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Means and Variances of Means and Variances of Random VariablesRandom Variables
Means and Variances of Means and Variances of Random VariablesRandom Variables
Activity 1: means of random VariablesActivity 1: means of random Variables
To see how means of random variables work, consider a random variable that takes values {1,1,2,3,5,8}. Then
do the following:
To see how means of random variables work, consider a random variable that takes values {1,1,2,3,5,8}. Then
do the following:1.Calculate the mean of the population:
2.Make a list of all the sample of size 2 from this population. You should have 15 subsets of size 2
3.Find the mean of the 15 x-bar in the third column and compare the result with the population mean.
4.Repeat steps 1-3 for a different (but still small) populations of your choice. Now compare your result with each other.
5.Write a brief statement that describes what you discovered.
1.Calculate the mean of the population:
2.Make a list of all the sample of size 2 from this population. You should have 15 subsets of size 2
3.Find the mean of the 15 x-bar in the third column and compare the result with the population mean.
4.Repeat steps 1-3 for a different (but still small) populations of your choice. Now compare your result with each other.
5.Write a brief statement that describes what you discovered.
Mean of a random VariableMean of a random Variable
The mean of a discrete
random variable X is a
weighted average of the
possible values that the
random variable can take.
The mean of a discrete
random variable X is a
weighted average of the
possible values that the
random variable can take.
exampleexampleA Tri-State Pick 3 game in New Hampshire,
Maine and Vermont let you choose three-digit
number and the state chooses three-digit
winning number at random and pays you $500 if
your number is chosen. Because there are 1000
three digit numbers, you have a probability of
1/1000 of winning. Taking X to be the amount
your ticket pays you, the probability distribution
of X is:
A Tri-State Pick 3 game in New Hampshire,
Maine and Vermont let you choose three-digit
number and the state chooses three-digit
winning number at random and pays you $500 if
your number is chosen. Because there are 1000
three digit numbers, you have a probability of
1/1000 of winning. Taking X to be the amount
your ticket pays you, the probability distribution
of X is:
What is the average payoff from the tickets?What is the average payoff from the tickets?
Payoff X $0 $500
Probability 0.999 0.001
($0 + $500)/2 = $250($0 + $500)/2 = $250
The long-run average pay off is: The long-run average pay off is:
$500(1/1000) + $0(999/1000)$500(1/1000) + $0(999/1000)= $0.50= $0.50
Unlike the sample mean of a group of observations, which gives each observation equal weight, the mean of a random variable weights each outcome xi according to its probability, pi. The common symbol for the mean (also known as the expected value of X) is , formally defined by
The mean of a random variable provides the long-run average of the variable, or the expected average outcome over many observations.
Unlike the sample mean of a group of observations, which gives each observation equal weight, the mean of a random variable weights each outcome xi according to its probability, pi. The common symbol for the mean (also known as the expected value of X) is , formally defined by
The mean of a random variable provides the long-run average of the variable, or the expected average outcome over many observations.
Benford’s LawBenford’s LawCalculating the expected first digitCalculating the expected first digit
First digit X 1 2 3 4 5 6 7 8 9
Probability 1/9 1/9 1/9 1/9 1/9 1/9 1/9 1/9 1/9
First digit V
1 2 3 4 5 6 7 8 9
Probability
0.301 0.176 0.125 0.097 0.079 0.067 0.058 0.051 0.046
Find the Find the of the X distributionof the X distribution
y= 3.4441y= 3.4441Find the Find the of the V distributionof the V distribution
x= 5x= 5
x=
5
x=
5
y=
3.4441
y=
3.4441
Variance of discrete Random VariablesVariance of discrete Random Variables
The variance of a discrete random variable X measures the spread, or variability, of the distribution, and is defined by:
The variance of a discrete random variable X measures the spread, or variability, of the distribution, and is defined by:
The standard deviation is the square root of the
variance.
The standard deviation is the square root of the
variance.
Gabby Sells Cars:Gabby Sells Cars:
Gabby is a sales associate at a large auto
dealership. She motivates herself by using
probability estimates of her sales. For sunny
Saturday in April, she estimates her car sales as
follows:
Gabby is a sales associate at a large auto
dealership. She motivates herself by using
probability estimates of her sales. For sunny
Saturday in April, she estimates her car sales as
follows:
Cars sold: 0 1 2 3
Probability 0.3 0.4 0.2 0.1
Let’s find the mean and variance of XLet’s find the mean and variance of X
Cars sold: X 0 1 2 3
Probability: P 0.3 0.4 0.2 0.1
0 0.3 0.0 (0-1.1)2 (0.3) =0.363
1 0.4 0.4 (1-1.1)2 (0.4) =0.004
2 0.2 0.4 (2-1.1)2 (0.2) =0.162
3 0.1 0.3 (3-1.1)2 (0.1) =0.361
1.1 0.890
Law of Large numberLaw of Large number
Rule for MeansRule for Means
Rules for VariancesRules for Variances
Example:Example:
Cars sold: X 0 1 2 3
Probability: P 0.3 0.4 0.2 0.1
Trucks and SUV: Y 0 1 2
Probability: P 0.4 0.5 0.1
Mean of X:
1.1 cars
Mean of X:
1.1 carsMean of Y:
0.7 T’s & SUV’s
Mean of Y:
0.7 T’s & SUV’s
At her commission rate of 25% of gross profit on each vehicle she sells, Linda
expects to earn $350 for each cars sold and $400 for each truck and SUV’s
sold. So her earnings are:
At her commission rate of 25% of gross profit on each vehicle she sells, Linda
expects to earn $350 for each cars sold and $400 for each truck and SUV’s
sold. So her earnings are:
Z= 350 X + 400YZ= 350 X + 400YCombining rule 1 and 2 her mean earnings
will be:
Combining rule 1 and 2 her mean earnings
will be:Uz= 350 UX + 400 UYUz= 350 UX + 400 UY
Uz= 350 (1.1) + 400 (.7)Uz= 350 (1.1) + 400 (.7)= $665 a day= $665 a day