Download Random Variables and Distributions Example x P r(X = x) 2

Random Variables and Distributions
A random variable X is a variable that takes numeri-
cal values associated with random outcomes of an experiment.
A probability distribution species the probabilities
(or chance) associated with each possible value the random variable X can take.
Example
If X represents the number of RED M&Ms in a vending machine size package, then perhaps the probability
distribution of X is something like the following:
x
5
P r(X = x) 282
6 7
2
28
8 9 10 11
5
28
7
28
5
28
4
28
3
28
We read this table by nding a value of X, say 8, then
reading the probability below it:
\The probability of observing 8 M&Ms in a package is 287 , meaning, if we open many many packages
of M&Ms, about 7 out of 28 will have exactly 8
M&Ms within."
This is an example of a discrete random variable.
Rules for Distributions
All probabilities are positive.
All probabilities sum to one.
Random variables that can assume any of the values in
given range of values are called continuous random
variables.
Example of a continuous random variable is the
weight of an ISU student.
The experiment that results in the values of this random
variable is the weighing of ISU students.
The probability distribution of a continuous random
variable is a smooth curve. The areas under the curve
between specied values of the random variable correspond to the probabilities.
To model the probability distribution of many populations we often use the Normal distribution. This involves
the assumption that the associated random variable
X , say, is distributed as a Normal distribution with mean
and standard deviation , both unknown. We say
N (; ).
Here and , are parameters of the population.
X
2
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Most population distributions are bell-shaped and can
be modeled by the Normal distribution. Probability
function of a Normal random variable X is
1 x; 2
1
f (x) = p e; 2 ( )
2
where
= mean : measures the center of the distribution.
= standard deviation : measures the spread.
= 3.1415. . . e = 2.71828. . .
We shall denote the Normal distribution by N (; 2)
where 2 = variance of the distribution. The Normal distribution is symmetric about the mean .
There are innitely many Normal curves, one for each
pair of values of and .
We need to compute the areas under the curve for many
of these for statistical inference.
Table IV lists the computed areas for the Normal distribution with = 0, = 1. We call this the standard
normal distribution and denote it by N (0; 1).
A random variable with a N (0; 1) distribution is usually
denoted by Z .
You need to learn to use Table IV to nd probabilities
(areas) for ranges of values of Z .
Note that Table IV only give probabilities (or areas) of
z -values greater than 0.
The entries in the table give the probability between 0
and a specied z .
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We shall use the facts that
1. Standard Normal Curve is symmetric about 0.
2. The total area under the curve is 1.
Example Find the following probabilities:
a). P (0 Z 2:0)
= 0.4772
[ = P (0 < Z < 2:0)]
b). P (Z > 1:64)
= 0:5 ; P (0 Z 1:64)
= 0.5 ; .4495 = .0505
c). P (Z > ;:74)
= 0:5 + P (;:74 < Z 0)
= 0:5 + P (0 Z < :74)
= 0:5 + :2704 = :7704
d). P (;1:33 < Z < 1:33)
= P (;1:33 < Z < 0) + P (0 < Z 1:33)
= P (0 < Z < 1:33) + P (0 < Z < 1:33)
= 2 .4082 = .8164
e). P (;1:26 Z 2:00)
= P (;1:26 Z 0) + P (0 Z 2:00)
= P (0 Z 1:26) + P (0 Z 2:00)
= 0:3962 + 0:4772 = 0:8734
f). P (1:00 < Z < 2:00)
= P (0 < Z < 2:00) ; P (0 < Z < 1:00)
= 0:4772 ; 0:3413 = 0:1359
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g). P (jZ j > 1:96)
= P (Z < ;1:96 or Z > 1:96)
= 2 P (Z > 1:96)
= 2 :0250 = :05
Example 5.6
Let X = length of time between charges of a calculator.
Assume X N (; 2) with = 100, = 15. Find the
probability that a charge will last for a period between 80
and 120 hrs.
First we need to convert X to a random variable Z that
has the standard Normal distribution, so that we may use
Table IV.
We do this by centering and scaling:
X ;
Z=
If X N (; 2) then Z N (0; 1).
To answer the question compute
the probability
0
80
;
100
X ; 100 120 ; 100 1CA
B
P (80 X 120) = P r @
15
15
15
= P (;1:33 Z 1:33)
= 2 :4082 = :8164
Note that when we do centering and scaling the probability remains the same.
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Given a standard Normal distribution, sometimes you
need to nd the z value that corresponds to a specied
probability.
Examples:
In the following, nd the value of z0 that satises the
given probability statement.
a). P (Z z0) = :67
, P (0 < Z z0) = :17
In Table IV, we look for the value closest to .17 and reado the corresponding z value. Here z0 ' :44.
b). P (Z z0) = :1
, P (0 < Z z0) = .4
In the table the closest value is .3997 giving z0 ' 1:28
c). P (;z0 < Z < z0) = :9
, P (0 < Z < z0) = :45
Thus z0 = 1:645
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