random variable

Report
Statistics
5.2
Quiz 6
A quick quiz consists of a true/false question
followed by a multiple-choice question with four
possible answers (a,b,c,d). An Unprepared
student makes random guesses for both
answers.
a. What is the probability of that both answers
are correct
b. Is guessing a good strategy?
Random Variables
Everyone look at figure 5-1 on page 204 of our book.
• A random variable is a variable (typically represented by x) that has a
single numerical value, determined by chance, for each outcome of a
procedure
Random Variables
Everyone look at figure 5-1 on page 204 of our book.
• A random variable is a variable (typically represented by x) that has a
single numerical value, determined by chance, for each outcome of a
procedure
• A probability distribution is a description that gives the probability for
each value of the random variable. It is often expressed in the format of
a graph, table, or formula.
Random Variables
Everyone look at figure 5-1 on page 204 of our book.
• A random variable is a variable (typically represented by x) that has a
single numerical value, determined by chance, for each outcome of a
procedure
• A probability distribution is a description that gives the probability for
each value of the random variable. It is often expressed in the format of
a graph, table, or formula.
Random Variables
• A discrete random variable has either a finite number of values or a
countable number of values.
Random Variables
• A discrete random variable has either a finite number of values or a
countable number of values.
• A continuous random variable has infinitely many values, and those
value can be associated with measurements on a continuous scale
without gaps or interruptions.
Random Variables
• A discrete random variable has either a finite number of values or a
countable number of values.
• A continuous random variable has infinitely many values, and those
value can be associated with measurements on a continuous scale
without gaps or interruptions.
Random Variables
Determine whether the given random variable is
discrete or continuous.
a. The total amount in (ounces) of soft drinks that you
consumed in the past year.
b. The number of cans of soft drinks that you consumed in
the past year.
c. The number of movies currently playing in U.S. theaters.
d. The running time of a randomly selected movie.
e. The cost of making a randomly selected movie.
Random Variables
We use probability histograms to graph a probability distribution
x
(number of peas
with green pods)
P(x)
0
0.001
1
0.015
2
0.088
3
0.264
4
0.396
5
0.237
Random Variables
We use probability histograms to graph a probability distribution
P(x)
0
0.001
1
0.015
2
0.088
3
0.264
4
0.396
5
0.237
Series 1
Axis Title
x
(number of peas
with green pods)
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
Axis Title
4
5
Random Variables
Requirements for a Probability Distribution
1.
() = 1
where x assumes all possible values.
2. 0 ≤   ≤ 1 for every individual value of x
Random Variables
Requirements for a Probability Distribution
1.
() = 1
where x assumes all possible values.
2. 0 ≤   ≤ 1 for every individual value of x.
Random Variables
Based on a survey conducted by Frank N. Magid Associates,
Table 5-2 lists the probabilities for the number of cell phones in
use per household. Does the table below describe a probability
Distribution?
x
P(x)
0
0.19
1
0.26
2
0.33
3
0.13
Random Variables

Does   = (where  can be 0, 1, 2, 3, or 4) determine a
10
probability distribution?
Random Variables

Does   = (where  can be 0, 1, 2, 3, or 4) determine a
10
probability distribution?
X
P(x)
0
0/10
1
1/10
2
2/10
3
3/10
4
4/10
Random Variables

Does   = (where  can be 0, 1, 2, 3, or 4) determine a
10
probability distribution?
X
P(x)
0
0/10
1
1/10
2
2/10
3
3/10
4
4/10
Total
10/10=1
Random Variables
Mean, Variance, and Standard Deviation
•  = [ ∙   ]
Mean for a probability distribution
Random Variables
Mean, Variance, and Standard Deviation
•  = [ ∙   ]
• 2 =
−
Mean for a probability distribution
2
∙ 
Variance for a probability distribution
Random Variables
Mean, Variance, and Standard Deviation
•  = [ ∙   ]
Mean for a probability distribution
• 2 =
−
2
• 2 =
2 ∙  
∙ 
− 2
Variance for a probability distribution
Variance for a probability distribution
Random Variables
Mean, Variance, and Standard Deviation
•  = [ ∙   ]
Mean for a probability distribution
• 2 =
−
2
• 2 =
2 ∙  
− 2
Variance for a probability distribution
• =
2 ∙  
− 2
Standard Deviation for a probability
distribution
∙ 
Variance for a probability distribution
Random Variables
Mean, Variance, and Standard Deviation
•  = [ ∙   ]
Mean for a probability distribution
• 2 =
−
2
• 2 =
2 ∙  
− 2
Variance for a probability distribution
• =
2 ∙  
− 2
Standard Deviation for a probability
distribution
∙ 
Variance for a probability distribution
Lets do example 5 in excel!, and then do problem 3 on the worksheet
Random Variables
Determine whether the following is a probability distribution and
if so find its mean and standard deviation . Groups of five
babies are randomly selected. In each group, the random
variable x is the number of babies with green eyes (0+ denotes a
positive probability value that is very small)
x
P(x)
0
0.528
1
0.360
2
0.098
3
0.013
4
0.001
5
0+
Random Variables
Determine whether the following is a probability distribution and
if so find its mean and standard deviation . Groups of five
babies are randomly selected. In each group, the random
variable x is the number of babies with green eyes (0+ denotes a
positive probability value that is very small)

()
 ⋅ ()
0
0.528
0
1
0.360
0.360
2
0.098
0.196
3
0.013
0.039
4
0.001
0.004
5
0+
0
Random Variables
Determine whether the following is a probability distribution and
if so find its mean and standard deviation . Groups of five
babies are randomly selected. In each group, the random
variable x is the number of babies with green eyes (0+ denotes a
positive probability value that is very small)

()
 ⋅ ()
0
0.528
0
1
0.360
0.360
2
0.098
0.196
3
0.013
0.039
4
0.001
0.004
5
0+
0
Total
1
.599
Random Variables
Determine whether the following is a probability distribution and
if so find its mean and standard deviation . Groups of five
babies are randomly selected. In each group, the random
variable x is the number of babies with green eyes (0+ denotes a
positive probability value that is very small)


()
 ⋅ ()
0
0.528
0
1
0.360
0.360
2
0.098
0.196
2 − .599
2
⋅ 0.098
3
0.013
0.039
3 − .599
2
⋅ 0.013
4
0.001
0.004
4 − .599
2
⋅ 0.001
5
0+
0
Total
1
.599=0.6
−
⋅ 
0 − .599
2
⋅0
1 − .599 2 0.360
5 − .599
2
⋅0
0.5269 = 0.725=0.7
Random Variables
Round off rule for ,   
Round results by carrying one more decimal place than the number of
decimal places used for the random variable x. If the values of x are
integers, round ,    2 to one decimal place.
Random Variables
Round off rule for ,   
Round results by carrying one more decimal place than the number of
decimal places used for the random variable x. If the values of x are
integers, round ,    2 to one decimal place.
Recall the range rule of thumb
Random Variables
Round off rule for ,   
Round results by carrying one more decimal place than the number of
decimal places used for the random variable x. If the values of x are
integers, round ,    2 to one decimal place.
Recall the range rule of thumb
   =  + 
Random Variables
Round off rule for ,   
Round results by carrying one more decimal place than the number of
decimal places used for the random variable x. If the values of x are
integers, round ,    2 to one decimal place.
Recall the range rule of thumb
   =  + 
   =  − 
Random Variables
Round off rule for ,   
Round results by carrying one more decimal place than the number of
decimal places used for the random variable x. If the values of x are
integers, round ,    2 to one decimal place.
Recall the range rule of thumb
   =  + 
   =  − 
Random Variables
Use the range rule of thumb to identify a range of values
containing the usual number of peas with green pods. Based on
this is it unusual to get only one pea with a green pod? Explain.
x (# of peas with green pods)
P(x)
0
0+
1
0+
2
0.004
3
0.023
4
0.087
5
0.208
6
0.311
7
0.276
8
0.100
Random Variables
Rare Event Rule for Inferential Statistics
If, under a given assumption(such that a coin is fair), the probability of a
particular observed event (such as 992 heads in 1000 tosses of a coin) is
extremely small, we conclude that the assumption is not correct.
Random Variables
Rare Event Rule for Inferential Statistics
If, under a given assumption(such that a coin is fair), the probability of a
particular observed event (such as 992 heads in 1000 tosses of a coin) is
extremely small, we conclude that the assumption is not correct.
• Unusually high number of successes: x successes among n trials is an
unusually high number of successes if the probability of x or more
successes is unlikely with a probability of 0.05 or less.
(  ) ≤ 0.05
Random Variables
Rare Event Rule for Inferential Statistics
If, under a given assumption(such that a coin is fair), the probability of a
particular observed event (such as 992 heads in 1000 tosses of a coin) is
extremely small, we conclude that the assumption is not correct.
• Unusually high number of successes: x successes among n trials is an
unusually high number of successes if the probability of x or more
successes is unlikely with a probability of 0.05 or less.
(  ) ≤ 0.05
• Unusually low number of successes: x successes among n trials is an
unusually low number of successes if the probability of x or fewer
successes is unlikely with a probability of 0.05 or less.
(  ) ≤ 0.05
Random Variables
Rare Event Rule for Inferential Statistics
If, under a given assumption(such that a coin is fair), the probability of a
particular observed event (such as 992 heads in 1000 tosses of a coin) is
extremely small, we conclude that the assumption is not correct.
• Unusually high number of successes: x successes among n trials is an
unusually high number of successes if the probability of x or more
successes is unlikely with a probability of 0.05 or less.
(  ) ≤ 0.05
• Unusually low number of successes: x successes among n trials is an
unusually low number of successes if the probability of x or fewer
successes is unlikely with a probability of 0.05 or less.
(  ) ≤ 0.05
Random Variables
x (# of peas
with green
pods)
P(x)
0
0+
1
0+
2
0.004
3
0.023
4
0.087
5
0.208
6
0.311
7
0.276
8
0.100
a) Find the probability of getting exactly 3
peas with green pods .
b) Find the probability of getting 3 or fewer
peas with green pods.
c) Which Probability is relevant to
determine whether 3 is an unusually low
number of peas with green pods: the result
from part (a) or part (b).
d) Is 3 and unusually low number of peas
with green pods? Why or why not?
Random Variables
x (# of peas
with green
pods)
P(x)
0
0+
1
0+
2
0.004
3
0.023
4
0.087
5
0.208
6
0.311
7
0.276
8
0.100
a) Find the probability of getting exactly 3
peas with green pods . 0.023
b) Find the probability of getting 3 or fewer
peas with green pods.
c) Which Probability is relevant to
determine whether 3 is an unusually low
number of peas with green pods: the result
from part (a) or part (b).
d) Is 3 and unusually low number of peas
with green pods? Why or why not?
Random Variables
x (# of peas
with green
pods)
P(x)
0
0+
1
0+
2
0.004
3
0.023
4
0.087
5
0.208
6
0.311
7
0.276
8
0.100
a) Find the probability of getting exactly 3
peas with green pods . 0.023
b) Find the probability of getting 3 or fewer
peas with green pods. 0.027
c) Which Probability is relevant to
determine whether 3 is an unusually low
number of peas with green pods: the result
from part (a) or part (b).
d) Is 3 and unusually low number of peas
with green pods? Why or why not?
Random Variables
x (# of peas
with green
pods)
P(x)
0
0+
1
0+
2
0.004
3
0.023
4
0.087
5
0.208
6
0.311
7
0.276
8
0.100
a) Find the probability of getting exactly 3
peas with green pods . 0.023
b) Find the probability of getting 3 or fewer
peas with green pods. 0.027
c) Which Probability is relevant to
determine whether 3 is an unusually low
number of peas with green pods: the result
from part (a) or part (b). Part (b)
d) Is 3 and unusually low number of peas
with green pods? Why or why not?
Random Variables
x (# of peas
with green
pods)
P(x)
0
0+
1
0+
2
0.004
3
0.023
4
0.087
5
0.208
6
0.311
7
0.276
8
0.100
a) Find the probability of getting exactly 3
peas with green pods . 0.023
b) Find the probability of getting 3 or fewer
peas with green pods. 0.027
c) Which Probability is relevant to
determine whether 3 is an unusually low
number of peas with green pods: the result
from part (a) or part (b). Part (b)
d) Is 3 and unusually low number of peas
with green pods? Why or why not? Yes
since   ≤ 3 = 0.027 ≤ 0.05
Random Variables
Expected Value
The expected value of a discrete random variable is denoted by E, and it
represents the mean value of its outcomes. It is obtained by finding the
value of [ ∙   ]
 = [ ∙   ]
Random Variables
Expected Value
The expected value of a discrete random variable is denoted by E, and it
represents the mean value of its outcomes. It is obtained by finding the
value of [ ∙   ]
 = [ ∙   ]
Random Variables
You are considering placing a bet on the number 7 in
roulette or red for roulette.
Random Variables
• If you bet $5 on the number 7 in roulette, the
probability of losing $5 is 37/38 and the probability
making a net gain of $175 is 1/38. Let’s find the
expected value if you bet on 7.
Random Variables
• If you bet $5 on the number 7 in roulette, the
probability of losing $5 is 37/38 and the probability
making a net gain of $175 is 1/38. Let’s find the
expected value if you bet on 7.
Event

()
 ⋅ ()
Lose
−5
37/38
−$4.87
$175
1/38
$4.61
Win(net)
Total
−$0.26
Random Variables
• If you bet $5 on red, the probability of losing $5 is
20/38 and the probability making a net gain of $5 is
18/38. Let’s find the expected value if you bet on
red.
Random Variables
• If you bet $5 on red, the probability of losing $5 is
20/38 and the probability making a net gain of $5 is
18/38. Let’s find the expected value if you bet on
red.
Event

()
 ⋅ ()
Lose
−$5
20/38
−$4.87
$5
18/38
$4.61
Win(net)
Total
−$0.26
Homework!!!
• 5-2: 1-17 odd ,21, 25, 27

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