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) = 1∫2x2. 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
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)
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.
