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)
0
3
4
1
2
5
P(x)
0.001
0.015
0.088
0.264
0.396
0.237

Random Variables
We use probability histograms to graph a probability distribution x
(number of peas with green pods)
0
3
4
1
2
5
P(x)
0.001
0.015
0.088
0.264
0.396
0.237
0,5
0,4
0,3
0,2
0,1
0
Series 1
0 1 2 3 4 5
Название оси

Random Variables
Requirements for a Probability Distribution
1. 𝑃(𝑥) = 1 where x assumes all possible values.
2. 0 ≤ 𝑃 𝑥 ≤ 1 for every individual value of x

Random Variables
Requirements for a Probability Distribution
1. 𝑃(𝑥) = 1 where x assumes all possible values.
2. 0 ≤ 𝑃 𝑥 ≤ 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?
1
2 x
0
3
P(x)
0.19
0.26
0.33
0.13

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

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

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

Random Variables
Mean, Variance, and Standard Deviation
• 𝜇 = [𝑥 ∙ 𝑃 𝑥 ] Mean for a probability distribution

Random Variables
Mean, Variance, and Standard Deviation
• 𝜇 = [𝑥 ∙ 𝑃 𝑥 ]
• 𝜎 2 = 𝑥 − 𝜇 2 ∙ 𝑃 𝑥
Mean for a probability distribution
Variance for a probability distribution

Random Variables
Mean, Variance, and Standard Deviation
• 𝜇 = [𝑥 ∙ 𝑃 𝑥 ]
• 𝜎 2 = 𝑥 − 𝜇 2 ∙ 𝑃 𝑥
• 𝜎 2 = 𝑥 2 ∙ 𝑃 𝑥 − 𝜇 2
Mean for a probability distribution
Variance for a probability distribution
Variance for a probability distribution

Random Variables
Mean, Variance, and Standard Deviation
• 𝜇 = [𝑥 ∙ 𝑃 𝑥 ]
• 𝜎 2 = 𝑥 − 𝜇 2 ∙ 𝑃 𝑥
• 𝜎 2 = 𝑥 2 ∙ 𝑃 𝑥 − 𝜇 2
Mean for a probability distribution
Variance for a probability distribution
Variance for a probability distribution
• 𝜎 = 𝑥 2 ∙ 𝑃 𝑥 − 𝜇 2 Standard Deviation for a probability distribution

Random Variables
Mean, Variance, and Standard Deviation
• 𝜇 = [𝑥 ∙ 𝑃 𝑥 ]
• 𝜎 2 = 𝑥 − 𝜇 2 ∙ 𝑃 𝑥
• 𝜎 2 = 𝑥 2 ∙ 𝑃 𝑥 − 𝜇 2
Mean for a probability distribution
Variance for a probability distribution
Variance for a probability distribution
• 𝜎 = 𝑥 2 ∙ 𝑃 𝑥 − 𝜇 2 Standard Deviation 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)
1
2 x
0
3
4
5
P(x)
0.528
0.360
0.098
0.013
0.001
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)
1
2
3 𝑥
0
4
5
𝑃(𝑥)
0.528
0.360
0.098
0.013
0.001
0+ 𝒙 ⋅ 𝑷(𝒙)
0
0.360
0.196
0.039
0.004
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)
1
2
3 𝑥
0
4
5
Total
𝑃(𝑥)
0.528
0.360
0.098
0.013
0.001
0+
1 𝒙 ⋅ 𝑷(𝒙)
0
0.360
0.196
0.039
0.004
0
.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)
1
2
3
4 𝑥
0
𝑃(𝑥)
0.528
0.360
0.098
0.013
0.001
𝒙 ⋅ 𝑷(𝒙)
0
0.360
0.196
0.039
0.004
𝒙 − 𝝁
𝟐
⋅ 𝑷 𝒙
0 − .599
2 ⋅ 0
1 − .599
2 0.360
2 − .599
2
⋅ 0.098
3 − .599
2 ⋅ 0.013
4 − .599
2 ⋅ 0.001
5
Total
0+
1
0
.599=0.6
5 − .599
2 ⋅ 0
0.5269
= 0.725
=0.7

Random Variables
Round off rule for 𝝁, 𝝈 𝒂𝒏𝒅 𝝈 𝟐
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 𝜇, 𝜎 𝑎𝑛𝑑 𝜎 2 to one decimal place.

Random Variables
Round off rule for 𝝁, 𝝈 𝒂𝒏𝒅 𝝈 𝟐
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 𝜇, 𝜎 𝑎𝑛𝑑 𝜎 2 to one decimal place.
Recall the range rule of thumb

Random Variables
Round off rule for 𝝁, 𝝈 𝒂𝒏𝒅 𝝈 𝟐
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 𝜇, 𝜎 𝑎𝑛𝑑 𝜎 2 to one decimal place.
Recall the range rule of thumb 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒖𝒔𝒖𝒂𝒍 𝒗𝒂𝒍𝒖𝒆 = 𝝁 + 𝟐𝝈

Random Variables
Round off rule for 𝝁, 𝝈 𝒂𝒏𝒅 𝝈 𝟐
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 𝜇, 𝜎 𝑎𝑛𝑑 𝜎 2 to one decimal place.
Recall the range rule of thumb 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒖𝒔𝒖𝒂𝒍 𝒗𝒂𝒍𝒖𝒆 = 𝝁 + 𝟐𝝈 𝒎𝒊𝒏𝒊𝒎𝒖𝒎 𝒖𝒔𝒖𝒂𝒍 𝒗𝒂𝒍𝒖𝒆 = 𝝁 − 𝟐𝝈

Random Variables
Round off rule for 𝝁, 𝝈 𝒂𝒏𝒅 𝝈 𝟐
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 𝜇, 𝜎 𝑎𝑛𝑑 𝜎 2 to one decimal place.
Recall the range rule of thumb 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒖𝒔𝒖𝒂𝒍 𝒗𝒂𝒍𝒖𝒆 = 𝝁 + 𝟐𝝈 𝒎𝒊𝒏𝒊𝒎𝒖𝒎 𝒖𝒔𝒖𝒂𝒍 𝒗𝒂𝒍𝒖𝒆 = 𝝁 − 𝟐𝝈

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)
4
5
2
3
6
0
1
7
8
0+
0+
0.004
0.023
0.087
0.208
0.311
0.276
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.
𝑃(𝑥 𝑜𝑟 𝑚𝑜𝑟𝑒) ≤ 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.
𝑃(𝑥 𝑜𝑟 𝑚𝑜𝑟𝑒) ≤ 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.
𝑃(𝑥 𝑜𝑟 𝑓𝑒𝑤𝑒𝑟) ≤ 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.
𝑃(𝑥 𝑜𝑟 𝑚𝑜𝑟𝑒) ≤ 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.
𝑃(𝑥 𝑜𝑟 𝑓𝑒𝑤𝑒𝑟) ≤ 0.05

x (# of peas with green pods)
0
1
4
5
2
3
6
7
8
Random Variables
P(x)
0+
0+
0.004
0.023
0.087
0.208
0.311
0.276
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?

x (# of peas with green pods)
0
1
4
5
2
3
6
7
8
Random Variables
P(x)
0+
0+
0.004
0.023
0.087
0.208
0.311
0.276
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?

x (# of peas with green pods)
0
1
4
5
2
3
6
7
8
Random Variables
P(x)
0+
0+
0.004
0.023
0.087
0.208
0.311
0.276
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?

x (# of peas with green pods)
0
1
4
5
2
3
6
7
8
Random Variables
P(x)
0+
0+
0.004
0.023
0.087
0.208
0.311
0.276
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?

x (# of peas with green pods)
0
1
4
5
2
3
6
7
8
Random Variables
P(x)
0+
0+
0.004
0.023
0.087
0.208
0.311
0.276
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 𝑃 𝑥 ≤ 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 [𝑥 ∙ 𝑃 𝑥 ]
𝐸 = [𝑥 ∙ 𝑃 𝑥 ]

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 [𝑥 ∙ 𝑃 𝑥 ]
𝐸 = [𝑥 ∙ 𝑃 𝑥 ]

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
Win(net)
Total 𝑥
−5
$175
𝑃(𝑥)
37/38
1/38 𝒙 ⋅ 𝑷(𝒙)
−$4.87
$4.61
−$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
Win(net)
Total 𝑥
−$5
$5
𝑃(𝑥)
20/38
18/38 𝒙 ⋅ 𝑷(𝒙)
−$4.87
$4.61
−$0.26

• 5-2: 1-17 odd ,21, 25, 27