Transcript P(x) - jmullenkhs

7.1 Discrete & Continuous Random
Variables
AP Statistics
Spring 2010
Discrete Random Variables
When we roll two dice, we can define
the sum to be a variable, X. X can
take on any value from 2 through 12.
Since we don’t know exactly what sum
will appear on a given roll, we call X a
random variable.
Terminology:
A random variable is a variable whose value is a
numerical outcome of a random phenomenon where
one and only one number is assigned to each outcome.
Discrete Random Variable—has a countable number
of possible values.
Probability Distribution—list of values and their
probabilities but can be in a table, graph, formula
which specifies the probability associated with each
possible value the RV can assume. probability mass
function (p.m.f.)
Example:
One form of a p.m.f. for x:
x
x1
x2
P(x) P(x1) P(x2)
…
…
xn
xn
Important things to note:
• Each of the events are mutually exclusive
• The sum of all the probabilities is equal to 1
Note: If “x” is not in the range, then P(X = x) = 0
Example 1: Consider rolling two dice. Define X to be the sum of the
two dice. Construct the probability distribution of X and display it with
a probability histogram.
x
P(x)
2
3
4
5
6
7
8
9
10
11
12
1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36
Example 2: Consider flipping a coin 4 times and recording H or T.
Define X to be the number of Heads flipped. Construct the probability
distribution of X and use it to answer the following questions:
x
0
1
2
3
4
P(x) 1/16 4/16 6/16 4/16 1/16
1.
2.
3.
4.
5.
P(X = 0) =
P(X = 0 or 1) =
P(X > 2) =
P(X < 3) =
P(X > 1) =
1/16
5/16
5/16
15/16
15/16
Example 3: (7.16 p. 705 2nd Ed. ) Weary of the low turnout in student
elections, a college administration decides to choose a sample of three
students to form an advisory board that represents student opinion. Suppose
that 40% of all students oppose the use of student fees to fund student
interest groups and that the opinions of the three students on the board are
independent. Then the probability is 0.4 that each opposes the funding of
interest groups.
(a) Call the three students A, B, and C. What is
the probability that A and B support funding
and C opposes it?
(0.6)(0.6)(0.4) = 0.144
(b) List all possible combinations of opinions that can
be held by students A, B, and C. Then give the
probability of each of these outcomes.
S = Support
O = Oppose
S = {SSS, SSO, SOS, OSS, SOO, OSO, OOS, OOO}
P(SSS) = (0.6)3 = 0.216
P(SSO) = P(SOS) = P(OSS) = (0.6)2(0.4) = 0.144
P(SOO) = P(OSO) = P(OOS) = (0.6)(0.4)2 = 0.096
P(OOO) = (0.4)3 = 0.064
(c) Let the random variable X be the number of
student representatives who oppose the funding
of interest groups. Give the probability
distribution of X.
x
P(x)
0
1
2
3
0.216 0.432 0.288 0.064
(d) Express the event “a majority of the advisory
board opposes funding” in terms of X and find
its probability.
P(x > 2) or P(x > 1)
= 0.288 + 0.064 = 0.352
Continuous Random Variables
Rolling dice and flipping coins result in random variables
whose outcomes are countable. Some situations result
in outcomes that can take on any value over a given
interval.
Terminology:
Continuous Random Variable—represent numerical
values that go on forever; ex: time, distance. The
values of these variables are intervals, not discrete
numbers.
Probability Distribution of a Continuous R. V.—
described by a density curve since we cannot list all
possible outcomes. We view the areas under the curve
as the probability values.
Special Notes:
• The total area under a density curve is 1.
• The probability of an individual outcome is 0.
Example 4: Draw a rectangular density curve whose
height is .5. What is the length?
0.5
Find:
2
P(x > .25) = 1.75 x .5 = 0.875
P(x < .5) = .5 x .5 = 0.25
P(.25 < x < .5) = 0.25 x 0.5 = 0.125
Example 5: Suppose that an opinion poll asks a random sample
of 1500 adults, “Do you happen to jog?” Suppose that the
population proportion who jog is 0.15 and that this distribution
is approximately normally distributed with mean µ = 0.15 and
standard deviation 0.0092.
N~(µ = 0.15, σ = .0092)
Find P( X ≥ .16) = .1385
x
.16  .15 


P z 
  P z 
  P  z  1.08696 
 
.0092 


Normalcdf(.16, 1E99, .15, .0092)
= .7229
Find P( .14 ≤ X ≤ .16)
Normalcdf(.14, .16, .15, .0092)
Example 6: Let the random variable X represent the profit made
on a randomly selected day by a certain store. Assume X is
normal with a mean of $360 and standard deviation $50. The
probability is approximately 0.6 that on a randomly selected day
the store will make less than x0 amount of profit. Find x0.
X ~ N(µ = 360, σ = 50)
P(z < x0) = 0.6
invNorm(0.6, 360, 50)
x0 = $372.67
Example 7: According to a recent AP poll, approximately 40% of American adults
indicated they used the internet to get news and information about political
candidates. Suppose 40% of all American adults use this method to get their political
information. What would happen if you randomly sampled a group of 1500 American
adults and asked them if they used the internet to get this information? Define X to
be the % of your sample that would respond that the internet was their primary
source.
Suppose we are told that the distribution of X is
approximately N(0.4, 0.01265). Use this information
to sketch the probability distribution of X and answer
the following questions:
What is P(X > 0.42)?
Normalcdf(.42, 1E99, 0.4, 0.01265)
= 0.0569
What is P(X < 0.35)?
Normalcdf(-1E99, 0.35, 0.4, 0.01265)
≈0
What is P(your result is within 5% of the actual
% who use the internet as a primary source)?
P(0.35 < x < 0.45) = 0.99992