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```					Mc100405351
Aqib Farooq

a) The probability density function of a random variable is given as
F(x) = 2(x-1)        1<x<2
0          Elsewhere
(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

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

Second moment about origin = E (x²)

E (x²) = -∞ ∫∞ x2. f(x) dx

E (x²) = 1∫2 x2. 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)

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

= 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
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

♪=√npq

♪ = √ 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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 views: 3 posted: 6/27/2012 language: pages: 4