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Session 06

Exercise: The number of accidents occurring in a city in a day is a Poisson variate with mean0.8. Find the probability that on a randomly selected day(i)There are no accidents(ii)There are accidents

Solutions:Let X: number of accidents per day.Then, X is P(λ=0.8).The p.m.f. is –

,....3,2,1,0,!

)8.0()(

8.0

xx

exp

x

(i) Probability that ob a particular day three are no accidents is ------------P[no accidents] = P[X=0]=p(0)

= 449.0!0

)8.0( 8.008.0

ee

(ii) P[accidents occur] = 1-P[no accidents]= 1-p(0) = 1-0.449 = 0.551

Exercise: The number of persons joining a cinema queue in a minute has Poisson distributionwith parameter 5.8. Find the probability that (i) no one joins the queue in a particular minute(ii)2 or more persons join the queue in the minute.

Solution:Let X : number of persons joining the queue in a minute Then, is P(λ=5.8).The p.m.f is -------------------

,....3,2,1,0,!

)8.5()(

8.5

xx

exp

x

(i) P[no one joints the queue] = P[X=0] = p(0)

=!0

)8.5( 08.5e

= 8.5e =0.003

(ii) P[two or more join] =

9796.00204.01

8.6003.01

8.511

!1

)8.5(

!0

)8.5(1

)}1()0({1

]2[1

]2[

8.5

18.508.5

e

ee

pp

XP

XP

Exercise: The average number of telephone calls booked at an exchange between 10-00 A.M. and 10-10 A.M. is Find the probability that on a randomly selected day 2 ormore calls are booked between 10-00 A.M. and 10-10 A.M. On how many days of ayear, would you expect booking of 2 or more calls during that times gap.

by: Sudheer paiPage 2 of 5

Solutions:Let X : number of telephone calls booked at the exchange during 10-00 A.M. to 10-

10 A.M. Then, X is P(λ=4).The p.m.f is ---

,...3,2,1,0,!

4)(

4

xx

exp

x

P[2 or more calls] = 1-P[less than 2 calls]= 1 – [p(0) +p(1)]= 1 – e-4[(40)/0! +(41)/1!]= 1 – 0.0183[1+4]= 1 – 0.0915 = 0.9085

An year has 365 days. Out of these N = 365 days, the number of days on which therewill be 2 or more calls is ---N *P[2 or more calls] = 365 * 0.9085 = 332

Exercise: 2 percent of the fuses manufactured by a firm are expected to be defective,Find the probability that a box containing 200 fuses contains(i) defective fuses(ii) 3 or more defective fuse

Solutions:2 percent of the fuses are defective. Therefore, probability that a fuses is defective isp = 2/100 = 0.02

Let X denote the number of defective fuses in the box of 200 fuses. Then, X is B(n =200, p = 0.02)Let X denote the number of defective fuses in the box of 200 fuses. Then, X is B (n =200, p = 0.02)Here, p is very small and n is very large. Therefore, X can be treated as Poissonvariate with parameter λ=np = 200 * 0.02 = 4.The p.m.f. is ----

,...3,2,1,0,!

4)(

4

xx

exp

x

P[box has defective fuses] = 1-P[no defective fuses]= 1 – p(0)= 1 – e-4[(40)/0!= 1 – 0.0183 =0.9817

P[3 or more defective fuses] = 1-P[less than 3 defective fuses]= 1 – [p(0)+p(1) +p(2)]= 1 – e-4[1+4+8]= 1 – 0.0183*13= 1 – 0.2379 =0.7621

Exercise: The probability that a razor blade manufactured by a firm is defective is1/500. Blades are supplied in packets of 5 each. In a lot of 10,000 packets, how manypackets would(i)Be free defective blades?

(ii) Contains exactly one defective blade?(e-0.01=0.99)

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Solution:Let X be the number of defective blades in a packet of 5 blades. Then, X is B (n = 5, p= 1/500)

Since p is very small and n is sufficiently large, X is treated as Poisson variatewith parameter λ=np = 5*(1/500) = 0.01

,...3,2,1,0,!

)01.0()(

01.0

xx

exp

x

(i) P[ no defective blades] = p(0)= e-0.01(0.01)0/0! = 0.99

The number of packets which will be free of defective blades is -----N * P[no defective blades] = 10000*0.99 = 9900(ii) P[one defective blade] = p(1)

The probability that a razor blade manufactured by a firm is defective is 1/500. Bladesare supplied in packets of 5 each. In a lot of 10,000 packets, how many packets would(i)Be free defective blades?

(ii) Contains exactly one defective blade?(e-0.01=0.99)

(ii) P[one defective blade] = p(1)= e-0.01(0.01)1/1! = 0.0099

The number pf packets which will have one defectives blade is ----N * P[one defective blade] = 10000*0.0099 = 99

Exercise:On an average, a typist mistakes while typing one page. What is theprobability that a randomly observed page in free of mistakes? Among 200 pages, inhow many pages would you expect mistakes?Solutions:Let X: number of mistakes in a page.Then, X is P(λ=3).The p.m.f. is ---

,....3,2,1,0,!

3)(

3

xx

exp

x

P[page is free of mistakes] = p(0)

0498.0!0

3 303

ee

P[page has mistakes] = 1-P[Page has no mistakes]= 1-0.0498=0.9502

Among 200 pages, the expected number of pages containing mistakes is ---N*P[page has mistakes] = 200*0.9502=190

Exercise: In a Poisson distribution P[X=2] = P[X=3]. Find P[X=4].Solution:Let λ be the parameterHere, P[X = 2] = P[X=3]

by: Sudheer paiPage 4 of 5

31

!3!23

32

ee

And so, λ=3The p.m.f. is ----

,....2,1,0,!

3)(

3

xx

exp

x

P[X=4] = p(4) =!4

323e

= 1681.024

810498.0

Exercise: For a Poisson variables 3 * P[X=2] = P[X=4]. Find standard deviation.Solution:Here, 3*P[X = 2] = P[X=4]

433

!4!23

2

42

ee

And so, λ2=36That is , λ=3Thus, the parameter is λ=6The standard deviation S.D.(X) = 449.26

Exercise: The following data relates to the number of mistakes in each page of a bookcontaining 180 pages.

No of mistakes per page: 0 1 2 3 4 5 or more Total

No. of Pages 156 16 5 2 1 0 180

Fit a Poisson distribution to the data. Obtain the theoretical frequencies

Solution:Let X denotes the number of mistakes per page. Then, X is a Poisson variate. Theparameter is ---

2.0180

36180

504132251160156

N

fxx

The p.m.f is ---

by: Sudheer paiPage 5 of 5

,....3,2,1,0,!

2.0)(

2.0

xx

exp

x

The frequency function is---

37.1478187.0180!0

2.0180

,....3,2,1,0,!

2.0180

02.0

0

2.0

eT

xx

eT

x

X

NORMAL DISTRIBUTION

A probability distribution which has the following probability density functions(p.d.f)is called Normal distribution

Here, the variable X is continuous and it is called Normal variate.

Note 1: The distribution has two parameters, namely, µ and σ.(Here, Π=3.14 ande=2.718.

Note 2 : This normal distribution has Mean E(X) = µ and Variance = V(X) = σ2.

S.D.(X)=σ.

Note 3: A normal variate with parameters µ and σ is denoted by N(µ,σ2)

Note 4: The normal p.d.f. can also be written as-

EXAMPLES FOR NORMAL VARIATE

Many of the variables which occur in nature have normal distribution. Someexamples are –1.Height of students of a college2.Weight of apples grown in an orchard3.I.Q(Intelligence Quotient) of a large group of children.4.Marks scored by students in an examination