------------------------------------------• One thousand tickets are sold at $3 each for a PC value at $1600.
What is the expected value of the gain if a person purchases one
ticket?
a) -$1.4
b) -$1.8
c) $1.8
X
0
1
2
P(X)
0.5
0.1
0.3
• The value of the mean for the previous probability
distribution is:
a) 0.7
b) 0.4
--------------------------------------------------------------------------• If X is discrete r.v with  XP(X )  2 and  X 2 P( X )  5.5
Then the variance equal:
a) 1.5
b) 2
•