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STA301 (Statistics & Probability)

Assignment No.3

Solved & Attempt By

Innocent Prince

 

Question 1:

  1. Flows in a certain type of drapery material appear on the average of three in 250 square feet. If we assume the poison distribution, fin the probability of at most one flow in 450 Square feet.

 

P(X = x) = e-2 2x                                       x = 0, 1, 2, 3, 4…..

                     x!

 

Answer:

Sign of at most is (≤)

So,

We have to find:

 

As he said (One flaw in 450)

So (x=1)

 

Now

P(x ≤ 1) = e-5.4 (5.4)1

                        1

              = (o.oo4) (5.4)

                          1

P(x ≤ 1) = 0.0216

 

[ùe-5.4 =          1             = 0.004 ] (See Example on page # 217)

                (2.71828) -5.4

 

  1. What is the mean and variance for the given poison distribution of a random variable X;

 

Answer:

 

X

X2

0

0

1

1

2

4

3

9

4

16

X=10

X2 = 30

 

Mean = X/n

            = 10/5

            = 2

 

So, Mean is 2

 

Now find variance

V = X2/n – (x/n) 2

V = 30/5 - (30/5) 2

V = 6 – 36

 

 

 

Question 2:

              I.      For the following probability distribution;

Show that E (5x + 3) = 5E(x) + 3

 

X

P(X)

0

0.3

1

0.6

2

0.1

 

 

Answer: (see page 178)

X

P(X)

XP(x)

0

0.3

0

1

0.6

0.6

2

0.1

0.2

 

P(x) = 1

XP(x) = 0.8

 

µ= E(X) = Xp(x) = 0.8

So

5(EX) + 3

5(0.8) + 3

4+3 = 7

 

Hence Prove that,

E (5x + 3) = 5E(x) + 3 are equal

 

B) Find the distribution function for the following density function.

F(x) = 1/8x               0 ≤ x ≤ 4

 

Answer: ( see lect 25 )

P [0 ≤ x ≤ 4] = 4f0 f(x) dx

                    = 4f0 2x dx

                    = 2 [x2/2]

                    = [42]

                    = 16

 

 

 

 

 

 

Question 3:

(See & read page # 189 for joint probability distribution)

 

Let X & Y have the joint probability distribution described as follows,

X

1

2

3

Y

 

 

 

1

1/12

1/6

0

2

0

1/9

1/5

3

1/18

¼

2/15

 

Find the two marginal probability distribution for X and Y.

 

Answer:

X

1

2

3

P(X = xi)

G(x)

Y

1

1/12 = 0.08

1/6 = 0.17

0

0.25

2

0

1/9 = 0.11

1/5 = 0.2

0.31

3

1/18 = 0.06

¼ = 0.25

2/15 = 0.13

0.44

P(Y = yj)

0.14

0.53

0.33

1

 

Marginal Probability Distribution for X & Y

So

 

F(x, y) = P (X = xi and Y = yj)

            = P (X = xi) (Y = yj)

           = 1. 1

           = g(x) h(y) = 1

 

 

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Attachments:

Replies to This Discussion

Question part 2.... is rong... correct solution upload soon... thnks

thnx............alot

See the attached file 

Attachments:

question 1 nd part b... 

mean n variance...

Mean is 2

 

Now find variance

E(X) = μ
E(X) = 2 
V(X) = σ2 = μ
V(X) = σ2 = 2

Dear Fellows

Please find attached herewith complete solution of STA301 assignment for reference and guidance only. The contributor holds no liability for any consequence arising out from copying this solution completely or partially.

Regards,
BlackMist 

Attachments:

q1 part b m variance m x nd x2 dnu ki value 30 q put ki hue h jb k x ki value to 10 h r x2 ki 30

yaar konsa theak hai. 1 he group me 2 ,2 solutions given hain. or confuse ker dya hai.

100% solution kon konsy question ka theak hai.

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