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5-Minute Check on section 7-2b
Large Numbers
1. The Law of ______
_________ says that as n increases x-bar  μ.
Given X and Y are normally distributed with μx = 4, μy = 7, σx = 2, and
σy = 3; find the following
2. Mean of 2X + 3Y
μ2X+3Y = 2μX + 3μY = 2(4) + 3(7) = 29
3. Variance of 2X + 3Y
σ²2X+3Y = 2²σ²X + 3²σ²Y = 4(4) + 9(3) = 43
4. Mean of 3X – 5
μ3X-5 = 3μX – 5 = 3(4) – 5 = 7
5. Variance of 3X – 5
σ²3X-5 = 3²σ²X = 9(4) = 36
6. Mean of 4X – Y
μ4X-Y = 4μX – μY = 4(4) – 7 = 9
7. Variance of 4X – Y
σ²4X-Y = 4²σ²X + σ²Y = 4²(4) + (9) = 73
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Lesson 7 - R
Review of
Random Variables
Objectives
• Define what is meant by a random variable
• Define a discrete random variable
• Define a continuous random variable
• Explain what is meant by the probability distribution
for a random variable
• Explain what is meant by the law of large numbers
• Calculate the mean and variance of a discrete
random variable
• Calculate the mean and variance of distributions
formed by combining two random variables
Vocabulary
• Nothing
AP Outline Fit:
III. Anticipating Patterns: Exploring random phenomena
using probability and simulation (20%–30%)
A. Probability
2. “Law of Large Numbers” concept
4. Discrete random variables and their probability distributions,
including binomial and geometric
6. Mean (expected value) and standard deviation of a random
variable, and linear transformation of a random variable
B. Combining independent random variables
1. Notion of independence versus dependence
2. Mean and standard deviation for sums and differences of
independent random variables
What we Learned
• Random Variables
– Recognize and define a discrete random variable,
and construct a probability distribution table and
a probability histogram for the random variable
– Recognize and define a continuous random
variable, and determine probabilities of events as
areas under density curves
– Given a Normal random variable, use the standard
Normal table or a graphing calculator to find
probabilities of events as areas under the
standard Normal distribution curve
What we Learned
• Means and Variances of Random Variables
– Calculate the mean and variance of a discrete
random variable. Find the expected payout in a
raffle or similar game of chance
– Use simulation methods and the law of large
numbers to approximate the mean of a
distribution
– Use rules for means and rules for variances to
solve problems involving sums, differences, and
linear combinations of random variables
Using your TI-83 calculator
We can use 1-Var-Stats to calculate the mean and
standard deviation of a discrete random variable given
it’s outcomes and probability
• Type in outcomes in L1
• Type in corresponding probabilities in L2
• Use 1-Var-Stats L1, L2 to get statistics
• Notes:
– Discrete Random Variables have countable (finite) values
– Continuous Random Variables have an interval of values
(infinite)
– Ranges of Random Variables are determined by minimum or
maximum values that they can take on
Discrete Random Variable - Mean
The mean, or expected value [E(x)], of a discrete
random variable is given by the formula
μx = ∑ [x ∙P(x)]
where x is the value of the random variable and
P(x) is the probability of observing x (multiply them
together and add all of them up)
Mean of a Discrete Random Variable Interpretation:
If we run an experiment over and over again, the
law of large numbers helps us conclude that the
difference between x and ux gets closer to 0 as n
(number of repetitions) increases
Discrete Random Variable - Variance
Variance and Standard Deviation of a Discrete RV:
The variance of a discrete random variable is given by:
σ2x = ∑ [(x – μx)2 ∙ P(x)] = ∑[x2 ∙ P(x)] – μ2x
and standard deviation is √σ2
Note: round the mean, variance and standard deviation
to one more decimal place than the values of the
random variable
Probability Laws
• Law of Large Numbers – True
– Sample mean, x, approaches population mean, μ,
as sample size increases
• Law of Small Numbers – False
– No such thing
– Random behavior in short term does not mimic
long-term behavior
• Law of Averages – Bad Statistics
– Eventually everything evens out
– Each trial is independent
Random Variables and Probability
• Area under the probability density function (PDF)
curve between the values of the random variable
determine the probability
• Without calculus the only continuous random
variable PDFs we can use are
– Normal (calculator and tables)
– Uniform (always forms a rectangle)
– Piece-wise linear (other known geometric areas)
• Discrete PDFs are calculated by summing up the
given (or calculated) probabilities
Means and Variances
• Rules for Means
– Means follow the rules for linear combinations (from Algebra)
– When you linearly combine two or more (rules give only the 2
case example) random variables, you combine their means in
the same manner
• E(a + X + bY) = a + E(X) + bE(Y)
• Rules for Variances
– Adding a number to a random variable does not change its
variance
– Multiply a random variable by a number changes the variance
by the square of that number
• V(a + X + bY) = V(X) + b²V(Y)
– When you combine random variables, you always add the
variances
• V(X - Y) = V(X) + V(Y) = V(X + Y)
Summary and Homework
• Summary
– Random Variables (RV)
• Discrete RV – finite outcomes
• Continuous RV – an interval outcomes (infinite)
– Mean, Variance and Standard Deviation of RV
•
•
•
•
Discrete RV – know the formulas
Continuous RV – memorize for each distribution we study
Use your calculator to do the computations
Linear Combinations Rules
– Adding a number changes mean, but not the variance
– Multiplying a number changes mean and variance
• Homework
– pg 505-7; 7.53-62
Problem 1a
The random variable X represents the number of people
that you have to wait behind in line when you go to the
post office to buy stamps at lunch time. The probability
distribution of X is provided below:
X=
Probability =
0
.1
1
.5
2
.3
3
.1
(a) Find the mean number of people that will be in front of
you in the stamp line. Use the definition and show
work.
Mean: ∑ [x ∙P(x)] = (.1)(0) + (.5)(1) + (.3)(2) + (.1)(3)
= 0 + .5 + .6 + .3 = 1.4
Problem 1b
The random variable X represents the number of people
that you have to wait behind in line when you go to the
post office to buy stamps at lunch time. The probability
distribution of X is provided below:
X=
Probability =
0
.1
1
.5
2
.3
3
.1
(b) Find the standard deviation for the number of people
in the line in front of you. Use the definition and show
work.
Var: ∑[x2 ∙ P(x)] – μ2x = ∑ [x2 ∙ P(x)] – μx2
= (0 + .5 + .3(4) + .1(9) ) – 1.96)
= 2.6 – 1.96
= 0.64
St Dev
= 0.8
Problem 2
From the previous problem, let f(X) = 2X + 0.5 represent
the amount of time (in minutes) required for the clerks to
process X people. Show your work and use the shortcut
methods (not the definitions) to find:
(i) The mean number of minutes that you will have to
wait.
μX = 1.4
so μf(X) = 0.5 + 2 μX = 0.5 + 2(1.4) = 3.3 minutes
(ii) The standard deviation of the number of minutes you
will have to wait.
σX = 0.8
so σ²f(X) = 2² σ²X = 4 (0.8)² = 2.56 minutes
σf(X) = 2.56 = 1.6 minutes
Problem 3
While you are at the post office you also need to pick up a
package. The random variable Y represents the number
of people you have to wait behind in the pickup line. The
probability distribution of Y is provided below:
Y=
Probability =
0
.2
1
.3
2
.5
(d) Use your calculator to find the mean number of people
that will be in front of you in this line
Mean: ∑ [x ∙P(x)] = 1.3
(e) Use your calculator to find the standard deviation of
the number of people in this line
Var: ∑[x2 ∙ P(x)] – μ2x = 0.781
Problem 4
Suppose that the numbers of people in the two lines are
independent of each other. Let Z = X + Y represent the
total number of people you will have to wait behind at the
post office. Use the rules we discussed in class to find:
(i) The mean or expected value of Z.
E(Z) = E(X) + E(Y) = 1.4 + 1.3 = 2.7
(ii) The standard deviation of Z.
V(Z) = V(X) + V(Y) = 0.8² + 0.781² = 1.25
σ(Z) = 1.25 = 1.118
Problem 5
The weight of eggs produced by a certain breed of hen
is normally distributed with mean of μ = 65 grams and
standard deviation of σ = 5 grams. What is the
probability that the weight of a dozen (12) randomly
selected eggs is between 750 grams and 800 grams?
E(D) = μD = μE1 + μE2 + … + μE12 = 12  μE
= 12  65 = 780 grams
V(D) = σ²D = σ²E1 + σ²E2 + σ²E3 … + σ²E12 = 12  σ²E
= 12  5² = 300 grams
σD = 300 = 17.32
normalcdf(750, 800, 780, 17.32) = 83.43%
750
780
800