Transcript Stat 200: LGM 7

Presentation 6.
Random Variables
Random Variables

A random variable assigns a number (or
symbol) to each outcome of a random
circumstance.
Example: Lets call our random variable X!
1. Let X = the number of spades in a random
sample of 4 cards from a deck.
2. Let X = the sum of 2 rolls of a six sided die.
3. Let X = the number of people with blue eyes in
a sample of 10 people.
4. Let X = the weight of a randomly chosen
person.
Two types of Quantitative R. Vs

Discrete Random Variables


Result in a countable set of possibilities (e.g. only integer
values.)
Cannot take any possible value in an interval (since there are
uncountable many values in an interval).
Examples:
1. X = the sum of 2 rolls of a six sided die. Outcomes: 2-12
2. X = number of tosses until the first “head”. Outcomes: 1,2,3,…

Continuous Random Variables

The outcome can be in any interval or collection of intervals.
Examples:
1. X = Time spent waiting for the bus.
2. X = the weight of a randomly chosen individual.
Outcomes: 120 lbs, 120.00001lbs, 183.12302 lbs,…
Discrete Random Variables
Probability Notation and Distributions


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
X = the random variable.
k = a number that the discrete r.v. could assume
(a possible outcome!)
P(X=k) is the probability that X equals k.
The probability distribution function (PDF) for a
discrete random variable is a table or rule that
assigns probabilities to the possible outcomes of
a random variable X.
Discrete Random Variables
Example
Assume the probability of a girl is ½. Let X = the
number of girls in a family with 3 children. What is
the probability distribution of X?
Possible Outcomes: 0, 1, 2, or 3 girls.
Event: BBB
Prob: 1/8
X:
0
PDF of X:
BBG
1/8
1
BGB
1/8
1
GBB
1/8
1
BGG
1/8
2
GBG
1/8
2
GGB
1/8
2
GGG
1/8
3
k
0
1
2
3
P(X=k)
1/8
3/8
3/8
1/8
Discrete Random Variables
Cumulative Distribution Function

The Cumulative Distribution Function for a random variable
X is rule or table that provides the probabilities P(X ≤ k).
The term ‘cumulative probability’ refers to the probability
that X is less than or equal to a particular value.
Example: Number of Girls
CDF of X:
k
0
1
2
3
P(X≤k)
1/8 4/8 7/8 1
Discrete Random Variables
Expectations

The expected value of a random variable is the mean
value of the variable X in the space of possible outcomes
(or population). Also can be interpreted as the mean
value from an infinite number of observations of the
random variable. It is also denoted with the Greek letter
μ.
E(X) = μ =
xp
i i
So the expected value of the number of girls is
E(X)= μ = 0*1/8 + 1*3/8 + 2*3/8 + 3*1/8
= 3/8+6/8+3/8
= 12/8 or 1.5 girls!
Special Case of Discrete Random Variables:
Binomial Random Variables
Binomial Experiment is defined by the following:
 There are n “trials” where n is determined in
advance.
 There are two possible outcomes on each trial, a
“success”, and a “failure”.
 The outcomes are independent from one trial to
the next.
 The probability of a “success” remains the same
from one trial to the next and is denoted by p.
Binomial Random Variables

If we have a Binomial experiment with n trials
and p probability of success, then X= number
of successes (out of n trials), is called a binomial
random variable.
What is the sample space of X?
Example: Drawing 10 cards from a deck with
replacement and counting the number of spades.
Then, we have the binomial random variable
X = ______________________________
sample space of X = ________________
PDF, Expectation, and Standard Deviation
of a Binomial Random Variable
n!
p k (1  p) n k
k!(n  k )!
P(X=k) =
E(X) = μ = np
S.D.(X) = σ = np(1  p)
Note: n! = 1x2x…xn, e.g. 3!=1x2x3=6
Example: Number of Girls
k
n!/[k!(n-k)!]
pk(1-p)n-k
P(X=k)
0
3!/[0!*3!]=1
.50(1-.5)3=1/8
1/8
1
3!/[1!*2!]=3
.51(1-.5)2=1/8
3/8
2
3!/[2!*1!]=3
.52(1-.5)1=1/8
3/8
3
3!/[3!*0!]=1
.53(1-.5)0=1/8
1/8
Example Binomial:
Number of Spades in 10 draws:
Probability (X=k)
0.25
0.20
0.15
0.10
0.05
0.00
0
1
2
3
4
5
6
7
K = Number of Spades
8
9
10
Special Case of Continuous Random Variables:
Normal Random Variables
0.12
0.10
0.08
0.06
0.04
Density/Probability
Example: Heights of Males
are Normally Distributed
Probability Density Function
for Heights of Males --->
0.02

In the class of continuous random variables, we are
primarily interested in NORMAL random variables.
These are a continuous random variable with a bell-shaped
distribution. These normal or bell-shaped variables occur
often in nature.
0.00

55
60
65
70
Height (inches)
75
80
Probabilities with Normal RVs
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When we consider Normal Random Variables (or any continuous
r.v.), we are interested in the probability that X falls into some
INTERVAL.
There are infinitely many normal pdf’s (curves). To fully describe a
normal curve, we need the location (mean, μ) and the spread
(s.d., σ).
When talking about the population mean and s.d. we use the
Greek letters μ and σ, when talking about the sample mean and
s.d. we use x and s.
Example: Suppose X is the height of a randomly chosen college
woman. Further suppose that the heights of college women can
be described as a normal, with μ = 65inches, and σ = 2.7 inches.
We might ask:
1.
What is the proportion of women that are shorter than 62 inches?
2.
What is the probability that X is between 65 and 67 inches?
Graphical Representation of
Probabilities
P(65<X<67)
55
60
65
70
Height (inches)
75
80
0.10
0.08
0.00
0.00
0.02
0.02
0.04
0.06
Density
0.08
0.06
0.04
Density
0.10
0.12
0.12
0.14
0.14
P(X<62)
55
60
65
70
Height (inches)
Note: The total area under the curve is equal to 1!
75
80
How to Calculate Probabilities
If you want P(X<x) where x equals some value, first compute a z-score!
z = (Value – mean)/(Standard Deviation)
z = (x-µ)/σ, P(X<x) = P(Z<z) for which we have tables!!
Examples:
1. P(X<62)
z = (62 – 65)/2.7 = -3/2.7 = -1.11
P(X<62) = P(Z < -1.11), now use Normal Table
(Table A.1, page 538 in your text)
P(Z< -1.11) = 13.4%
2. P(65<X<67)
z1 = (65 – 65)/2.7 = 0, z2 = (67 – 65)/2.7 = 1.11
P(65<X<67) = P(0<Z<1.11) = P(Z<1.11) – P(Z<0)
P(Z<1.11) – P(Z<0) = .867 – .5 = .367 or 36.7%
3. P(X>62) = 1- P(X<62) = 1-0.134 = 0.866
Some notes for calculating probabilities
The normal table provides probabilities of the form
P(Z<“number”).
For continuous random variables the probability of being
less than or less of equal than a number are the same.
(e.g. P(Z<2.1) = P(Z  2.1))
It is always a good idea to draw a normal curve and
shade the area corresponding to the probability of
interest.
If c, c1, c2 are some numbers, then
1.
2.
3.
4.

P(Z>c) = 1 – P(Z<c)

P(c1<Z<c2) = P(Z<c2) - P(Z<c1)
Draw a normal curve and shade the areas corresponding to the above probabilities.
Can you see why we have these equalities?
Example: Suppose verbal SAT scores of high-school freshman
are normally distributed with a mean of 500 and a standard
deviation of 50.
 What is the probability of a randomly chosen individual
having a score greater than 600?
z-score = [600-500]/50 = 2
P(X>600) = P(Z>2) = 1- P(Z  2)= 1-P(Z<2) = 1.9772 = 0.228
0.008
0.006
0.004
Density
0.000
300
400
500
Verbal SAT Score
600
700
P(Z>2)
0.002
0.004
0.002
P(X>600)
0.000
Density
0.006
0.008
Note that the only difference in the two graphs below is the
scale on the tow axes. However, the shaded areas are
equal…since the total area under any of this curves is one.
-4300
400
500
600
-2
0 Score 2
Verbal SAT
4
700
What is the probability of a randomly chosen individual having a
score between 400 and 500?
We want P(400<X<500).
z-score1 = z1 = [400-500]/50 = -2
z-score2 = z2 = [500-500]/50 = 0
P(400<X<500) = P(-2 < Z < 0)
= P(Z<0) – P(Z<-2) = .5-.228 = .4772 (from Table)
0.2
0.1
Density
0.3
P(-2<Z<0)
0.0

-4
-2
0
Z-Score
2
4
That is the probability of a randomly chosen student having a score
between 400 and 500 is about .48 or 48%.

What is the probability of a randomly chosen individual having a score
between 350 and 450?
z-score1 = z1 = [350-500]/50 = -3
z-score2 = z2 = [450-500]/50 = -1
P(350<X<450) = P(-3 < Z < -1)
= P(Z<-1) – P(Z<-3)
= .1587-.0013 = .1574 (from normal Table)
0.2
0.0
0.1
Density
0.3
P(-3<Z<-1)
-4
-2
0
2
4
Z-Score
That is the probability of a randomly chosen student having a score
between 350 and 450 is about .16 or 16%.
Approximating binomial distribution
probabilities
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Suppose X is a binomial distribution with
trials n and success probability p.
np >10 and n(1  p) >10
(i.e. the expected number of both,
successes and failures, in the sample is
greater than 10.)
X is approximately a normal random
variable with mean m= np and standard
deviation s= np(1  p)
Normal appr. of binomial variables
Let X be a binomial distribution with n=20 and p=0.5.

Check if the approximation rules are satisfied?
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What is the mean and the s.d. of X?
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Compute P(X≤10) by two ways:
1.
2.
Using minitab to compute P(X≤10) where X is a
binomial r.v. with n=20 and p=0.5, we get
P(X≤10) =0.5881
Using normal approximation: for X a normal r.v.
with μ=______ and σ=________ we get
P(X≤10)=0.5
Summary
Definitions and theory for binomial rv’s

If X is a r.v. representing the number of successes in n independent,
identical trials, with probability of success p remaining constant from trial
to trial, then is called a binomial r.v. with parameters n and p.

The cumulative density function (cdf) of X is P(X ≤ k), for all k.

For k integer between 0 and n we have that
P(X < k) = P(X ≤ k-1)
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Note that is not true for discrete random variables in general!!! Binomial
random variables can take only the integer values 0,1,…,n (since is the
number of successes out of n). If X is not a binomial variable it might be
the case that the possible values of X are 2, 2.5, 3, 3.5 and 4. Then, in
this case P(X < 3) = P(X ≤ 2.5).
For a binomial random variable X with n number of trial and p prob. of
success
μ = E(X) = np
σ = np(1  p)
So…for binomial random variables you do not need to use the general
formula for the expected value of discrete r.v.’s!
Summary
Definitions and theory for Normal r.v’s.

Knowing μ and σ, specifies the particular normal distribution out of the class of
all normal distributions. (Similarly, knowing n and p, specifies a particular
binomial distribution.)
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The pdf of any normal r.v X, also called normal curve, is symmetric, bell

The standard normal random variable has mean 0 and standard deviation
1. We denote it with Z.

shaped and centered at the mean, μ.
We have the tables for all the probabilities of the form P(Z ≤ z). So, for any
normal r.v X, with mean μ and standard deviation σ, we can obtain any
probabilities of interest using the following “standardization theorem’.
If X has a normal distribution with mean μ, and standard deviation σ, then
{(X- μ)/ σ } has a normal distribution with mean 0, and standard deviation 1,
P(X ≤ x) = P [(X- μ)/ σ ≤ (x- μ)/ σ] = P[Z ≤ (x- μ)/ σ] = P(Z ≤ z),
Where z = (x- μ)/ σ, is called the z-score of x.
Summary
Finding Probabilities of X
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First find the z-score of x (or x’s if more than one) to be able to use the tables.
Think what is the area under the curve that corresponds to this probability.
Figure out how you can get this probability using probabilities rules and
values form the tables.
Have in mind that the normal curve is symmetric and that the total area under
the curve is equal to 1.
The empirical rule for the standard deviation on page 44, is valid for all bellshaped distributions (μ ± σ, μ ± 2σ, μ ± 3σ, approximate intervals), but it
is EXACTLY RIGHT in the case of normal distribution. i.e.
P(-1<Z<1) = _______, P(-2<Z<2) = _______, P(-3<Z<3) = ________
How can we find percentiles?
i.e. for a normal r.v. X with mean μ and standard deviation σ , how can we
find x (a value of X), such that P( X ≤ x) = α, where α is α known probability.
e.g. if α = 95% (we want to find the 95th percentile of X).
First we get the α-th percentile for Z,
P(Z≤ z) = 0.95, then z = 1.64.
and we get x using
x= σ z + μ