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Integrtn Nd Binomial Distribution

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National Institute of Technology Durgapur DEPARTMENT OF MANAGEMENT STUDIES 2012-2014 PROJECT By Group 5(44-54)
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Page 1: Integrtn Nd Binomial Distribution

National Institute of Technology Durgapur

DEPARTMENT OF MANAGEMENT STUDIES2012-2014

PROJECT By

Group 5(44-54)

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Exercise 10(I) Integration

Q1. Define Integration as inverse process of differentiation. Give three illustrations.

Ans1. Integration ,the inverse process of differentiation .

If f(x) be a given function of x and if another function F(x) be obtained such that its differential coefficient with respect to x is equal to f(x),then F(x) is defined as an integral ,or more properly an indefinite integral of f(x) with respect to x.

The process of finding an integral of a function of x is called and the operation is indicated by

writing the integral sign “ ” before the given function and dx after the given function ,the symbol dx indicating that x is the variable of integration. The function to be integrated,viz f(x) is called the integral.

Symbolically,if F(x)=f(x) then .Where is called an indefinite integral of f(x)

with respect to x.It will be shown later that if is condition the exits.Thus considered as

symbols of operation, and are inverse to each other.

As we can see the example.

Derivatives Integral(Anti derivatives)

Q2.What do you mean by the constant of integration? Is its value fixed?

Ans 2. Constant of Integration

It may be noted that if F(x)=f(x),then we also have

,

Where is an arbitrary constant.Thus ,if ,a general value of the indefinite integral

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In other words , in finding the indefinite integral of a function ,an arbitrary constant is to be added to the result to make it general .This is the reason why the integral is referred to as an indefinite integral.

The arbitrary constant is usually referred to as the constant of integration.Addition of constant to the

and not any non constant function is explained below.

We know that if two function is defined on the closed interval [a,b] are such that

,

For all then is constant on [a,b].Thus it is possible to get the indefinite integral of the same function in different forms by different processes,but ultimately these forms can atmost differ from each other by constant quantities only.Hence an arbitrary constant added to the indefinite integral of a given function obtained by any process makes the result perfectly general.

Q3.State for what value of n,the formula ,where is constant ,is not true.

Ans 3. For =-1,the formulae is not true.

As for n=-1, so for this we have

And we write our formulae as

4. (i) To show

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We know or

Here

So ,therefore

(ii) To show

(iii)

(iv)

(v)

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5.

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

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6

(i)

(ii)

(iii)

(iv)

7.

(i)

(ii)

(iii)

(iv)

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8

(i)

(ii)

(iii)

(iv)

(v)

9

(i)

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(ii)

(iii)

10

(i)

(ii)

(iii)

(iv)

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Exercise 10(II)Integration-A

1. Evaluate:

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2. Integrate:

3. Evaluate:

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4. Evaluate:

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6. Evaluate:

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7. Integrate:

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Integration-B1. Evaluate

2. Evaluate

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3. Integrate

4. Evaluate

5. Evaluate

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6. Evaluate 7. Evaluate

EXERCISE 10(III)-A

1. EVALUATE:

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2. Evaluate

3. EVALUATE

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4. EVALUATE

5. EVALUATE

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6. Evaluate

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Exercise 10(iii) B

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Exercise 11(I) Definite Integrals

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2.(i) by first principle.

Let f(x ) be a continuous real valued function, defined in a closed interval [a,b]. Then we define.

= h [f (a) +f (a+h) +f (a+2h) +……..+f [a+ (n-1) h]

Where nh=b-a

Let in this case f(x)= and a =0; b=1, nh=b-a =1

= h [f (0) +f (0+h) +f (0+2h) +……..+f [0+ (n-1) h]

= h[0+ ]

= h[2 { }]

Now we know that

=

As h 0 and nh =1

=23 ans.

2.(ii)

, by first principle.

Let and a=0; b=1 nh= b-a=1

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From formula;

2.(iii)

Let a=0 b=1 nh=1

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5.(i)

5.(ii)

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5.(iii)

Taking log both side we get

Differentiating both side w.r.t to x

Also when

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5.(iv)

6.(i)

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6.(ii)

6.(iii)

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6.(iv)

7.(i)

On dividing 1-x by 1+x

7.(ii)

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7.(iii)

let

And at

7.(iv)

Let

And at

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Question 3(1) ∫a

b

x❑dx

Answer =

∴∫a

b

x❑dx=1/2¿

= 1/n+1[b1+1−a1+1¿

= 1/n+1[b2−a2 ¿

Question no :3(2) ∫1

5

3 x dx

Solution :

= ∫1

5

3 x dx

=) 3∫1

5

xdx

= ) 3∫1

5

¿¿

=) 3/2[52−12¿¿

=) 36

Question no :3(3) ∫2

3

(2x2+5)dx

Solution :

=2∫2

3

2 x2+5∫2

3

dx

=2∫2

3

¿¿

= 2∫2

3

¿¿

= 2/3∫2

3

¿¿

Page 41: Integrtn Nd Binomial Distribution

=) 2/3[33−23¿23+5 [31−21 ¿2

3

= 53/3

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8.(i)

Let

Also at

8.(ii)

Let

Also at

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8.(iii)

For this problem,

Integrating by parts.

Let x be 1st function and exbe 2nd

8.(iv)

Let

Also at

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8.(v)

Let logx=1st function

x2 =2nd function

Now integration by part we get

8.(iv)

1st function=logx

2nd function=x

Integration by parts

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9.(i)

Put

Also

9.(ii)

let

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Also at

By using partial fraction:-

We can write

Let

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9.(iii)

Put

Also

9.(iv)

Put

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Also

10.(i)

Let

Also at

10.(ii)

Let

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Also at

10.(iv)

Let

Also

10.(v)

Solving by partial fraction.

Page 50: Integrtn Nd Binomial Distribution

let

10.(vi)

Using partial fraction

Let

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11.(i)

11.(ii)

Using partial fraction

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Let

11.(iii)

Using partial fraction

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Let

12.(i)

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Let

Also at

Dividing z3by 1+z we get

Let

Also

Therefore

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Question 13(1) ¿ lim ¿n→∞ (1n+m

+ 1n+2m

+ 1n+3m

+…+ 1n+nm

)¿

Answer =

¿ lim ¿n→∞ (1n+m

+ 1n+2m

+ 1n+3m

+…+ 1n+nm

)¿

¿ lim ¿n→∞1n¿¿

= lim ¿n→∞1n∑r=1

n 1

1+m. rn

¿

= lim ¿n→0h∑r=1

n 11+mrh

¿ [where 1n=h] [

= ∫0

1 dx1+mx

= 1m∫0

1 m1+m x

dx = 1

m¿

= 1m¿

Page 56: Integrtn Nd Binomial Distribution

Question 13(2) lim ¿n→∞¿¿ )

Solution:

lim ¿n→∞¿¿)

lim ¿n→∞1/n {¿¿

lim ¿n→∞1/n∑r=1

n

¿¿¿

lim ¿n→∞1/n∑r=1

n

¿¿¿

∫0

1

xmdx

=[ xm+1

m+1 ] =

1m+1

Question13(5)

¿ lim ¿n→∞ (1n+1

+ 1n+2

+ 1n+3

+…+ 1n+n

)¿

Evaluate

¿ lim ¿n→∞ (1n+1

+ 1n+2

+ 1n+3

+…+ 1n+n

)¿

¿ lim ¿n→∞ (+(1n )

1+ 2m

+( 1n )

1+ 3m

+…+( 1n )

1+ nn)¿

¿ lim ¿n→∞∑r=1

n 1

1+. rn

¿

Page 57: Integrtn Nd Binomial Distribution

¿∑r=1

n 11+rh

¿∫0

1 11+r

dx

=[log(1+x]

=log2 .

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14.

15.(i)

Let

Also

Taking exp both side

Also at

Solving by part 1st function =t 2, and 2nd function=e t

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Again solving by parts.

15.(ii)

Solving x+1

(x+2)2 by partial fraction

Let

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By using the formula

15.(iii)

Solving i.e. dividing then partial fraction

Therefore

Now p.f of

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Let

15.(iv)

Let

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Also at

Using p.f

Let

15.(v)

Let

Also

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Solving by P.F

Let

16.

The given curve is

The given lines are x-axis x=1 and x=3.

The required area between the above curve and area ABCD(the back region) in the fig

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17.(i)

If f’(x) =3 x2+2

Now given that,

17.(ii)

If f’(x) =logx

Solving by parts by taking

Page 65: Integrtn Nd Binomial Distribution

Given

18.

Evaluating the first integral by parts

Also given that;

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Exercise 11(II)

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Page 70: Integrtn Nd Binomial Distribution

3

3 30

3

3 30

3

3 30 0 0

33

3 30

0

0

4(3)

.................... (1)

....................(2) ( ) ( )

(1) (2)

2

2 1

2

2 0

.2

a

a

a a a

a

a

a

xdxa x x

xdxSol Ia x x

a xdxI f x dx f a x dxx a x

After adding

x a xI dx

a x x

I dx

I x

I a

aI Ans

∫ ∫ ∫

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13 2 5

1

3 2 5

3 2 5

13 2 5

1

5(1) (1 )

( ) (1 )

( ) (1 ) ( )

( )

(1 ) 0 . ( ) 0, ( )

n

a

a

x x dx

Sol Here f x x x

Now f x x x f x

f x is anodd f

x x dx Ans f x dx when f x is anodd

∫ ∫

22

2

2

2

22

2

5(2) 4

( ) 4

( ) 4 ( )

( )

4 0 ( ) 0, ( )

n

a

a

x x dx

Sol Here f x x x

f x x x f x

f x is an odd f

x x dx f x dx when f x is an odd

∫ ∫

22 2

2

2 2 22 4 2 4

2 2 2

2 23 5

2 2

6(1) (1 )

( )

1 1(8 8) (32 32)3 5 3 5

112 .15

x x dx

x x dx x dx x dx

x x

Ans

∫ ∫ ∫

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1 2

6 61

1 12 2

3 2 3 2 3 3 3 31 1

1 2

3 3 3 3 31

1 12 2

3 3 3 3 31 1

1 13 3 2

31 1

6(2) ( 1)

( ) ( ) ( )( )

1 12 ( ) ( )

12 ( ) ( )

1 1 1 12 3 3 3

16

x dx aa x

x dx x dxa x a x a x

x dxa a x a x

x dx x dxa a x a x

dz dz a x z so x dx dza z z

a

∫ ∫

∫ ∫

∫ ∫

11 ,3 1 1

11 3 3

3 , 3 3 31 1

3

3 3

3

3 3

log( ) log( )

1 1log( ) log6 6

1 12log6 1

1 1log .3 1

z z

z a xa z a a x

aa a

a Ansa a

5

550

5

550

5

5 50

5 5

550

0 0

7(2)

............... 1

........................ 2

1 2

2

2 1

2

a

a

a

a

aa

x dxx a x

x dxSol Ix a x

a x dxAgainwecanwrite I

a x x

After adding

x a x dxI

x a x

I dx x

aI Ans

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4

4 4

2

10 102

2 2

5 2 5 2 5 5 5 52 2

2 5

5 5 5 5 52

2 2 25 5 5 5

5 5 5 5 5 5 5 5 5 5 52 2 2

2

52

7(3) 2

( ) ( ) ( )( )

1 12

1 1 12 2 2

1 1 12 5 2

x dx aa x

x dx x dxa x a x a x

x dxa a x a x

x x x dx x dxdxa a x a x a a x a a x

dza z

∫ ∫

∫ ∫ ∫

25 5 4 5 5 ,

5 ,2

2 2, 5 5 5 55 52 2

5 5 5 55

5 55

1 15 5

1 1log( ) log( ) log( ) log( )10 10

1 log( 32) log 32 log( 32) log( 32)10

1 .2 log( 32) log 3210

dz a x z then x dx dz same a x za z

z z a x a xa a

a a a aa

a aa

5

5 5

1 ( 32)log .5 ( 32)

a Ansa a

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BS( Assignment-9)Probability distribution(Binomial and poisson )

1. A binomial variable x satisfies the relation 9 P(X=4)= P(X=2) when n=6. Find the value of the parameter p.

2. Five unbiased coins are tossed 3200 times; find the frequencies of the distribution of heads and tails and tabulate the results.

3. Suppose 5% of the inhabitants of Calcutta are cricket fans. Determine the probability that a sample of 100 inhabitants will contain at least 3 cricket fans.

4. A die is rolled 6 times. Find the probability that each of the six faces appears exactly once.5. A department in a works has 10 machines, which may need adjustments from time to time

during the day. Three of these machines are old, each; having a probability of 1/11 of needing adjustment during the day, and 7 are new, having corresponding probabilities of 1/21. Assuming that no machine needs adjustment twice on the same day, determine the probabilities that on a particular day (i) just 2 old and no new machines need adjustment, (ii) if just 2 machines need adjustment, they are of the same type.

6. An irregular six-faced die is thrown and the expectation that in 10 throws it will give five even numbers is twice the expectation that it will give four even numbers. How many times in 10,000 sets of 10 throws would you expect it to give no even numbers?

7. Seven coins are tossed and number of heads noted. The experiment is repeated 128 times and the following distribution is obtained:

No.of heads 0 1 2 3 4 5 6 7 Total

Frequencies 7 6 19 35 30 23 7 1 128

Fit a Binomial distribution assuming (i) the coin is unbiased. (ii) the nature of the coin is not known.

8. A car hire firm has two cars, which it hires out day by day. The number of demands for a car on each day is distributed as poisson variate with mean 1.5. Calculate the proportion of days on which (i) neither car is used, (ii) one car remains idle (assuming that each car is used equally) (iii) no demand is refused nor any car remains idle and (iv) some demands are refused. Also calculate proportion of demands refused.

9. A manufacturer of cotter pins known that 5% of his product is defective. If he sells cotter pins in boxes of 100 and guarantees that not more than 10 pins will be defective, what is the approximate probability that a box will fail to meet the guaranteed quality?

10. If X and Y are independent Poisson variates such that P (X=1) = P (X=2) and P(Y=2) =P(Y=3) . Find the variance of X-2Y.

11. A skilled typist kept a record of mistakes made per day during 300 working days of a year:Mistakes per day (x) : 0 1 2 3 4 5 6

Number of days (frequency fx) : 143 90 42 12 9 3 1

Fit an appropriate Poisson distribution to the above data.

12.A telephone operator receives an average of 3 calls during three days within 11-00 to 11-05 AM. What is the probability that (i) an operator receives no call during that interval?

(ii) receives exactly one call in three days during that interval?

13. If X is a poisson variate such that P (X=2)=9 P(X=4) + 90 P(X=6)

Find mean and variance of Poisson Distribution and calculate P(X>2).

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3. Solution

Probability of inhabitant of Calcutta being a cricket fan=

= 0.05

Sample size = 100

Required probability = Atleast 3 are cricket fans.

Let x => Random variable = No. of cricket fans

Then x is the binomial variate where the parameters n = 100, p = 0.05

Its is = P(x=i) =

P (atleast 3 cricket fans) = P(x 3)

= 1-P(x<3)

= 1-[P(x=0) +P(x=1) +P(x=2)]

=1-[0.0059+0.031+0.081]

=0.8821

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9. A manufacturer of cotter pins known that 5% of his product is defective. If he sells cotter pins in boxes of 100 and guarantees that not more than 10 pins will be defective, what is the approximate probability that a box will fail to meet the guaranteed quality?

Soln. Let the random variable X denote the number of defectives in the box.

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Given: .

And n=100

The probability that a box will fail to meet the guaranteed quality

=P(it contains more than 10 pins defective)

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12. A telephone operator receives an average of 3 calls during three days within 11-00 to 11-05 AM. What is the probability that (i) an operator receives no call during that interval? (ii) receives exactly one call in three days during that interval?

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13. If X is a poissonvariate such that P (X=2)=9 P(X=4) + 90 P(X=6).Find mean and variance of Poisson Distribution and calculate P(X>2).


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