group4-randomvariableanddistribution-151014015655-lva1-app6891.pdf

Chapter 4: Random Variables and
Distribution
Statistics

2
Where We’re Going
 Develop the notion of a random
variable
 Numerical data and discrete random
variables
 Discrete random variables and their
probabilities

4.1: Two Types of Random
Variables
 A random variable is a variable hat
assumes numerical values associated
with the random outcome of an
experiment, where one (and only one)
numerical value is assigned to each
sample point.
3

Variables
 A discrete random variable can assume a
countable number of values.
 Number of steps to the top of the Eiffel Tower*
 A continuous random variable can
assume any value along a given interval of
a number line.
 The time a tourist stays at the top
once s/he gets there
*Believe it or not, the answer ranges from 1,652 to 1,789. See Great Buildings
4

Variables
 Discrete random variables
 Number of sales
 Number of calls
 Shares of stock
 People in line
 Mistakes per page
 Continuous random
variables
 Length
 Depth
 Volume
 Time
 Weight
5

4.2: Probability Distributions
for Discrete Random Variables
 The probability distribution of a
discrete random variable is a graph,
table or formula that specifies the
probability associated with each
possible outcome the random variable
can assume.
 p(x) ≥ 0 for all values of x
 p(x) = 1
6

4.2: Probability Distributions
for Discrete Random Variables
 Say a random variable
x follows this pattern:
p(x) = (.3)(.7)x-1
for x > 0.
 This table gives the
probabilities (rounded
to two digits) for x
between 1 and 10.
x P(x)
1 .30
2 .21
3 .15
4 .11
5 .07
6 .05
7 .04
8 .02
9 .02
10 .01
7

4.3: Expected Values of
Discrete Random Variables
 The mean, or expected value, of a
discrete random variable is
( ) ( ).
E x xp x
   
8

 The variance of a discrete random
variable x is
 The standard deviation of a discrete
random variable x is
2 2 2
[( ) ] ( ) ( ).
E x x p x
  
   

2 2 2
[( ) ] ( ) ( ).
E x x p x
  
   

9

)
3
3
(
)
2
2
(
)
(
























x
P
x
P
x
P
Chebyshev’s Rule Empirical Rule
≥ 0  .68
≥ .75  .95
≥ .89  1.00
10

11
 In a roulette wheel in a U.S. casino, a $1 bet on
“even” wins $1 if the ball falls on an even number
(same for “odd,” or “red,” or “black”).
 The odds of winning this bet are 47.37%
9986
.
0526
.
5263
.
1
$
4737
.
1
$
5263
.
)
1
$
(
4737
.
)
1
$
(












lose
P
win
P
On average, bettors lose about a nickel for each dollar they put down on a bet like this.
(These are the best bets for patrons.)

4 Properties of Binomial Distribution
1. Fixed number of Trials (n)

2. Two outcomes in a trial, SUCCESS or FAILURE

3. Trials are independent

3. Trials are independent
4. Probability of success (p) remains constant

Tree Diagram
X ~ B(n,p)
X – number of
successes in a trial
X ~ B(3, 1/6)

Is there a formula for calculating
Binomial Probabilities rather than draw
a tree diagram?

There are five things you need to do to
work a binomial story problem.
1. Define Success first. Success must be for a single
trial. Success = "Rolling a 6 on a single die"
2. Define the probability of success (p): p = 1/6
3. Find the probability of failure (q): q = 5/6
4. Define the number of trials: n = 3
5. Define the number of successes out of those trials (r)

group4-randomvariableanddistribution-151014015655-lva1-app6891.pdf

The General Binomial
Probability Formula
r – number of successes out of those trials
n – number of trials
p – probability of success
q – probability of failure
Where: q = 1 - p

The General Binomial
Probability Formula

In the old days, there was a probability of 0.8 of success
in any attempt to make a telephone call. Calculate the
probability of having 7 successes in 10 attempts.

4.5: The Poisson Distribution
39
 The Poisson distribution is a discrete probability
distribution for the counts of events that occur randomly in
a given interval of time (or space).
 The Poisson distribution can be used to calculate the
probabilities of various numbers of "successes" based on
the mean number of successes. In order to apply the
Poisson distribution, the various events must
be independent.

40
Many experimental situations occur in which we observe the
counts of events
within a set unit of time, area, volume, length etc. For example,
 The number of cases of a disease in different towns
 The number of mutations in set sized regions of a chromosome
 The number of dolphin pod sightings along a flight path through
a region
 The number of particles emitted by a radioactive source in a
given time
 The number of births per hour during a given day

41
FORMULA
The formula for the Poisson probability mass function is
where
•e is the base of natural logarithms (2.7183)
•μ is the mean number of "successes"
•x is the number of "successes" in question

42
 EXAMPLE
 The average number of homes sold by the Acme Realty company is 2
homes per day. What is the probability that exactly 3 homes will be
sold tomorrow?
 Solution: This is a Poisson experiment in which we know the following:
 μ = 2;
 x = 3;
 e = 2.71828; since e is a constant equal to approximately 2.71828.
 We plug these values into the Poisson formula as follows:
 P(x; μ) = (e-μ) (μx) / x!
P(3; 2) = (2.71828-2) (23) / 3!
P(3; 2) = (0.13534) (8) / 6
P(3; 2) = 0.180
 Thus, the probability of selling 3 homes tomorrow is 0.180 .

43
EXAMPLE
Suppose you knew that the mean number of calls to a fire station
on a weekday is 8. What is the probability that on a given weekday
there would be 11 calls?
 μ = 8;
 x = 11;
 e = 2.71828; since e is a constant equal to approximately
2.71828.

44
Changing the size of the interval
Suppose we know that births in a hospital occur randomly at an average rate of
1.8 births per hour.
What is the probability that we observe 5 births in a given 2 hour interval?
Well, if births occur randomly at a rate of 1.8 births per 1 hour interval
Then births occur randomly at a rate of 3.6 births per 2 hour interval
Let Y = No. of births in a 2 hour period
P(Y=5) = (e-3.6)(3.65) / (5!)
= 0.13768

45
Sum of two Poisson variables
Now suppose we know that in hospital A births occur randomly at an average rate
of 2.3 births per hour and in hospital B births occur randomly at an average rate
of 3.1 births per hour.
What is the probability that we observe 7 births in total from the two hospitals
in a given 1 hour period?
So if we let X = No. of births in a given hour at hospital A
and Y = No. of births in a given hour at hospital B
P (X + Y = 7) = (e-5.4)(5.47) / (7!)
= 0.11999

46
Cumulative Poisson Probability
A cumulative Poisson probability refers to the
probability that the Poisson random variable is
greater than some specified lower limit and less
than some specified upper limit.

47
Example 1
een on a 1-day safari is 5. What is the probability that tourists will
see fewer than four lions on the next 1-day safari?
Solution: This is a Poisson experiment in
Suppose the average number of lions s which we know the
following:
 μ = 5;
 x = 0, 1, 2, or 3;
 e = 2.71828; since e is a constant equal to approximately
2.71828.
To solve this problem, we need to find the probability that tourists will see
0, 1, 2, or 3 lions. Thus, we need to calculate the sum of four probabilities:
P(0; 5) + P(1; 5) + P(2; 5) + P(3; 5). To compute this sum, we use the Poisson
formula:
P(x < 3, 5) = P(0; 5) + P(1; 5) + P(2; 5) + P(3; 5)
P(x < 3, 5) = [ (e-5)(50) / 0! ] + [ (e-5)(51) / 1! ] + [ (e-5)(52) / 2! ] + [ (e-5)(53) / 3!
]
P(x < 3, 5) = [ (0.006738)(1) / 1 ] + [ (0.006738)(5) / 1 ] + [ (0.006738)(25) / 2
] + [ (0.006738)(125) / 6 ]
P(x < 3, 5) = [ 0.0067 ] + [ 0.03369 ] + [ 0.084224 ] + [ 0.140375 ]
P(x < 3, 5) = 0.2650
Thus, the probability of seeing at no more than 3 lions is 0.2650.

4.6: The Hypergeometric
Distribution
 In the binomial situation, each trial was
independent.
 Drawing cards from a deck and replacing
the drawn card each time
 If the card is not replaced, each trial
depends on the previous trial(s).
 The hypergeometric distribution can be
used in this case.
48

Distribution
49
 Randomly draw n elements from a set
of N elements, without replacement.
Assume there are r successes and N-r
failures in the N elements.
 The hypergeometric random variable
is the number of successes, x, drawn
from the r available in the n selections.

Distribution
50



























n
N
x
n
r
N
x
r
x
P )
(
where
N = the total number of elements
r = number of successes in the N elements
n = number of elements drawn
X = the number of successes in the n elements

Distribution
51



























n
N
x
n
r
N
x
r
x
P )
(
)
1
(
)
(
)
(
2
2





N
N
n
N
n
r
N
r
N
nr



Distribution
44
.
22
.
2
)
2
(
)
2
(
)
2
or
2
(
22
.
45
)
1
)(
10
(
2
10
2
2
5
10
2
5
)
2
(
)
2
(









































F
P
M
P
F
M
P
F
P
M
P
 Suppose a customer at a pet store wants to buy two hamsters
for his daughter, but he wants two males or two females (i.e.,
he wants only two hamsters in a few months)
 If there are ten hamsters, five male and five female, what is the
probability of drawing two of the same sex? (With hamsters,
it’s virtually a random selection.)
52

Continuous Random Variable
53
The normal distribution refers to a family
of continuous probability
distributions described by the normal
equation.
• Normal Distribution

54
z = (X - μ) / σ
where X is a normal random variable, μ is the
mean of X, and σ is the standard deviation of X

55
Example
An average light bulb manufactured by the Acme
Corporation lasts 300 days with a standard deviation
of 50 days. Assuming that bulb life is normally
distributed, what is the probability that an Acme light
bulb will last at most 365 days?
Solution:
The value of the normal random variable is 365 days.
•The mean is equal to 300 days.
•The standard deviation is equal to 50 days.
z = (X - μ) / σ = (365-300)/50
z= 1.3
The answer is: P( X < 365) = 0.90. Hence, there
is a 90% chance that a light bulb will burn out
within 365 days.

56
• Standard Normal Distribution
The standard normal distribution is a special case of the normal
distribution. It is the distribution that occurs when a normal random
variable has a mean of zero and a standard deviation of one.
Standard Score (aka, z Score)
The normal random variable of a standard normal distribution is called
a standard score or a z-score. Every normal random variable X can be
transformed into a z score via the following equation:
z = (X - μ) / σ
where X is a normal random variable, μ is the mean of X, and σ is the
standard deviation of X.

57
The standard normal distribution is a special case of the normal
distribution. It is the distribution that occurs when a normal random
variable has a mean of zero and a standard deviation of one.
Standard Score (aka, z Score)
The normal random variable of a standard normal distribution is called
a standard score or a z-score. Every normal random variable X can be
transformed into a z score via the following equation:
z = (X - μ) / σ
where X is a normal random variable, μ is the mean of X, and σ is the
standard deviation of X.

58
Example
Molly earned a score of 940 on a national achievement test. The mean test
score was 850 with a standard deviation of 100. What proportion of
students had a higher score than Molly? (Assume that test scores are
normally distributed.)
(A) 0.10
(B) 0.18
(C) 0.50
(D) 0.82
(E) 0.90

59
•First, we transform Molly's test score into a z-score, using the z-score
transformation equation.
z = (X - μ) / σ = (940 - 850) / 100 = 0.90
•Then, using the standard normal distribution table, we find the cumulative
probability associated with the z-score. In this case, we find P(Z < 0.90) =
0.8159.
•Therefore, the P(Z > 0.90) = 1 - P(Z < 0.90) = 1 - 0.8159 = 0.1841.
Thus, we estimate that 18.41 percent of the students tested had a higher
score than Molly.

group4-randomvariableanddistribution-151014015655-lva1-app6891.pdf

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