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STAT 111 Introductory Statistics
Lecture 6: Random Variables
May 26, 2004
Today’s Topics
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•
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•
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Special random variables
Expected value of a random variable
Variance of a random variable
Conditional Probability
Multiplication Rule and Independence
Review: Random Variables
• Recall that a random variable is shorthand
notation to denote possible numerical outcomes
of a random phenomenon.
• For repetitions of this phenomenon, we would
expect to get different values for this random
variables.
• Random variables can be either discrete or
continuous.
Discrete Random Variables
• Suppose we have an experiment with only two
possible outcomes, success and failure.
• Examples of this type of experiment:
– Getting heads on a coin flip
– Rolling a 6 on a standard die
• Suppose we have a random variable X such that
– X = 1 if the experiment produces a success
– X = 0 if the experiment produces a failure
• Then X is a Bernoulli random variable.
Discrete Random Variables (cont.)
• If X is a Bernoulli random variable, then the
probability distribution is given by
– P(X = 1) = p
– P(X = 0) = 1 – p
• Going back to our two aforementioned examples:
– P( successful coin flip ) = 1/2
– P( successful roll of the die ) = 1/6
Discrete Random Variables (cont.)
• Suppose we have n independent and identically
distributed Bernoulli random variables Xi.
• Then the sum X of these n Bernoulli random
variables is what we call a binomial random
variable.
• Examples:
– Total number of heads from flipping a coin 10 times
– Total number of 6’s from rolling a die 100 times
Discrete Random Variables (cont.)
• Our binomial random variable X can take on any
values between (and including) 0 and n.
• The probability of X = k, where k = 0, 1, …, n, is
given by
n k
nk
P( X  k )    p (1  p)
k 
• Here,
n
n!
  
 k  k!(n  k )!
Discrete Random Variables (cont.)
• The binomial random variable is fairly important
when we start dealing with sample and
population proportions.
• Suppose the number of trials n becomes very
large. Then, for some fixed number λ,
 k
n k
e

nk
  p (1  p)
k 

k!
• A random variable X that has the probabilities on
the right for k = 0, 1, …, n, is a Poisson random
variable.
Discrete Random Variables (cont.)
• The Poisson distribution was originally used to
approximate the binomial distribution for very
large numbers of trials, but the Poisson was also
found to be good at fitting a variety of random
phenomena.
• One of the earliest phenomena that was fit using a
Poisson random variable was data on Prussian
cavalry.
Discrete Random Variables (cont.)
• During the latter part of the 19th century, Prussian
officials gathered information on the hazards that
horses posed to cavalry soldiers.
• The number of fatalities due to kicks, X, were
recorded for each year (20) and each corps (10).
• The table shows observed counts and also
expected counts obtained by letting λ = 0.61 and
multiplying the probabilities by the total number
of corps-years observed (200).
Discrete Random Variables (cont.)
# of
Deaths
Observed # of
Expected # of
Corps-Years
Corps-Years
0
109
108.7
1
65
66.3
2
22
20.2
3
3
4.1
4
1
0.6
200
199.9
Discrete Random Variables (cont.)
• The Poisson distribution has turned out to be an
excellent model for describing phenomena as
disparate as
–
–
–
–
–
–
–
Radioactive emission
Outbreak of wars
Positioning of stars in space
Telephone calls from pay phones
Traffic accidents along given stretches of road
Mine disasters
Misprints in books
Discrete Random Variables (cont.)
• Go back to the scenario of independent Bernoulli
trials, but instead of asking how many successes
we obtain in n trials, we ask how many trials we
need in order to obtain the first success.
• So if X is a random variable representing the trial
number of the first success, then for x = 1, 2, …,
P(X = x) = p * (1 – p)x – 1
• A random variable with this probability
distribution is a geometric random variable.
Discrete Random Variables (cont.)
• Suppose now that instead of looking at the
number of trials required for the first success, we
want to look at the number of trials required for
the r-th success.
• The random variable in this scenario comes from
the negative binomial distribution.
• The geometric distribution is the special case
where r = 1.
 x  1 r
 p (1  p) x r
P( X  x)  
 r  1
Continuous Random Variables
• Previously discussed continuous random
variables:
– Uniform on an interval [a, b]
– Normal with mean µ and standard deviation σ
• One other continuous probability distribution that
is commonly used to model observed phenomena
is the exponential distribution.
Continuous Random Variables (cont.)
• If a random variable X has the exponential
distribution with parameter λ > 0, then the height
of its density curve at any x > 0 is given by
λe-λx
• A random variable X that has this density curve is
an exponential random variable with parameter λ.
Continuous Random Variables (cont.)
• The exponential distribution is useful for
describing random phenomena such as
– Distance traveled by a molecule of gas before it
collides with another molecule
– Lifetime of mechanical and electrical equipment
(equivalently, failure time)
– Demand for a product
Expected Values and Variances
• Suppose X is a discrete random variable whose
distribution is given as follows:
Value of X
x1
x2
x3
…
xk
Probability
p1
p2
p3
…
pk
• Then the mean of X, or the expected value, is
k
E ( X )   X   x i  pi
i 1
Expected Values and Variances
• The variance of X under the same probability
distribution is
k
Var ( X )   X2  E[( X   X ) 2 ]   ( xi   x ) 2  pi
i 1
• The standard deviation σX of X is the square root
of the variance.
• There is an identical expression for the variance
that may be somewhat faster to calculate in many
cases.
Example: Roll a die
• Let X be the number we obtain as a result of our
roll. The distribution of X, then, is
Value of X
1
2
3
4
5
6
Probability
P(X)
1/6
1/6
1/6
1/6
1/6
1/6
• Determine the expected value and variance of X.
Example: Pick 3 Ticket
• The payoff X of a $1 ticket in the Tri-State Pick 3
Game is $500 with probability 1/1000 and 0 the
rest of the time. Calculate the mean and variance
of the payoff.
Rules for Means
• Let X and Y be (not necessarily independent)
random variables and let a and b be constants.
Let Z = a + bX be a linear transformation of X.
• Rule 1
E(Z) = µZ = E( a + bX ) = a + b E(X) = a + b µX
• Rule 2
E( a X + b Y ) = a E(X) + b E(Y) = a µX + b µY
• On a side note, if X and Y are independent, then
E(XY) = E(X) E(Y)
Rules for Variances
• Let X and Y be random variables, and let a and b
again be constants. Again, let Z = a + b X be a
linear transformation of X.
• Rule 1
Var(Z) = Var(a + b X) = b2 Var(X)
• Rule 2 If X and Y are independent, then
Var(X ± Y) = Var(X) + Var(Y)
• Rule 3 If X and Y have correlation ρ, then
Var(X ± Y) = Var(X) + Var(Y) ± 2ρ σX σY
Rules for Variances (cont.)
• The quantity ρσXσY is also known as the
covariance of X and Y.
• The covariance is a measure of relationship.
• Covariance can also be calculated as follows:
Cov(X, Y) = E(XY) – E(X) E(Y)
• For X and Y independent, the covariance is 0.
• Covariance depends upon the units of the
variables, so what is considered a large
covariance varies.
Example: Pick 3 Ticket
• Suppose you buy a $1 Pick 3 ticket on each of
two different days. The payoffs X and Y on the
two tickets are independent. Let X + Y be the
total payoff. Calculate
– Expected value of the total payoff
– Variance of the total payoff
– Standard deviation of the total payoff
Example: Heights of Women
• The height of young women between 18 and 24
in America is approximately normally distributed
with mean µ = 64.5 and s.d. σ = 2.5.
• Two women are randomly chosen from this age
group.
• What are the mean and s.d. of the difference in
their heights?
• What is the probability that one is at least 5”
taller than the other?
• What is the IQR of heights in this age group?
Conditional Probability
• The probability of an event can change if we
know some other event has occurred.
• The conditional probability of an event gives us
the probability of one event under the condition
that we know the outcome of another event.
• Let A and B be any two events such that P(B) > 0.
The conditional probability of A assuming that B
has already occurred is written P(A | B):
P( A  B)
P( A | B) 
P( B)
Example: Rolling Dice
• Let A be the event that a 4 appears on a single
roll of a fair 6-sided die, and let B be the event
that an even number appears.
– Find P(A | B) and P(B | A).
• Suppose we add another (different-colored so we
can distinguish between the two) die to the mix,
and let C be the event that the sum of the two
dice is greater than 8.
– Find P(A | C) and P(C | A).
Example: Gender of Children
• Suppose we have a family with two children.
Assume all four possible outcomes ({older boy,
younger girl},…) are equally likely. What is the
probability that both are girls given that at least
one is a girl?
• Suppose instead that we ignored the age of the
children and distinguished only three family
types. How would this change the above
probability?
Example: Drawing Cards
• Draw 2 cards off the top of a well-shuffled deck.
• What is the probability that the second card is an
Ace, given that the first card was an Ace?
• On the other hand, consider only the first card for
a minute. Suppose you do not see what the card
is, and your friend tells you the card is a King.
What is the probability that the card is a
diamond?
Multiplication Rule
• The probability that both event A and event B
occur is given by
P(A and B) = P(A) P(B | A) = P(B) P(A | B)
• Here, P(A | B) and P(B | A) have the usual
meaning of being conditional probabilities.
Example: Home Security
• House security experts estimate that an untrained
house dog has a 70% probability of detecting an
intruder – and, given detection, a 50% chance of
scaring the intruder away.
• What is the probability that Fido successfully
thwarts a burglar? (The probability of a trained
watchdog detecting and running off an intruder is
estimated to be around 0.75)
Example: Drawing Chips from an Urn
• An urn contains 5 white chips and 4 blue chips.
Two chips are drawn sequentially and without
replacement. What is the probability of obtaining
the sequence (W, B)?
• The multiplication rule can be extended to higherorder intersections. For example, suppose we
throw 3 red chips and 5 yellow chips into our urn.
Five chips are drawn sequentially and without
replacement. What is the probability of obtaining
the sequence (W, R, W, B, Y)?
Independence
• Recall that two events A and B are independent if
knowing one occurs does not change the
probability that the other occurs.
• When two events are independent, we have that
P(B | A) = P(B) and P(A | B) = P(A)
• Recall our example about the probability of a
single card draw being a diamond given that we
are told it is a King.
Tree Diagrams
• A tree diagram is often helpful for solving more
elaborate calculations, and in particular,
problems that have several stages.
• In a tree diagram, each segment in the tree
represents one stage of the problem.
• Each complete branch shows a possible path.
• Tree diagrams combine both the addition and
multiplication rules.
Example: Tossing a Coin Twice
P(HH) = P(H)P(H)
= (1/2)*(1/2) = 1/4
H
H HH
Second flip
T HT
H TH
First flip
Second flip
T
T
P(HT) = P(H)P(T)
= (1/2)*(1/2) = 1/4
P(TH) = P(T)P(H)
= (1/2)*(1/2) = 1/4
TT
P(TT) = P(T)P(T)
= (1/2)*(1/2) = 1/4
Many Independent Events
• Suppose we have n independent events
A1, A2, …, An. Then the multiplication rule is
P( A1  A2  ...  An )  P( A1 ) P( A2 ) ... P( An )
Example: Ten Dice Rolls
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Roll a dice ten times.
Find P(2 appears 10 times).
Find P(2 appears at least once).
Find P(2 appears less than 5 times).
Example: Height of Women
• Randomly select 8 American young women aged
18 to 24.
• Find P(less than 6 women are taller than 65”).
• Find P(at least 3 women between 62” and 67”).