# STA301 Assignment No. 3 solution

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STA301 IDEA SOLUTIOON
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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.
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.
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
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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.

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?
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).

STA301 IDEA SOLUTIOON
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STA301 IDEA SOLUTIOON
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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.

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?
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)
P(x = 2) = 35/100 = 0.35
ii)    E(X)
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
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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).
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

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
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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

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
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
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