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Title: Probability and Statistics II tutorial
Description: Well comprehensive notes on Probability and Statistics II tutorial
Description: Well comprehensive notes on Probability and Statistics II tutorial
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PROBABILITY & STATISTICS II TUTORIAL
(1) Let the joint distribution function of X1 and X2 be given by
(
8x1 x2, 0 < x1 < x2 < 1
f (x1, x2 ) =
0,
elsewhere
find (i) E[X2 |X1 ]
(ii) Var[X2 |X1 ]
(2) The joint probability density function of two random variables X and Y is given by
(
3
(4 − 2x − y), x > 0, y > 0, 2x + y < 4
16
f (x, y) =
0, elsewhere
Determine (i)the conditional distribution of Y given X (ii) Pr(Y ≥ 2|X = 0
...
(4) Let X1 , X2 , X3 have joint probability distribution function
(
k(x21 + x2x3 ), 0 ≤ x1 ≤ 1, 0 ≤ x2 ≤ 1, 0 ≤ x3 ≤ 1
f (x1, x2 , x3) =
0, elsewhere
Find
(i) the value of k
...
Find the correlation matrix of Y
...
Let Z1 = X1 + 2X2 − 4X3 and Z2 = X1 + X2 + X3 , determine
(i) E[Z], Z = (Z1 , Z2 )
0
(ii) the values of x and y such that Z1 and Z2 are independent and V ar(Z2 ) = 10
...
, Xn be n independent and identically distributed random variables, each having
Poisson distribution with rate parameter λ
...
(9) Let X1 , X2 ,
...
Use characteristic function technique to find the
Pn
i=1 Xi
distribution of X =
...
What can you say about the variables?
2
(11) Let X1 , X2 ,
...
P
Further let Zk = ki=1 Xi
...
, Xk be n independent random variables , each having probability distribution
function
( −λ xi
e i λi
,
xi = 0, 1, 2,
...
(15) Let X1 , X2 ,
...
(16) Among fourty-five patients, nineteen had mild disease status, twelve had moderate disease
and the rest had severe disease status
...
(17) The average life-length of a certain kind of electric bulb is classified into four categories
as follows:(< 500) hrs, (500− < 1000)hrs, (1000− < 1500)hrs and (> 1500) hrs
...
15, 0
...
2 and 0
...
Find the probability that of 20 such
bulbs, three will have an average life-length (< 500)hrs, six will have average life-length
(500− < 1000) hrs, four will have average life-length (1000− < 1500) hrs and seven will
have average life-length (> 1500)hrs
...
3)x1 (0
...
4)x3 (0
...
15et1 + 0
...
3et3 + 0
...
1et5 ]30
Determine the probability distribution function of X
...
Hence obtain
E[(X1 , X3 , X5 ) | X2 = 8, X4 = 6] and Var[(X1 , X3 , X5 ) | X2 = 8, X4 = 6]
(iii) the characteristic function of the distribution of (X2 , X3 , X4 )
...
Inspectors assign a rating of high, medium or low to each component inspected
...
5 , Pr(medium) = 0
...
1
...
(21) Let X1 , X2 , X3 , X4 , X5 be independent random variables having Poisson distributions with
respective rate parameters α1 = 3
...
4, α3 = 2
...
6, α3 = 1
...
Determine the
P
conditional distribution of (X1 , X2 , X3 ) given 5i=1 = 30
(22) Let Y = [Y1 , Y2, Y3 ]0 have tri-variate normal distribution with mean vector µ = [1, 3, 2]0 and
variance-covariance matrix
5 3 1
V ar(X) = 3 7 2
1 2 8
Determine the probability distribution of Z = [Z1 , Z2]0 where Z1 = Y1 + 2Y2 + 4Y3 and
Z 2 = Y1 − Y2 − Y3
0
(23) Let X = (X1 , X2 , X3 ) has a tri-variate normal distribution given by
f (x1 , x2, x3) =
1
exp{−Q}
k
where
Q = 2x21 + 4x22 + 6x23 + 4x1 x2 − 12x1 − 20x2 − 18x3 + d
such that k and d are constants to be determined
...
(ii) the characteristic function of U = Y1 + Y2 + 3Y3 + 2Y4
(ii) the probability distribution distribution of Z = [Z1 , Z2 ] where Z1 = 2Y1 − Y2 + Y3 + Y4
and Z2 = 3Y1 + Y2 − 2Y3 − Y4
(iii) the conditional distribution of Y2 , Y4 given (Y1 , Y3 )
(25) Let X1 , X2 ,
...
2 and standard deviation 1
...
Find P (21
...
4, 0
...
328)
...
8
marks and standard deviation of 8
...
Suppose a random sample of 81 students is
taken and a sample mean X and a sample variance S 2 of their marks are calculated
...
9 < X < 69
...
703 < S 2 < 115
...
612 <
X < 0
...
(28) Let X be the sample mean of a random sample of size 81 drawn from population that has
probability density function
(
1 2
1
xe− 8 x ,
0
f (y) =
0, elsewhere
Find Pr(1
...
02)
...
4 < X < 0
...
(31) Let X1 , X2 , X3 ,
...
, Xn)
...
Hence find its mean and variance values
...
, Xn be a random sample from the distribution
f (x) = 5(1 − x)4
0
Further let Z = min(X1 , X2 ,
...
Find the probability density function of Z and hence compute the mean of Z
(33) The joint probability density of X1 , X2 , X3 is
e−(x1 +x2 +x3 )
f (x1 , x2, x3) =
0
6
xi > 0 i = 1, 2, 3
otherwise
(i) Using the change of variable technique, determine the joint distribution of Y1 = X1 ;Y2 =
X1 + X2 ; Y3 = X1 + X2 + X3
(ii) Using result of part(i), find the marginal distribution of Y3 and the marginal distribution
of Y1 , Y2
(34) Let X1 , X2 , X3 be three random variables with joint probability density function
(
1 −(4x1 +x2 +x3 )
e
, xi > 0, i = 1, 2, 3
4
f (x1 , x2, x3) =
0, elsewhere
Find the probability density function of Y = X1 + X2 + X3
(35) Let X1 , X2 , X3 be independent random variables each having probability density function
f (xi ) = e−xi
xi > 0
(i) Find the joint probability density function of Y1 =
Y3 = X1 + X2 + X3
...
(ii) Find the marginal probability density of Z1
Title: Probability and Statistics II tutorial
Description: Well comprehensive notes on Probability and Statistics II tutorial
Description: Well comprehensive notes on Probability and Statistics II tutorial