sta301 3 Statistics & Probability assignment solution

Question

1:

a) 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;

Solution:

(a)

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.

Solution

We have 3 flaws per 250 square feet its mean

In this problem we are interest in 450 square feet and we can

find the value of through divided 450 square feet by 250 square feet

which is given below.

T= = =1.8

T=1.8

=3*1.8=5.4

=5.4

According to the question we want to the probability of at most one

flaw in 450 square feet

By putting the value of x

= 0.004517+0.0244

=0.02892

(b)

What is the mean and variance for the given Poisson distribution of a

random variable X;

SOLUTION:

We want to find the mean and variance from the given

Poisson distribution

AS we know that the formula of Poisson distribution

And according to the property of Poisson distribution with the

parameter, then its mean and variance are given by and

In given Poisson distribution we know that therefore and we also know

that therefore

Question

2:

a) For the following probability distribution;

Show that E (5X+3) = 5E(X)

X P(X)

X	P(X)
0	0.3
1	0.6
2	0.1

b) Find the distribution function for the following density function.

SOLUTION:

According to the question we want to find first of all we find the

calculation of which is given below:

S# X P(X) XP(X)

X	P(X)
0	0.3
1	0.6
2	0.1

SUM = 0.8

= = =0.8

Hence E(X) =0.8 now we find the expected value of random variable

5X+3, then

computations:

E (5X+3) = (5X+3)*P(X) we carry out the following

S# X P(X) 5X (5X+3) P(X)*(5X+3)

X	P(X)	XP(x)
0	0.3	0
1	0.6	0.6
2	0.1	0.2
	P(x) = 1	XP(x) = 0.8

SUM = 15 24 7

Hence =7, it should be noted that

For proving the both sides are equal then we put the values in both

sides which is given below:

7 = 5(0.8) +3

7 = 4+3

7 = 7

(b)

Find the distribution function for the following density function.

SOLUTION:

P [0 ≤ x ≤ 4] =1/8  4f0 f(x) dx

                    = 1/8  4f0 2x dx

                    =  1/8  [x²/2]

                    = 1/8  [4²_/2]

                    = 1/8*16/2

                    =  1

Question 3:

Let X and Y have the joint probability distribution described as follows,

X	1	2	3
Y
1	1/12	1/6	0
2	0	1/9	1/5
3	1/18	¼	2/15

Find the two marginal probability distribution for X and Y.

SOLUTION:

For finding the marginal distribution we first arrange table and find out

the joint probability distribution.

Joint probability :

Y

x

                         1             2                    3             P(X= )

                                                                                  g()

1                   1/12            0                  1/18             5/36

2                   1/6           1/9                  1/4             19/36

3                   0               1/5                2/15             1/3

P(Y= )        1/4           14/45           79/180          1.00

Marginal probability distribution :

X              

1         2           3

g(x)     

5/36    19/36      1/3

Y          

1         2            3h(y)    1/4    14/45    79/180