Transcript 4-2

Random Variables
A random variable is a variable (usually we
use x), that has a single numerical value,
determined by chance, for each outcome of
a procedure
A probability distribution is a graph, table, or
formula that gives the probability for each
value of the random variable.
Example of a probability distribution
and probability histogram
Outcome of the toss of two coins
0
1
2
¼
½
¼
Probability
x (number of heads) P(x)
0.6
0.5
0.4
0.3
0.2
0.1
0
0
1
2
Number of heads in 2
coin tosses
This is very similar to what we did in Chapter 2,
except in Chapter 2 we were constructing
frequency distributions and histograms based
on observed data.
In this chapter, we are doing similar things, but
to describe what we expect to happen.
Requirements for a Probability
Distribution
 P(x) 1
Where x takes on every possible
value.
Means: Something has to
happen, so sum of all
probabilities is 1.
0  P( x)  1
for each x.
Means: the probability of any
event is between 0 and 1
(inclusive)
Center
For a probability distribution,
  x  P(x)
This gives the mean of the probability
distribution. In other words, it tells us what the
average (mean) outcome is. Notice how each
outcome is weighted by how likely it is to occur.
This is also called the Expected Value of the
distribution, denoted E or E(x).
Example
Outcome of the toss of two coins
(number of heads)
x
P(x)
x . P(x)
0
1
2
¼
½
¼
0.¼=0
1 .½ = ½
2.¼=½
   x  P( x)  0  1 2  1 2  1
Example
Number of matching digits
(of a guess to a random 2-digit number)
x
P(x)
x . P(x)
0
0.81
0.18
0.01
0(0.81) = 0
1
2
1(0.18) = 0.18
2(0.01) = 0.02
  x  P( x)  0  0.18  0.02  0.2
So on average, we will match 0.2 digits. So over 1000 tries,
we would expect to match about 200 digits total.
Variance
For a probability distribution
   x  P( x) 
2
2
2
This gives the variance of the probability
distribution. In other words, it tells us how
spread out we expect the outcomes to be.
As before, Standard Deviation can be found
from the variance:
 
2
Example
Outcome of the toss of two coins (number
of heads)
x
P(x)
x . P(x)
x2 . P(x)
0
¼
0 . ¼ = 0 02 . ¼ = 0
1
½
1 . ½ = ½ 12 . ½ = ½
2
¼
2 . ¼ = ½ 22 . ¼ = 1
   x  P( x)  0  1 2  1 2  1


 2   x 2  P( x)  2  0  1 2  1  12  1 2
Example
Number of matching digits
(of a guess to a random 2-digit number)
x
P(x)
x . P(x)
x2 . P(x)
0
0.81
0.18
0.01
0(0.81) = 0
02(0.81) = 0
1
2
1(0.18) = 0.18 12(0.18) = 0.18
2(0.01) = 0.02 22(0.01) = 0.04
  x  P( x)  0  0.18  0.02  0.2
   x  P( x)   0  .18  .04  .2  0.18
2
2
2
2
You try
Number of Girls out of 4 children
x
P(x)
0
1/16 = 0.0625
1
¼ = 0.25
2
3/8 = 0.375
3
¼ = 0.25
4
1/16 = 0.0625
Homework Part 1
4.2: 3, 5, 7
Read the rest of section 4.2, especially
“Identifying Unusual Results with
Probabilities”, pg 189.