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)

       u_{1 =} E (x) ……… first moment

       u_{2 =} E (x) ………. second moment

          u_{3 =} E (x)…………third moment   

       u_{4 =} 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 = E (x²­)

  E (x²­) = _-∞∫^∞ x^2.   f(x) dx

  E (x²­) = ₁∫²x^2. 2 (x+1) dx

  E (x²­) = 2₁∫² x². (x-1) dx

  E (x²­) = 2₁∫² (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

 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

   = ₀∫¹ x (1+ 3y²) /4 .dy

   = ₀ ∫¹ x (1+ 3y²) /4 .dy

   = 1/4 ₀∫¹ (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)

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 

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

Z = X - µ

         ♪

Z= X-30

Which means

P (x ≥ 25)

Continuity correction

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.