11bs201 prob and stat - ch33

11-BS201: Probability and Statistics Department of Mathematics

K L University, Vaddeswaram Page 59

CHAPTER - 3

Probability Probability Distributions 3.17 Binomial Distribution

A random variable X is said to follow a binomial distribution if it assumes only non

negative integral values and its probability mass function is given by

b(x; n, p) = , for x = 0, 1, , n, p + q = 1

Where

n is the number of independent trails

p is the probability of success of each trail

q is the probability of failure of each trail

x is the number of success

Note

1) X ~ b(n, p) to denote that X follows binomial distribution with parameters n and p

2) Each trial has two possible outcomes, one called success and the other failure.

3) The outcomes are mutually exclusive and collectively exhaustive.

4) The probabilities of success p and of failure 1 p remain the same for all trials.

5) The outcomes of trials are independent of each other.

6) The mean of the binomial distribution is np

7) The variance of the binomial distribution is npq

8) The standard deviation of binomial distribution is

9) In binomial distribution, the mean is always greater than the variance

Example 3.37

It has been claimed that 60% of solar heat installations that utility bill is reduced by at

least one-third. Accordingly, what are the probabilities that the utility bill will be

reduced by at least one-third in (i) 4 of 5 installations (ii) at least 4 of 5 installations?

Solution

(i) Here, x = 4, n = 5, p = 0.60 and q = 0.40

n

xp qx n x



b(x; n, p) =

b(4; 5, 0.60) =

(ii) P(X4) = b(4; 5, 0.60)+b(5; 5, 0.60) =

Example 3.38

Of a large number of mass-produced articles, one-tenth is defective. Find the probabilities that a

random sample of 20 will obtain

(a) exactly two defective articles;

(b) at least two defective articles.

Solution

Let X be the number of defective articles in a random sample of 20. X b(20, 10

1)

(a) 28517.010

9

10

1

2

20)2(

182

XP

(b) 60825.027017.012158.1

10

9

10

1

1

20

10

9

10

1

0

201

)1()0(1)2(

19200

XPXPXP

Example 3.39

A test consists of 6 questions, and to pass the test a student has to answer at least 4

questions correctly. Each question has three possible answers, of which only one is

correct. If a student guesses on each question, what is the probability that the student

will pass the test?

Solution

Let X be the no. of correctly answered questions among 6 questions. X b(6, 3

1)

xxx x x

xXPXP

66

4

6

43

23

16

)()4(

n

xp qx n x



10014.03

23

16

6

32

31

5

6

32

31

4

6 061524

Example 3.40

A packaging machine produces 20 percent defective packages. A random sample of ten

packages is selected, what are the mean and standard deviation of the binomial

distribution of that process?

Solution

Let X be the no. of defective packages in a sample of 10 packages. X b(10, 0.2)

Its mean is = np = (10)(0.2) = 2

Its standard deviation is 265.1)8.0)(2.0)(10( npq

3.18 Poisson Distribution

A random variable X is said to follow a Poisson distribution if it assumes only non


f(x; ) = for x = 0, 1, 2, , >0

Note

1) X ~ P(x, ) to denote that X follows Poisson distribution with parameter

2) In a Poisson distribution mean and variance are equal, which is equal to the

parameter

3) A Poisson experiment has the following properties:

The number of successes in any interval is independent of the number of

successes in other interval.

The probability of a single success occurring during a short interval is

proportional to the length of the time interval and does not depend on the

number of successes occurring outside this time interval.

The probability of more than one success in a very small interval is negligible.

4) The following are the some of the examples of random variables following Poisson

distribution:

The number of customers arrived during a time period of length t.

e

x

x

!



The number of telephone calls per hour received by an office.

The number of typing errors per page.

The number of accidents occurred at a junction per day.

5) Poisson approximation to the binomial distribution: If n is large and p is near 0 or near

1.00 in the binomial distribution, then the binomial distribution can be approximated by

the Poisson distribution with parameter np.

General speaking, the Poisson distribution will provide a good approximation to

binomial when

(i) n is at least 20 and p is at most 0.05; or

(ii) n is at least 100, the approximation will generally be excellent provided p< 0.1.

Example 3.41

It is known that 5% of the books bound at a certain bindery have defective bindings.

Find the probability that 2 of 100 books bound by this bindery will have defective

bindings using (i) the formula for the binomial distribution (ii) the Poisson approximation

to the binomial distribution

Solution

(i) Here x=2, n =100, and p = 0.05,

b(x; n, p) =

b(2; 100, 0.05) =

(ii) Here x =2 and = np = 100*0.05 = 5

f(x; ) =

f(2; 5) =

= 0.084

Example 3.42

The average number of radioactive particles passing through a counter during 1

millisecond in a laboratory experiment is 4. What is the probability that 6 particles enter

the counter in a given millisecond?

n

xp qx n x

e

x

x

!



Solution

Let X be the no. of particles entering the counter in a given millisecond. X P(4)

1042.0!6

4)6(

64

e

XP

Example 3.43

Ships arrive in a harbour at a mean rate of two per hour. Suppose that this situation can

be described by a Poisson distribution. Find the probabilities for a 30-minute period that

(a) No ships arrive;

(b) Three ships arrive.

Solution

Let X be the no. of ship arriving in a harbour for a 30-minute period. X Po( 122 )

(a) 3679.0!0

1)0(

01

e

XP

(b) 0613.0!3

1)3(

31

e

XP

Example 3.44

If the prob. that an individual suffers a bad reaction from a certain injection is 0.001,

determine the prob. that out of 2000 individuals, more than 2 individuals will suffer a

bad reaction.

Solution

Using Poisson distribution:

P(x=0 suffers) = = np = 2

P(x=1 suffers) = 2

21 2

!1

2

e

e

P(x=2 suffers) =

2

0

10 2

2

e

e

!

2

2!

22 2

2

e

e



Then the required probability =

Example 3.45

Two percent of the output of a machine is defective. A lot of 300 pieces will be

produced. Determine the probability that exactly four pieces will be defective.

Solution

Let X be the no. of defective pieces among 300 pieces. X b(300, 0.02)

1338.0)98.0()02.0()4( 29644300 CXP

By Poisson Approximation:

6)02.0)(300( np

1338.0!4

6)4(

46

e

XP

3.19 Geometric Distribution

A random variable X is said to follow a geometric distribution if it assumes only non


Where p is probability of success, which is the parameter of the distribution.

Note

The mean of geometric distribution is

Example 3.46

If the probability is 0.05 that a certain kind of measuring device will show excessive drift,

what is the probability that the 6th measuring device tested will be the first to show

excessive dirft?

Solution

Here p = 0.05 and x = 6,

By using geometric distribution formula we get

15

0 3232

e

.



3.20 Normal Distribution

The Normal Distribution (N.D.) was first discovered by De-Moivre as the limiting form of

the binomial model in 1733, later independently worked Laplace and Gauss.

The Normal distribution is the most important distribution in statistics. It is a probability

distribution of a continuous random variable and is often used to model the distribution

of discrete random variable as well as the distribution of other continuous random

variables. The basic from of normal distribution is that of a bell, it has single mode and is

symmetric about its central values. The flexibility of using normal distribution is due to

the fact that the curve may be centered over any number on the real line and it may be

flat or peaked to correspond to the amount of dispersion in the values of random

variable.

3.20.1 Definition

A random variable X is said to follow a Normal Distribution with parameter mean () and

variance ( ) if its density function is given by the probability law

Symbolically we can represent the normal variate as X ~ N(, 2)

3.21 The Normal Probability Curve

The graph of the normal distribution depends on two factors the mean and the

standard deviation. The mean of the distribution determines the location of the center

of the graph, and the standard deviation determines the height and width of the graph.

When the standard deviation is large, the curve is short and wide; when the standard

deviation is small, the curve is tall and narrow. All normal distributions look like a

symmetric, bell-shaped curve, as shown below.



The curve on the left is shorter and wider than the curve on the right, because the curve

on the left has a bigger standard deviation.

3.22 Standard Normal Distribution

If X is a normal variate with mean () and standard deviation () then

is a

standard normal variate with mean (0) and standard deviation (1).

The probability density function of the standard normal variate Z is given by the

probability law

Symbolically we can represent the standard normal variate as Z ~ N(0, 1)

A graph representing the density function of the Normal probability distribution is also

known as a Normal Curve or a Bell Curve (see Figure below). To draw such a curve, one

needs to specify two parameters, the mean and the standard deviation. The graph

below has a mean of zero and a standard deviation of 1, i.e., ( =0, =1). A Normal

distribution with a mean of zero and a standard deviation of 1 is also known as the

Standard Normal Distribution.

3.23 Characteristics of Normal distribution and normal curve:

The normal probability curve with mean and standard deviation is given by the

equation

and has the following properties

i) The mean, median and mode of the normal distribution coincide, i.e., mean = median

= mode = . (The height of normal curve is at its maximum at the mean. Hence the mean



and mode of normal distribution coincides. Also the number of observations below the

mean in a normal distribution is equal to the number of observations above the mean.

Hence mean and median of N.D. coincides.)

ii) The curve is bell shaped

iii) The normal curve is symmetrical about the line x=

iv) As x increases numerically, f(x) decreases rapidly, the maximum probability occurring

at the point x = , and given by

v) Linear combination of independent normal variates is also a normal variate

vii) The total area under the normal curve (

is distributed as follows

covers 68.26% of the area

covers 95.44% of the area

covers 99.74% of the area, and it can be represented

as follows

Note

1) The standard normal distribution, N (0, 1), is very important because

probabilities of any normal distribution can be calculated from the probabilities

of the standard normal distribution.

2) If X is a normal random variable with mean and standard deviation , then

is a standard normal random variable and hence

3) Suppose Z ~ N(0, 1) is standard normal variate then by using the standard normal

distribution area tables, we can calculate the various probabilities as explained

below:

ZX

P x X x Px

Zx

( ) ( )1 21 2



i) P(Zb) = this probability can be represented by using the following graph

of standard normal distribution and it cannot be obtained directly from

the tables

P(Z>b)=1- P(Zb), where P(Zb) available directly from table.

iii) P(aZb) = this probability can be represented by using the following

graph of normal distribution

P(aZb) = P(Zb) P(Za), where P(Zb) and P(Za) available directly

from the tables.

4) The Normal Approximation to the Binomial Distribution

Given X is a random variable which follows the binomial distribution with

parameters n and p, then the limiting form of the distribution function of

standard normal variate is

provided, if n is large



and p is not close to 0 or 1. If both np and nq are greater than 5, the

approximation will be good.

Example 3.47

If a random variable having standard normal distribution, find the probabilities that it

will take on a value: (a) less than 1.50; (b) greater than 2.16; (c) less than -1.20; (d)

greater than -1.75; (e) between 0.87 and 1.28; (f) between -0.34 and 0.62

Solution

Given that a random variable is having standard normal distribution, i.e., Z ~ N(0, 1),

then we have to find the probability that it will take on a value

a) Less than 1.50 i.e., P(Z2.16) = 1 - P(Z2.16) = 1 0.9846 = 0.0154 [ from area tables and by using 3(ii)]

c) Less than -1.20 i.e., P(Z-1.75) = 1 - P(Z-1.75) = 1 0.0401 = 0.9599 [ from area tables and by using 3(ii)]

e) Between 0.87 and 1.28 i.e.,

P(0.87Z1.28) = P(Z1.28) P(Z0.87) = 0.8997 0.8078 = 0.0919 [ from area tables

and by using 3(iii)]

f) Between -0.34 and 0.62 i.e.,

P(-0.34Z0.62) = P(Z0.62) P(Z-0.34) = 0.7324 0.3669 = 0.3655 [ from area

tables

and by using 3(iii)]

Example 3.48

If the amount of cosmic radiation to which a person is exposed while flying by jet across

the US is a random variable having normal distribution with mean = 4.35mrem and

standard deviation = 0.59mrem. Find the probabilities that the amount of cosmic



radiation to which a person will be exposed on such flight is (i) between 4.00 and

5.00mrem and (ii) atleast 5.50mrem.

Solution

Given that X - be (the amount of cosmic radiation to which a person is exposed) a

normal random variable with mean = 4.35mrem and standard deviation =

0.59mrem.

i.e. X ~ N( 4.35, = 0.59)

We have to find the probabilities that the amount of cosmic radiation to which a person

will be exposed on such flight is

(i) between 4.00 and 5.00mrem

i.e.,P(4.00



Let X be the balance in the charge account. X N(80, 230 )

(a) )125

()125(

XPXP

0668.0)5.1()30

80125(

ZPZP

(b)

9565)9565(

XPXP

30

8095

30

8065ZP

3830.0

)5.05.0(

ZP

Example 3.50

A process yields 10% defective items. If 100 items are randomly selected from the

process, what is the probability that the number of defective exceeds 13?

Solution

Let X be the no. of defective in a random sample of 100 items. X b(100, 0.1)

10)1.0)(100( np , 3)9.0)(1.0)(100( npq

)5.13'()13( XPXP by normal approximation

121.0)167.1(3

105.135.13'

ZPZP

XP

Example 3.51

A multiple-choice quiz has 200 questions each with four possible answers of which only

one is the correct answer. What is the probability that sheer guesswork yields from 25

to 30 correct answers for 80 of the 200 problems about which the student has no

knowledge?

Solution

Let X be the no. of correct answers for 80 with sheer guesswork. X b(80, 0.25)

20)25.0)(80( np , 15)75.0)(25.0)(80( npq



)5.30'5.24()3025( XPXP by normal approximation

1196.000336.01230.0)71.216.1(15

205.30

15

205.24

ZPZP



3.24 Exponential Distribution

The exponential distribution (exponential p.d.f) often arises, in practice, as being the

distribution of the amount of time until some specific value of the variable (event)

occurs. For example: the time until a new car breaks down, time until an arrival at

emergency room, ... etc.

3.24.1 Definition

A continuous random variable X is said to have an exponential distribution with

parameter (=

> 0 then the probability density function of exponential distribution is

given by

3.25 Properties of Exponential Distribution

Mean: = 1/

Variance: 2 = 1/2,

Standard Deviation = 1/

3.26 Cumulative Distribution Function of Exponential Distribution

Let X be exponential random variable with f(x) as its probability density function then its

cumulative distribution is given by having parameter (=

> 0

P(X a) = 1 ea

Example 3.52

Suppose that the length of a phone call in minutes is an exponential random variable

with parameter = 1/10. If someone arrives immediately ahead of you at a public

telephone booth, find the probability that you will have to wait (i) more than 10

minutes, and (ii) between 10 and 20 minutes.

Solution

Let X be the be the length of a phone call in minutes by the person ahead of you is an

exponential random variable with = 1/10.



i) P(more than 10 minutes) = P(X >10) = ea = e1 =0.368

(ii) P( between 10 and 20 minutes) = P(10 < X < 20) = e1 e2 = 0.233

Example 3.53

The amount of time, in hours, that a computer functions before breaking down is an

exponential random variable with = 1/100.

(i) What is the probability that a computer will function between 50 and 150 hours

before breaking down?

(ii) What is the probability that it will function less than 100 hours?

Solution

Given that the amount of time (hrs) that a computer functions before breaking down is

an exponential random variable with = 1/100.

i) The probability that a computer will function between 50 and 150 hours before

breaking down is given by

P(50 X 150) = e50/100 e150/100 = e1/2 e3/2 =0 .384

ii) The probability that it will function less than 100 hours is given by

P(X



(i)both will be alive (ii)only the man will be alive (iii)only the woman will be

alive (iv)none will be alive (v)at least one of them will be alive.

3.6 A problem in statistics is given to two students A and B.The odds in favour of A

solving the problem are 6 to 9 and against B solving the problemare 12 to 10.If

both A and B attempt.Find the probability of the problem being solved.

3.7 A problem in statistics is given to three students A,B and C whose chances of solving

it are

,

and

respectively.Find the probability that the problem will be solved if

they all try independently.

3.8 A piece of equipment will function only when all the three components A,B and C are

working. The probability of A failing during one year is 0.15,that of B failing is

0.05 and that of C failing is 0.10.What is the probability that the equipment will

fail before the end of one year?

3.9 The probability that a management trainee will remain with a company is 0.60.The

probability that an employee earns more than Rs.10,000 per month is 0.50.The

probability that an employee is a management trainee who remained with the

company or who earns more than Rs.10,000 per month is 0.70.What is the

probability that an employee earns more than Rs.10,000 per month,given that he

is a management trainee who stayed with the company?

3.10 A box of 100 gaskets contain 10 gaskets with type A defects,5 gaskets with type B

defects and 2 gaskets with both types of defects.Find the probabilities that (i)a

gasket to be drawn has a type B defect under the condition that it has a type A

defect.(ii)a gasket to be drawn has no type B defect under the condition that it

has no typed A defect.

3.11 There are 12 balls in a bag, 8 red and 4 green.Three balls are drawn successively

without replacement.What is the probability that they are alternately of the

same colour?

3.12 If the probability is 0.05 that a certain wide- flange column will fail under a given axial load, what are the probabilities that among 16 such columns

(i) At most two will fail (ii) At least four will fail?

3.13 During one stage in the manufacture of integrated circuit chips, a coating must be

applied. If 70% of chips receive a thick enough coating, find the probabilities that, among 15 chips:



(i) atleast 12 will have thick enough coating (ii) atmost 6 will have thick enough coating (iii) exactly 10 will have thick enough coating

3.14 Find the mean and variance of a binomial distribution of the number of heads

obtained in 3 flips of a balanced coin

3.15 In a given city, 6% of all drivers get at least one parking ticket per year. Use the

Poisson approximation to the binomial distribution to determine the

probabilities that among 80 drivers (i) 4 will get at least one parking ticket in any

given year (ii) at least 3 will get atleast one parking ticket in any given year (iii)

anywhere from 3 to 6, inclusive, will get atleast one parking ticket in any given

year.

3.16 If 0.8% of the fuses delivered to an arsenal are defective, use the Poisson

approximation to determine the probability that 4 fuses will be defective in a

random sample of 400.

3.17 Prove that for the Poisson distribution

for x = 0,1,2,..

3.18 If X is normally distributed with mean 12 and standard deviation is 4. Find the

probability of the following : (i) X20 (ii) X20 (iii) 0X12

3.19 The mean yield for one-acre plot is 662kilos with a s.d. of 32kilos. Assuming normal

distribution, how many one-acre plots in a batch of 1000 plots would you expect

to have yield (i) over 700kilos, (ii) below 650kilols (iii) what is the lowest yield of

the best 100 plots.

3.20 The local authorities in a certain city install 10,000 electric lamps in the streets of

the city. If these lamps have an average life of 1,000 burning hours with a s.d. of

200 hours, assuming normality, what number of lamps might be expected to fail

(i) in the first 800 burning hours (ii) in between 800 and 1200 burning hours?

After what period of burning hours would you expect that (i) 10% of the lamps

would fail (ii) 10% of the lamps would be still burning?



3.21 In a distribution exactly normal, 10.03% of the items are under 25 kilogram weight

and 89.7% of the items are under 70 kilogram weight. What are the mean and

standard deviation of the distribution?

11bs201 prob and stat - ch33

Documents

trail x

random variable x

probability of success

probability of failure

probabilities of success

poisson distribution

probability mass function

number of defective