Download Chapter 5 - CLSU Open University

Special Probability
Distributions
5
5.1 Some Discrete Probability Distributions
5.2 Some Continuous Probability Distributions
5.3 Normal and Poisson Approximations to the Binomial Distritution
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Learning Objectives
It is expected that you will be able to do the following:
1. Describe the characteristics of a given probability distribution
function.
2. Compute probabilities using a given probability distribution
function.
3. Describe the characteristics of the Normal probability
distribution
4. Determine the probability of an observation in a given interval
on a Normal probability distribution
5. Use the Normal probability distribution to approximate the
Binomial distribution
6. Use the Poisson probability distribution to approximate the
Binomial probability distribution,
A group or family of probability distributions refers to a
collection of probability distributions that is indexed by a quantity
called a parameter. A parameter is a numerical characteristic of a
probability distribution.
Very often, observations generated by different random
experiments behave similarly, and as a result, the random variables
associated with these experiments can be expressed essentially by the
same probability distribution and therefore can be represented by a
common formula. Let us now first look at some of these random
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experiments that generates discrete random variables, and their
corresponding probability functions.
5.1 Some Discrete Probability Distributions
5.1.1 Discrete Uniform Distribution
Consider the experiment of randomly selecting a number from a
set consisting of the numbers x1, x2, …, xN. If we let X to be the
number selected, and assuming that each of the numbers are equally
likely to be chosen, then the probability of each number being selected
is equal to 1/N. A special case is when the numbers x1, x2, …, xN
coincides with the numbers 1, 2, … , N. In this specific case, the
random variable X is called a discrete uniform random variable with
parameter N.
Definition 5.1 Discrete Uniform Distribution A random variable X
having a probability mass function given by
1

f(x)  f(x, N)   N
0

for x  1,2,..., N  1
  I 1,2,..., N ( x )
 N
otherwise
where the parameter N ranges over the positive integers, is
called a discrete uniform random variable.
Theorem 5.1 If X has a discrete uniform distribution, then
E[X] =
N 1
2
and variance
var[X] =
N
 1
12
2
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Proof:
N
E[X]
 i
i 1
var[X]  E[X 2 ]  ( E[X]) 2
1
N
1  N  1
 i


N  2 
i 1
N

1 N
i
N i 1
1  N ( N ! ) 


N
2

N 1

2

2
2
N2  1

12
Example 1 In the die rolling experiment, each possible outcome in
the sample space  = {1,2,3,4,5,6} can occur with probability
1/6. Here, X, the number of dots in the upturned face of the die
is a discrete random variable, with probability mass function
f(x;6) = 1/6
for
x = 1,2,3,4,5,6.
The expected value of X is E[X] = 3.5
and its variance is Var[X] = 35/12
///
In example 1, X is a discrete uniform random variable since its
possible values are from 1 to 6, and each of these values are equally
likely to occur with probability 1/6.
5.1.2 Bernoulli Distribution
Now, consider an experiment which may result only into any of
the two possible outcomes, either a success or a failure. Such an
experiment is called a Bernoulli trial. If we define a random variable
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X to be equal to 1 when the trial’s result is a success, and 0 if
otherwise, then X is called a Bernoulli random variable. Given the
probability of success p as a constant value, then formally this
distribution is defined as follows:
Definition 5.2 Bernoulli Distribution
A random variable X is
defined to have a Bernoulli distribution if the probability mass
function of X is given by
p x (1 - p)1-x
f x (x)  f x (x; p)  
0
for x  0 or 1
x
1-x
  p (1 - p) I 0,1( x )
otherwise

where the parameter p satisfies 0< p < 1. 1-p is often denoted
by q.
Theorem 5.2
Proof:
If X has a Bernoulli distribution, then E[X] = p and
variance var[X] = pq.
E[X] =0q + 1p = p
Var[X] = E[X2] – (E[X])2
= 02q + 12p – p2
= p - p2
= p(1-p)
= pq
5.1.3 Binomial Distribution
Consider now n performances of the Bernoulli trial, that is, n
performances of an experiment consisting of only two possible
outcomes labeled as a success or a failure, with the probability of
success in each trial equal to a constant p, and the probability of
failure is q = 1-p. This type of experiment is called a Binomial
random experiment, and the random variable X which denotes the
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number of successes out of the n trials is called a binomial random
variable.
The binomial experiment has the following characteristics:
(1) The experiment consists of n trials.
(2) The n trials are independent.
(3) Each trial may result only into two possible outcomes,
either a success or a failure.
(4) The probability of a success denoted by p is constant from
trial to trial.
The binomial experiment is similar to sampling or selecting n
objects with replacement, wherein the group being sampled consists
of two types of items, say types A and B, so that randomly selecting
an object of type A is a success, and getting an object of type B is a
failure. Since before each draw, the object is replaced back into the
group, then the probability of success p (which is the proportion of A
objects in the group) is constant in each draw.
Definition 5.3 Binomial Distribution A random variable X is
defined to have a Binomial distribution if the probability mass
function of X is given by
 n  x

n-x
 p (1 - p) for x  0, 1,2,..., n 
f x (x)  f x (x; n, p)   x 

0

otherwise


n
   p x (1 - p) n-x I 0,1,2, ... n( x )
x
n
The   in the probability function of the binomial distribution
x
 
is the number of combinations out of n trials which has exactly x
successes and n-x failures.
Theorem 5.3
If X has a Binomial distribution, then
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E[X] = np
and
var[X] = npq
Example 2
Find the probability of obtaining exactly 4 heads in 7
tosses of a fair coin.
Solution:
This is a binomial random experiment. There are n=7
tosses (trials) and the probability of obtaining a head in each
toss is equal to a constant p, which we may assume to be equal
to 1/2 since the coin is fair, and there are only two possible
outcomes in each toss; the coming out of a head is regarded as a
success, so that the random variable of interest X is defined as
the number of successes or heads out of 7 tosses. The possible
values of X are 0,1,2,3,4,5,6, and 7, and we want to find the
probability that X = 4.
1
P(X = 4) = f  4;7,  =
2

7  1   1 
    1 - 
 4  2   2 
4
4
7! 1
1
=     
7- 4
3
 3!4!  2   2 
= 0.2734
///
Example 3
An archer can hit the target bull’s eye 80% of the
time. If he is allowed to shoot 5 arrows, find the probability
that he will get 3 bull’s eyes.
Solution
n = 5, x =3, and p = 0.8
 5
P(X=3) = f 3;5,0.8 =   0.83 0.2 2
3
 
= 0.2048
///
92
Example 4 In example 4, the probability distribution of X is given by
the following table:
Value of X
P(X=x)
5
0
F 0;5,0.8 =   0.80 0.2 5
0
1
F 1;5,0.8 =   0.81 0.24
1
2
F 2;5,0.8 =   0.82 0.23
2
3
F 3;5,0.8 =   0.83 0.2 2
3
4
F 4;5,0.8 =   0.84 0.2 1
4
5
F 5;5,0.8 =   0.85 0.2 0
5
 
 5
 
5
 
 5
 
5
 
 5
 
= 0.00032
= 0.0064
= 0.0512
= 0.2048
= 0.4096
= 0.32768
and the cumulative distribution function of X is
or
a
5 
Fx (a)    (0.8) x (0.2) 5 x I 0,1, 2,3, 4,5 ( x)
x 0  x 
x0
0
0.00032
0  x 1

0.00672
1 x  2

2x3
Fx(x) = 0.05792
0.26272
3 x  4

4x5
0.67232
1
x 5

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5.1.4 Poisson Distribution
The Poisson random variable X, defined as the number of
outcomes occurring during a given time interval or in a specified
region appears in many natural phenomena. Some examples are the
following:
(1) number of bacteria in a given culture
(2) number of radioactive particle emissions per unit of time
(3) number of telephone calls per minute
(4) number of typing errors per page
A Poisson experiment has the following characteristics:
(1) The numbers of occurrences in the different time intervals or
specified regions are independent, and intervals/regions do
not overlap.
(2) The probability that exactly one happening will occur during
a very short time interval or in a small region is proportional
to the length of the time interval or the size of the region and
does not depend on the number of outcomes occurring
outside this time interval or region.
(3) The probability that more than one outcome will occur in a
short interval or region is negligible.
For a Poisson random variable, the probabilities depend only on
, the average number of outcomes occurring in the given time
interval or specified region. The Poisson distribution is defined as
follows:
Definition 5.4 Poisson Distribution A random variable X is defined
to have a Poisson distribution if the probability mass function of
X is given by
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 e - x

f x (x)  f x (x;  )   x!
0

for x  0, 1,2,...,  e - x
I 0,1,2, ... ( x )

x
!

otherwise
where the parameter  satisfies  > 0.
Theorem 5.4
If X has a Poisson distribution, then
E[X] = Var[X] =  .
Example 5 If a secretary makes on the average  = 6 mistakes per
page, the probability that she will make exactly X = 4 mistakes
on a given page is f(4) = f(4;6) =
e 6 6 4
= 0.1338.
4!
///
Example 6 In example 5, what is the probability that the secretary
will have at least 4 mistakes in a given page?
Solution:
P(X > 4) = 1 – P(X < 4)
= 1 – [P(X=0) + P(X=1) + P(X=2) + P(X=3)]
= 1 – f(0) – f(1) – f(2) – f(3)
=1-
e 6 6 0
e 6 61
e 6 6 2
e 6 63
0!
1!
2!
3!
= 1 – 0.00248 – 0.01487 – 0.04462 – 0.08924
= 0.84879
///
5.1.5 Hypergeometric Distribution
Sampling without replacement is associated with the
Hypergeometric distribution. Consider a box containing M objects, of
which K items are labeled success and M – K items are labeled
failure. From this box, n objects will be selected without replacement.
The random variable X defined to be the number of success items
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drawn out of n items selected is of interest. X has a Hypergeometric
distribution.
Definition 5.5 Hypergeometric Distribution A random variable X
is defined to have a Hypergeometric distribution if the probability
mass function of X is given by
  K  M - K 

  
x
n
x





f x (x)  f x (x; M, K, n)  
M
 

n 

0

  K  M - K 
 

for x  0, 1,2,..., n   x  n - x 
I 0,1,2, ... n  ( x )

M

 

n 

otherwise
where M is a positive integer, K is a nonnegative integer that is
at most M, and n is a positive integer that is at most M.
Theorem 5.5 If X has a Hypergeometric distribution, then
E[X] = n 
K
and variance
M
var[X] = n 
K MK Mn


.
M
M
M 1
Example 7 A truck carries 50 boxes of toys of which 10 are
defective. Eight boxes will be delivered to the first costumer.
What is the probability that the first delivery will include 3
defective toys?
Solution
M =50, K=10, M-K =40, n=8, x=3, n-x=5
 K  M - K 
 

x  n - x 

P(X=3) = f(3;50,10,8) =
=
M
 
n 
10  40 
  
 3  5  = 0.1471 ///
 50 
 
8 
96
5.2
Some Continuous Probability Distributions
Continuous random variables and their associated probability
density functions result from experiments which are defined over a
continuous sample space. Three important continuous distributions
will be discussed: the continuous uniform distribution, exponential
distribution, and the normal distribution.
5.2.1 Continuous Uniform Distribution
The uniform is the simplest distribution and is used for
modeling situations in which events of equal length in an interval [a,b]
are equally likely to occur.
Definition 5.6
Continuous Uniform Distribution If
the
probability density function of a random variable X is given by
f x (x)  f x (x; a, b) 
1
I [a, b] ( x )
ba
where the parameters a and b satisfy - < a < b < , then the
random variable X is defined to be uniformly distributed over
the interval [a,b].
Theorem 5.6 If X has a continuous uniform distribution, then
E[X] =
ab
2
and variance
var[X] =
(b  a ) 2
12
Proof:
97

b
x2 1
 1 
E[X] =  xf ( x )dx =  x
dx =
2 ba
ba

a 
b
=
a
ab
2
Var[X] = E[X2] – (E[X])2

a  b
1 
a  b
2
=  x f ( x )dx - 
 =x 

dx - 
 2  a ba
 2 

2
b
2
2
x3 1
=
3 ba
b
a
a  b
b 2  ab  a 2
=

3
 2 
- 
2
a  b  (b  a ) 2
 =
12
 2 
- 
2
The continuous uniform distribution has the following
characteristics:
1. The possible values X are restricted to some interval [a,b] of
real numbers.
2. Within the interval [a,b], any value is as likely to occur as
any other value.
Example 8 A dispenser is designed to give out 6 to 8 ounces of
beverage in plastic glasses. If in fact the amount X dispensed is
a uniform random variable, what is the probability that the
amount released in a glass is
(a) less than 7 ounces
(b) more than 7.5 ounces
(c) exactly 7.5 ounces
Solution
The values of X ranges from 6 to 8 ounces and its probability
density function is
f(x) = f(x;6,8) = 1 I [6,8] ( x )
2
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7
1
76 1
1
(a) P(X < 7) =  dx = x =
= = 0.5
2
2
2
2
6
6
7
8
1
1
(b) P(X>7.5) =  dx = x
2
2
7.5
8
=
7.5
8  7.5 0.5
= = 0.25
2
2
(c) P(X = 7.5) = 0
///
5.2.2 Exponential Distribution
Random variables denoting the time until some prescribed
event occurs, such as the amount of time until a piece of equipment
breaks down, the time until a light bulb burns out, or the time until an
accident occurs usually follows an exponential distribution.
Definition 5.7
Exponential Distribution
If the probability
density function of a random variable X is given by
f x (x)  f x (x; )  e-x I[0,) (x)
where >0, then X is defined to have an exponential
distribution.
Theorem 5.7 If X has an exponential distribution, then
E[X] =
1

and variance
var[X] =
1
2
Proof:
99
E[X] =

E[X2] =

0

xe x dx =  xe x 

0

0
x 2e x dx =  x 2 e x 
Var[X] = E[X2] – (E[X])2 =

+

0

0
+
e x dx =0-


0
e  x


0
2xe x dx =0 +
=
1

2
2
E[X] = 2


2
1
1
- 2 = 2
2



///
The most interesting property of this distribution is its
“memoryless” property, which means that if the lifetime of an item is
exponentially distributed, then an item which has been in use for some
hours is as good as a new item with regard to the amount of time
remaining until the item fails. Only the exponential distribution
possesses this property.
Example 9 Let the exponential random variable X denote the time
until a small meteorite first lands anywhere in the desert.
Assuming that the expected value of X is 10 days and the time
is currently midnight. What is the probability that a meteorite
first lands some time between 6 in the morning and 6 in the
evening of the first day?
Solution
X is measured in days and E[X] =
1
1
= 10. Thus,   .

10
If a day starts at midnight and ends at midnight, then 6 am and
6 pm are equivalent to ¼ and ¾ day lengths respectively. Thus,
the desired probability is
3/ 4
P(¼<X< ¾) =

3/ 4
1/ 4
1 x /10
e
dx =  e  x / 10
10
= e-1/40 – e-3/40 = 0.0476
1/ 4
100
5.2.3 The Normal Distribution
One of the most used continuous probability distribution is
called the normal probability distribution. Many variables are
approximately normally distributed and therefore can be represented
by the normal distribution. The graph of this distribution has the
following characteristics:
1. It is bell-shaped and has a single peak at the center of the
distribution.
2. The mean, median, and mode are at the center of the
distribution.
3. It is symmetric about the mean.
4. It is continuous, asymptotic (never touches the x-axis) curve.
5. The total area under the curve is equal to 1 or 100 %
6. The position of the normal distribution on the x-axis is
determined by the mean, , and the spread of the distribution is
determined by the standard deviation, .
Since the normal distribution is very often used,
notation is used for it. If the random variable X is
distributed, with mean  and variance 2, it is written as X
The notation , ( x ) is used for the density function of X
2
a special
normally
N(,2).
N(,2),
and  , ( x ) for the cumulative distribution function.
2
Definition 5.8
Normal Distribution
A random variable X is
defined to be normally distributed if its density is given by
 ,2 ( x ) 
2
2
1
e -(x- ) / 2 
2
101
where the parameters  and 2 satisfy - <  <  and 2 > 0.
Theorem 5.8
If X is a normal random variable,
E[X] =  and variance
var[X] = 2
Below are some examples of graphs of normal distributions.
These graphs are called normal curves. Figure 5.1 is a sketch of two
normal curves having the same variances but different means. The
two curves have the same form but are located at different positions
along the horizontal axis. In Figure 5.2 are two normal curves with
the same mean but different variances, thus they are centered at the
same position on the horizontal axis, but have different forms. The
normal curve with the larger variance is lower and has a wider spread.
Figure 5.3 sketches two normal curves with different means and
different variances.
Figure 5.1 Normal curves with 1  2 and 12   22
102
Figure 5.2 Normal curves with 1 = 2 and 12   22
Figure 5.3 Normal curves with 1  2 and 12   22
5.2.3.1 Areas Under the Normal Curve
The graph of any continuous probability distribution may be
constructed so that the areas under the curve, bounded by the lines at
X = x1 and X = x2 may be obtained. This area is actually equal to the
probability that the random variable will assume a value between x 1
and x2. Thus for the normal curve in figure 5.4, the shaded area
represents the probability P(x1 < X < x2).
Figure 5.4 Shaded area is equal to P(x1 < X < x2).
103
Once the parameters  and 2 are specified, the graph of the
probability density function N(,2) is completely determined. If a
table of probabilities is available for the normal distribution under
study, it would be easier to make use of this table than to use integral
calculus. However, it would be a very tedious and hopeless task to
attempt to set up separate tables or curves for every conceivable pair
of  and 2. Fortunately, every normal random variable X may be
transformed to a new set of observations Z that is also normally
distributed but with mean 0 and variance equal to 1. Z is called the
standard normal random variable or standard score. In symbols,
ZN(0,1),
and any value of a normal random variable X with mean  and
variance 2 may be transformed to its standard score or standard
normal value Z using the formula Z =
X 
.

Table A1 in the
appendix shows the areas under the standard normal curve
corresponding to P(0 < Z < z).
Definition 5.9 Standard Normal Distribution
The distribution
of a normal random variable with mean zero and variance equal
to 1 is called a standard normal distribution.
The probability that the normal random variable X with mean 
and variance 2 will assume a value between x1 and x2, where x1 < x2,
is equal to the probability that the standard normal random variable Z
will assume a value in the interval z1 to z2, where z1 =
z2 =
x2  
, that is,

x1  
and

P(x1 < X < x2) = P(z1 < Z < z2) =  0,1 (z 2 )   0,1 (z1 ) .
104
Below are some examples on determining probabilities of
normal random variables by finding equivalent areas under the
standard normal curve, using Table A1.
Example 10 Let Z be a random variable with the standard normal
distribution. Find
(a) P(0 < Z < 1.34)
(b) P(-0.65 < Z < 0)
(c) P(-1.38 < Z < 2.56)
(d) P(0.75 < Z < 1.45)
(e) P(-2.32 < Z < -0.34)
(f) P(|Z| < 2.34)
(g) P(|Z| > 1.13)
Solution
(a) P(0 < Z < 1.34) is equal to the
area under the standard
normal curve between 0 and
1.34. Thus from table A1,
look down the first column
until 1.3 then continue right to
column 4,
The entry is
0.4099. Therefore,
P(0 < Z < 1.34) = 0.4099
(b) Since the normal curve is
symmetric, then
P(-0.65 < Z < 0)
= P(0 < Z < 0.65)
= 0.2422
105
(c) P(-1.38 < Z < 2.56)
= P(-1.38 < Z < 0) + P(0 < Z < 2.56)
= P(0 < Z < 1.38) + P(0 < Z < 2.56)
= 0.4162 + 0.4968
= 0.9130
(d) P(0.75 < Z < 1.45)
= P(0 < Z < 1.45) - P(0 < Z < 0.75)
= 0.4265 – 0.2734
= 0.1531
(e) P(-2.32 < Z < -0.34)
= P(-2.32 < Z < 0) - P(-0.34 < Z < 0)
= P(0 < Z 2.32) - P(0 < Z < -0.34)
= 0.4898 – 0.1331
= 0.3567
(f) P(|Z| < 2.34)
= P(-2.34 < Z < 2.34)
= P(-2.34 < Z < 0) + P(0 < Z < 2.34)
= P(0 < Z < 2.34) + P(0 < Z < 2.34)
= 0.4904 + 0.4904
(g) P(Z > 1.13)
= 0.5 - P(0 < Z < 1.13)
= 0.5 – 0.3708
= 0.1292
///
106
Example 11 Let X be a normal random variable with mean 5 and
variance 16. Find
(a) P(5 < X < 10)
(b) P(X > 18)
Solution
 = 5 and  = 4
Z=
X  X 5


4
P(x1 < X < x2) = P(z1 < Z < z2)
10  5
55
<X<
) = P(0 < Z < 1.25) = 0.3944
4
4
18  5
(b) P(X<18) = P(Z<
) = P(Z<3.25) = 0.5 + P(0 < Z < 3.25)
4
(a) P(5<X<10) = P(
= 0.5 + 0.4994 = 0.9994
///
Example 12 Let X be a normal random variable with mean  = 20
and standard deviation  = 5, find the value of x for the
following probabilities:
(a) P(20 < X < x) = 0.4948
(b) P(X > x) = 0.9382
Solution
(a) P(20<X<x) = P(
20  20
x  20
Z
) = P(0< Z < z) = 0.4948
5
5
107
From Table A1,
P(0 < Z < 2.56) = 0.4948
x  20
= 2.56, and x = 32.8
5
x  20
x  20
(b) P(X > x) = P(Z >
) = 0.9382 and z =
5
5
Thus,
The standard score z should be at the left side of zero.
From Table A1, P(0 < Z < 1.54) = 0.4382, which, by symmetry
is equal to P(-1.54 < Z < 0). Thus, the standard value that we
are looking for is z = -1.54.
x  20
= -1.54
5
x = (-1.54)(5) + 20 = 12.3
///
5.2.3.2 Applications of the Normal Distribution
Below are some problems in which the distribution of the data,
or the distribution of the random variables under consideration are
approximated closely with the normal distribution.
Example 13
A certain town in Nueva Ecija received an average
rainfall, according to PAGASA, of 9.32 centimeters for the
month of April. Assuming a normal distribution with a
standard deviation of 2.85 centimeters, find the probability that
next April, the town receives
108
(a) less than 11.92 centimeters of rain
(b) more than 4 centimeters but not over 8 centimeters or rain
(c) more than 14.2 centimeters of rain
Solution
X = rainfall in cm
X N(9.32, 2.852)
(a) P(X < 11.92) = P(Z <
11.92  9.32
) = P(Z < 0.32)
2.85
= P(Z<0) + P(0<Z<0.32) = 0.5 + 0.1255
= 0.6255 (shaded area)
(b) P(4 < X < 8) = P(
4  9.32
8  9.32
Z
) = P(-1.87 < Z < -0.46)
2.85
2.85
= P(0.46<Z<1.87) = P(0<Z<1.87) – P(0<Z<0.46)
= 0.4693 – 0.1772
= 0.2921
(c) P(X > 14.2) = P(Z >
14.2  9.32
) = P(Z > 1.71)
2.85
= 0.5 – P(0<Z<1.71) = 0.5 – 0.4564
= 0.0436
///
Example 14
The weights of pineapples received by DOLE have
a mean of 1.4 kilos and a standard deviation of 0.12 kilo.
(a) What percentage of all these pineapples is heavier than 1.6
kilos?
(b) What percentage of the pineapples is between 1.25 and
1.55 kilos?
109
Solution
X = weight of a pineapple received by DOLE
X  N(1.4,0.122)
(a) P(X > 1.6) = P(Z >
1 .6  1 .4
) = P(Z > 1.67)
0.12
= 0.5 – P(0<Z<1.67) = 0.5 – 0.4525
= 0.0475
Hence 4.75% of all these pineapples is heavier than 1.6 kilos
(b) P(1.25<X<1.55) = P(
1.25  1.4
1.55  1.4
<Z<
)
0.12
0.12
= P(-1.25<Z<1.25)
= P(-1.25<Z<0) + P(0<Z<1.25)
= 2P(0<Z<1.25) = 2(0.3944)
= 0.7888
Thus, 78.88% of the pineapples is between 1.25 and 1.55 kilos.
///
Example 15
The IQ of the 2000 first year student applicants to
CLSU are approximately normally distributed with a mean or
115 and a standard deviation of 10. If the university requires an
IQ of at least 95, how many of these students will not be
accepted on this basis with other qualifications disregarded?
Solution
X = IQ of a first year student applicant
X  N(115,100)
P(X < 90) = P(Z<
95  115
) = P(Z<-2) = P(Z>2)
10
= 0.5 – P(0<Z<2) = 0.5 – 0.4772
= 0,0228
110
2.28% of the student applicants’ IQ is below 95. Therefore,
0.0228 x 2000 = 46 students will not be accepted based on IQ
qualification.
///
Example 16
The indicated amount of hot tea which a dispenser
puts into 6-ounce disposable styro-cups varies from cup to cup,
and follows a random variable having a normal distribution
with a standard deviation of 0.1 ounce. If only 10 percent of
the cups are to contain less than 6 ounces of hot tea, what must
the average fill of the cups be?
Solution
X = amount of hot tea dispensed by a machine
X  N(, 0.12)
P(X < 6) = P(Z <
6
)
0.1
= P(Z < z) = 0.10
Obviously, z is at the left of 0, it is a negative value.
P(Z < z) = P(Z > -z) = 0.10
P(Z < z) = P(Z > -z) = 0.50 – P(0<Z<-z) = 0.10
Thus, P(0<Z<-z) = 0.40
From table A1, -z = 1.28 (the closest value)
Solving for  from -z =
6
:
0.10
 = 6 + (0.10)(1.28) = 6.128 ounces.
///
111
5.3
Normal and Poisson Approximations to the
Binomial
5.3.1 Normal Approximation to the Binomial
Distribution
When n, the number of trials, is very large, and p, the
probability of a success on an individual trial, is close to 0.5, the
normal distribution can be used to closely approximate the binomial
distribution. This normal approximation to the binomial distribution
is useful, especially where there is a need to use the binomial
probability function repeatedly to obtain the desired probability.
Making use of the normal approximation gives a very close answer
and simplifies the solution.
Theorem 5.9
If X is a binomial random variable with mean  =
np and variance 2 = npq, then the limiting form of the
distribution of
Z
X  np
npq
as n   , is the standard normal distribution.
Example 17
A balanced coin is flipped 12 times.
(a) What is the exact probability of getting 5 heads?
(b) Find the normal approximation to the probability in (a).
Solution
(a) Using p = 0.5,
12 
P(X = 5) =  (0.5) 5 (0.5) 7 = 792  (0.5)12 = 0.1934
5


(b) To find the normal distribution approximation to this
probability, 5 is represented by the interval from 4.5 to 5.5, to
correct continuity.
112
The mean  = np = 12(0.5) = 6,
and  = np (1  p)  12(0.5)(0.5) = 1.732.
P(4.5 <X<5.5) = P(
4.5  6
5. 5  6
Z
) = P(-0.87 < Z < -0.29)
1.732
1.732
= P(0.27 < Z < 0.87)
= P(0<Z<0.87) – P(0<Z<0.27)
= 0.3078 – 0.1064
= 0.2014 (approximate probability which differs
only by 0.008 from the exact probability)
///
Example 18
Suppose in the preceding example, the fair coin is
tossed 100 times, and what is asked is the probability of
obtaining at least 40 heads?
Solution
To solve this problem using the formula for the binomial
distribution, we would have to find the sum of probabilities
corresponding to 40, 41, 42, … and 100 heads, or subtract from
1 the sum of probabilities of 0,1,2,…, 39 heads. This will be a
lot of work, but if the normal approximation is used, the area to
the right of 39.5 need only to be found. For continuity, 40 is
represented by the interval from 39.5 to 40.5, 41 is represented
by the interval 40.5 to 41.5, and so on.
Since  = np = 100(0.5) = 50
and  = np (1  p)  100(0.5)(0.5) = 5
P(X>39.5) = P(Z >
39.5  50
) = P(Z > -2.1)
5
= P(-2.1<Z<0)+ P(Z >0)
= 0.4821 + 0.5
= 0.9821
113
5.3.2 Poisson Approximation to the Binomial
Distribution
If the parameter n is very large or approaches infinity and p
approaches 0 in such a way that np remains constant, say equal to ,
then the Poisson distribution very closely approximates the Binomial
distribution.
For a fixed x,
n x
 p (1 - p) n-x
x

e - x
.
x!
Example 19
MMDA records show that the probability is
0.00007 that a car will have a flat tire while driving through the
EDSA – Aurora Boulevard crossing. Find the probability that
among 25000 cars passing through this crossing, at least three
will have a flat tire.
Solution
It will be very difficult to use the formula for the binomial
distribution to answer this problem. Even a scientific calculator will
not be able to give us the combination to 25000 taken 3 at a time. The
only way to answer this is through the poisson approximation:
 = np = 25000(0.00007) = 1.75
X=3
P(X>3) = 1 – P(X<3) = 1 – [P(X=0) + P(X=1) + P(X=2)]
e -1.75 (1.75) 0 e -1.75 (1.75)1 e -1.75 (1.75) 2
=10!
1!
2!
= 1 – (0.174) – (0.304) – (0.266)
= 1 – (0.744)
= 0.256
///
114
Actitivities:
For numbers 1 - 5 , identify the distribution of the random variable,
find the mean and variance, and solve/answer the question asked in
the problem.
1. Consider the experiment of rolling a fair die and define X as the
random variable which assigns 1 if the number of dots that appears
is even and 0 if the number of dots that appears is odd.
a. What are the possible values of X?
b. Find P(X = 1) and P(X = 0).
2. A fitness club has 15 members, 10 of which prefer the exercise
bicycle and 8 prefer the aerobic stepper. Suppose 7 members are
selected at random, find the probability that at most 2 use the
bicycle.
3. If 15% of the patients who take a certain medication get a
headache, find the probability that if 6 people take the medication,
2 will get a headache.
4. A 700 paged book has 140 typographical errors randomly
distributed in the pages. Find the probability that any given page
has exactly 2 errors.
5. Suppose the time until a certain bulb fails to light up follows an
exponential distribution with a mean of 8000 hrs. Find the
probability that a bulb burns out some time between 6000 to 7000
hours.
Solve the following problems completely. Use Table A1 for problems
involving the normal distribution.
6. Find z if the area under the normal curve
(a) Between 0 and z is 0.3531
(b) To the right of z is 0.0197
115
(c) To the right of z is 0.9265
(d) Between –z and z is 0.8444
7. A employee travels by train every day going to his office. Being a
crossword puzzle buff, he makes use of his 30 minutes on the train
answering a puzzle published in either newspaper A or B. Timing
himself in this activity, he knows that it takes him an average of
25.5 minutes to solve a puzzle from newspaper A with a standard
deviation of 4 minutes, while a puzzle from newspaper B takes him
also 25.5 minutes on the average with a standard deviation of 2
minutes. What is the probability that he will complete the puzzle if
she buys
(a) newspaper A
(b) newspaper B
8. The kalamansi trees in an orchard have a mean height of 5.34 feet
with a standard deviation of 0.5 feet. Assuming that the
distribution of these trees is approximately normal, find
(a) what percentage of the trees are less than 5 feet tall
(b) what percentage are at least 5.5 feet tall
(c) what percentage of the trees are between 5 and 5.5 feet fall
9. In number (8), if there are 3000 kalamansi trees in the orchard,
how many are
(a) less than 5 feet tall
(b) at least 5.5 feet tall
(c) between 5 and 5.5 feet tall
10.A youth advocate organization has conducted annual walkaton for
charitable purposes. They have established that on the average,
6% of the participants fail to finish the walk. What is the
probability that fewer than 20 walkers out of 1000 walkers will fail
to finish?
11. The registrar of a university assigns its students to sections
following a rule that a class should be equally populated by male
and female students. Suppose a professor in a class randomly pick
116
a student to answer a question using drawlots, what is the
probability that out of 50 questions, a male student will be called
upon
(c) more than 30 times
(d) fewer than 20 times
117