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STA301 Assignment No. 3 solution

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STA301 Assignment No. 3 solution Powered By Docstoc
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                                STA301 IDEA SOLUTIOON
                                 Created and written by Ali



Question 1:                                                          Marks:2+3+3+3+4=15
a) If Z is a standard normal random variable with mean 0 and variance 1, then find
the lower quartile.
Answer; -
        Mean , µ = 0
Variance б2 = 1
Q1 = µ - 0.6745 б
Q1 = 0 - .6745 (1)
Q1 = -.6745
b) A random variable ‘x’ is said to be uniformly distributed such that c  x  k
(i) Write down the density function of ‘x’
(ii) Prove that total area between the given limits is unity.
Answer; -
This is an example of this Question it is not correct answer;
        Probability Density Function

The probability density function of a continuous random variable is a function
which can be integrated to obtain the probability that the random variable takes a
value in a given interval.

More formally, the probability density function, f(x), of a continuous random
variable X is the derivative of the cumulative distribution function F(x):



Since                       it follows that:
         f (x )dx  F (b )  F (a)  p (a  x  b )
If f(x) is a probability density function then it must obey two conditions:

   a. that the total probability for all possible values of the continuous random
      variable X is 1:

                                STA301 IDEA SOLUTIOON
                                 Created and written by Ali
                                                  Generated by Foxit PDF Creator © Foxit Software
                                                  http://www.foxitsoftware.com For evaluation only.
                               STA301 IDEA SOLUTIOON
                                Created and written by Ali




    b. that the probability density function can never be negative: f(x) > 0 for all x.

(ii) Prove that total area between the given limits is unity.


Answer; -

    c.          the total probability for all possible values of the continuous random
         variable X is 1:




c) What is the probability that a poker hand of 5 cards contain exactly 1 ace?
Answer; -
First, find the probability that the first card is an Ace and the 2nd to 5th are not:

(4/52) * (48/51) * (47/50) * (46/49) * (45/48).

Next, find the probability that the 2nd card is an Ace and the first, 3rd, 4th and 5th are
not:

(48/52) * (4/51) * (47/50) * (46/49) * (45/48).

Next, the 3rd card is an Ace; the others are not:

(48/52) * (47/51) * (4/50) * (46/49) * (45/48).

Next, 4th card is an Ace; not the others:

(48/52) * (47/51) * (46/50) * (4/49) * (45/48).

Lastly, the 5th card is an Ace; not the others:

(48/52) * (47/51) * (46/50) * (45/49) * (4/48).

Adding these, the answer is:

                               STA301 IDEA SOLUTIOON
                                Created and written by Ali
                                             Generated by Foxit PDF Creator © Foxit Software
                                             http://www.foxitsoftware.com For evaluation only.
                             STA301 IDEA SOLUTIOON
                              Created and written by Ali




0.29947... which is just under 30%.

d) A certain type of storage battery lasts on the average 3.0 years, with a standard
deviation of 0.5 year. Assuming that the battery lives are normally distributed, find
the probability that a given battery will last less than 2.3 years.
Answer; -




e) The incident of occupational disease is such that the workers have 20 percent
chance of suffering from it. What is the probability that out of six more than 4
workers will come in contact of the disease?
Answer; -
   P(4 < x <6) = 20 /100 = 0.2



Question 2:                                                           Marks:3+3+4+5=15
   a) The Probability distribution of X is given below:
Compute,
    i)     P(x=2)
Answer; -
           P(x = 2) = 35/100 = 0.35
    ii)    E(X)
Answer; -
           E(X) = ∑(x) p(X) = 1.36 because
(x) p(X)
0
0.5
0.7
0.12
0.04
∑ (x) p(X) = 1.36


                             STA301 IDEA SOLUTIOON
                              Created and written by Ali
                                                       Generated by Foxit PDF Creator © Foxit Software
                                                       http://www.foxitsoftware.com For evaluation only.
                                   STA301 IDEA SOLUTIOON
                                   Created and written by Ali



  X                     0              1           2            3           4

                        1           1             35           4           1
 P(X)                  10           2            100          100         100


b) The p. d. f. of a random variable is given
        f(x) = (2x-1)              1<x < 2
                   0               Elsewhere
Calculate P(X<1.5).
Answer; -
        1
                                   1
         (2x  1)dx   2x  x 
        0
                                   0
                                         1 solve your self also i think it is the answer



c) The given information is obtained from a joint distribution of X and Y. Calculate
Correlation coefficient (  ) between X and Y.


                  21           34            49            33               86
        EX        , E Y   , E  XY      E ( X 2 )  and E  Y 2  
                  15           15            15            15               15

Answer; -

         Cov (X ,Y )
xy 
        V er (x ) V er ( y )
Cov(X,Y) = E(XY) - E(X) E(Y)
         = 3.26-(1.4)(2.26)=3.26-3.164=0.096
0.096 Now,
                               2
Var(x) = E (X )2   E (X )  0.24

Var(y) = E (Y ) 2  [E (Y )]2  0.6

                                   STA301 IDEA SOLUTIOON
                                   Created and written by Ali
                                                  Generated by Foxit PDF Creator © Foxit Software
                                                  http://www.foxitsoftware.com For evaluation only.
                                     STA301 IDEA SOLUTIOON
                                     Created and written by Ali



           0.096
xy                  0.25
          0.24 x 0.6


d) Find the value of A, such that the function f(x, y) is a density function.
          f  x, y   Ay 1  x          0  x  1,     0  y 1
                   0                       elsewhere

   Answer; -

    1 1

      ay (1  x )dxdy
    0 0
                           1

     1    1
   a  (  y (1  x )dx )dy  1
     0    0
     1    1
   a  (  ( y  yx )dx )dy  1
     0    0

     1                      1
               x2    
   a   yx  y    dy   1
     0
                2     0
   after adding limit we get
     1
           y 
   a   y   dy  1
     0 
            2
                       1
     y2 y2
    A          1
      2      4 0
      2 1 
    A       1
      2 
      1
    A   1
      4
    A  4 i think it is the answer solve your self also




                                     STA301 IDEA SOLUTIOON
                                     Created and written by Ali

				
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Description: STA301 Assignment No. 3 solution (thanks Ali, to share this with vusolutions)