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STA301 Assignment Idea Solution june 2012

Answer No .1

a)      The probability density function of a random variable is given as
F(x) = 2(x-1)               1<x<2
0                                        Elsewhere
      Calculate second moment about origin=?
      (As we know the moment about origin)
       u1 = E (x) ……… first moment
       u2 = E (x) ………. second moment
          u3 = E (x)…………third moment   
       u4 = E (x)…………fourth moment

        As per rave
        P (z≥-1.1)
        -1.1 se bare

      In the second diagram 0.5+0.5=1
      Shaded area is raga

     In the third diagram
     P (z≥ -1.1) =0.5+0.3643
     =0.8643 Answer  

   Second Moment about Origin

   Second moment about origin = E (x²­)

  E (x²­) = -∞x2.   f(x) dx
 
  E (x²­) = 12 x2. 2 (x+1) dx

  E (x²­) = 212 x2. (x-1) dx

  E (x²­) = 212 (x3 – x2) dx

  E (x²­) = 2 [x4/5 – x3/3]2

  E (x²­) = 2 [(2)4/4 – (2)3/3   -   ( (1)4/4 – (1)3/3 ) ]

  E (x²­) = 2 (48-32/12) - (3-9/12)

   E (x²­) = 2 [16/12 + 1/12]
      

   E (x²­) = 2 (17/12)


 E (x²­) = 17/6 Answer

 Part b)
 
 F (x,y) = x (1+3y2) / 4

 = 0 , elsewhere

    0 < x < 2 , 0 < y < 1
Find the marginal probability density function of x =?

 g(x) = -∞ f (x,y) dy

 0 < y < 1

   = 01 x (1+ 3y2) /4 .dy

   = 01 x (1+ 3y2) /4 .dy

   = 1/4 01 (x + 3xy2) dy
                                   Y=1
   = 1/4 [ xy+ 3xy3 /3]10

    = 1/4 [xy +xy3]1

    =1/4 [x(1) + x(1)3 – (x(0) + x(0)3]

   =1/4 [x+x-0]

   =1/4 (2x)

G(x) = x/2       for (0 < x < 2)

Answer No.2

a)      what is the probability that a poker hand of 5 cards contain exactly 2 Aces ?

n (S) = 52C5

Let a donate exactly 2 aces



             (4C2)      (48C3)
P(A) = ______________
              (52C5)

P(A)=  6 (17296)
            2598960
         

           103776
P(A)= ______         Answer
           2598960 

Part b)
A pair of dice is rolled 180 times. By using the normal approximation to binominal distribution , find the probability of a total of 7 occurs at least 25 times ?

Probability of 7 is 

(1,6) (2,5) (3,4) (4,3) (5,2) (6,1)

P = 6/36 = 1/36

Now

n=180
p= 1/6
q= 1-p = 1-1/6  = 5/6

µ = n p =180*1/6 =30

 ♪ = √ n p q

 ♪ = √ 180*1/6*5/6

 ♪ = 5

Now

Z = X - µ
        

Z= X-30
       5
At least 25 times
Which means

P (x ≥ 25)
Continuity correction

                         Working Notes
when we go to continuous form discrete we add or less 0.5

In this case he said at least 25 times which means 25 is include that is why we less 0.5 from 25 for continuity correction.

P ( x ≥ 24.5)
Now

Z = X - µ
        
Z = X-30
         5

Z = 24.5 – 30
            5
Z = -1.1 

Now
Go to page # 225 of Sta301 handouts and see the value of 1.1 in 0 column.

Because it is 1.10 last pay zero ha agar 1.12 rate ho to 1.1 ki value column 2 me dekhein.
Now 1.1   in column zero = 0.3643
Is ka matlab ha k origin say -1.1 ka fasla 0.3643 ha.

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