ID # mc110401469 Name: Nasir Mehmood Assignment No.3 (STA301)
Solution
Question No: 1a) Flaws in a certain type of drapery material
appear on the average of
three in 250 square feet. If we assume the Poisson distribution,
find the probability of at most one flaw in 450 Square feet.b) What
is the mean and variance for the given Poisson distribution of a
random variable X;
e 2 2 x P ( X = x) = x! Solution: e(5.1) 1 P ( X 1) = 1 =
x = 0,1, 2,3, 4,.........
( .004 ) ( 5.4 )1
= .0216
1 = = .004 5.4 (2.71828) Now we will know mean and variance for
the given Poisson distribution of a random variable Mean = 2
Variance = let x = 3 so the variance will be [(x X(bar))2]/n = 1/n
Question No: 2 a) For the following probability distribution; Show
that E (5X+3) = 5E(X) + 3
X 0 1 2 b)
P(X) 0.3 0.6 0.1
Find the distribution function for the following density
function. 1 f ( x) = x 0 x4 8 Solution: a):
= E(X) = 0.8 = 5[EX] +3 = 7 E (5X+3) = or 5(EX) + 3 = 5(0.8) + 3
= 4 + 3 = 7 Hence these are equal. Or E (5X+3) = 5E(X) +3 b) x2 P[0
x 4] = f(x)dx= 2xdx=2 = 42 = 16 2 0 04 4
Question No: 3 Let X and Y have the joint probability
distribution described as follows, 1 X Y 1 2 3 0 1/5 1/18 1/9 1/4
1/12 0 2 1/6 3
2/15 Find the two marginal probability distribution for X and
Y.
Solution:
Find the two marginal probability distribution for X and Y. f(x,
y) = P(X = xi and Y = yj) = P(X = xi). P(Y = yj) = g(x) h(y) = 1.1
= 1
