Probability and Statistics Homework Solutions and Answers

This document presents a solved Probability and Statistics assignment, addressing various questions related to probability distributions, expected values, conditional probabilities, and statistical analysis. The solutions include detailed calculations and explanations for each question, covering topics such as joint probability distributions, marginal distributions, and the application of probability concepts to real-world scenarios. The assignment demonstrates the application of key statistical formulas and theorems to arrive at numerical answers, providing a comprehensive guide for students studying probability and statistics. The document also includes a bibliography of relevant academic resources.

Running head: PROBABILITY AND STATISTICS
Probability and Statistics
Name of the student:
Name of the University:
Author note:

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PROBABILITY AND STATISTICS
Table of Contents
Answer to the question 1............................................................................................................2
Answer to the question 2............................................................................................................3
Answer to the question 3............................................................................................................3
Answer to the question 4............................................................................................................4
Answer to the question 5............................................................................................................5
Answer to the question 6............................................................................................................5
Answer to the question 7............................................................................................................5
Answer to the question 8............................................................................................................5
Answer to the question 9............................................................................................................6
Answer to the question 10..........................................................................................................6
Answer to the question 12..........................................................................................................7
Answer to the question 13..........................................................................................................7
Answer to the question 14..........................................................................................................8
Answer to the question 15..........................................................................................................8
Bibliography...............................................................................................................................9

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PROBABILITY AND STATISTICS
Answer to the question 1
Given
fX, Y (x, y) = c 0≤ x ≤ 5
0≤ y ≤ 9
P (X ≤ d) where d=2
Now
∬f ( x , y ) dx dy = 1
Or ∫
0
5
∫
0
9
c dx dy = 1
Or c (5-0) (9-0) = 1
Or c= 1/45
f X (x)= ∫
y
❑
c dy
=∫
0
9
1/ 45 dy
= 1/45 (9-0)
= 1/5
P(X ≤ 2) = ∫
0
2
1/5 dx
= 1/5 (2-0)
= 2/5
= 0.4

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PROBABILITY AND STATISTICS
Answer to the question 2
Y= aX2+b
a=1 and b= 2
E (X) =0 and V(X) =1
E (Y) = E (aX2+b)
= a E(X2) +b (expectations of constant is constant)
Now
V(X) = E(X2) – ((E(X))2
Or 1 = E(X2) – 0
E(X2) = 1
Now
E (Y) = 1* 1+2
=1+2
= 3
Answer to the question 3
Getting head = 1
Not getting head= 0
P= 0.5
Let X denote the total number of flip

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PROBABILITY AND STATISTICS
E(X) = 0/0.5+1/0.5
=2
Answer to the question 4
Given
fX, Y (x, y) = (x +y)/3 0 ≤ x ≤ 1
0≤ y ≤ 2
∬f ( x , y ) dx dy = ∫
0
1
∫
0
2
( x+ y)/3 dx dy
= 1/3∫
0
1
∫
0
2
(x+ y) dx dy
= 1/3((x2/2)01 + (y2/2)02
=1/3 ((1/2)+(4/2))
= (1/3) (5/2)
= 5/6
f X (x)= ∫
y
❑
( x + y ) /3 dy
=1/3∫
0
2
( x+ y ) dy
= 1/3 (∫
0
2
xdy+∫
0
2
( y ) dy)
= 1/3(2x+2)

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PROBABILITY AND STATISTICS
P (X> 0.5) = ∫
0.5
1
( 2 X+ 2)/3 dx
= 2/3 ( ∫
0.5
1
¿ ¿ dx + ∫
0.5
1
dx)
= 2/3 ((1/2-0.25/2) + (1-0.5)
= 2/3 (0.375+0.5)
= 0.58333
Answer to the question 5
Given
fX, Y (x, y) = 1/15 0≤ x ≤ 5
0≤ y ≤ 3
f Y (y)= ∫
x
❑
1/15 dx
= ∫
0
5
1/15 dx
= 1/15 (5-0)
= 1/3
P (Y< 1.6) = ∫
0
1.6
1/3 dx
= 1/3 (1.6-0)
= 0.5333

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PROBABILITY AND STATISTICS
Answer to the question 6
Max (X,Y) ≤ 5 that is PX, Y (x, y) = 3/28+6/28+12/28+1/28+2/28+4/28
=1
Answer to the question 7
From the question it is clear that P(X=1) =0.3
Answer to the question 8
fX, Y (x, y) = 2 0 ≤ x ≤ y ≤1
f W (w)=∫
a
b
∫
g
h
2 dy dx
w= 0.5, W= x+y
a and b are constant and g and h potential limit.
At X=0
f y (y)=∫
a
b
∫
g
h
2 dx
x+y =0.5, y =0.5
Hence h (x) =0.5
Answer to the question 9
Given
fX, Y (x, y) = 4xy 0 ≤ x ≤1, 0≤y
f Y (y)= ∫
x
❑
4 xy dx

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PROBABILITY AND STATISTICS
= 4 y∫
0
1
x dx
= 4y (1/2)
= 2y
P (Y< 0.7) = ∫
0
0.7
2 y dy
= 2 ((0.7)2/ 2)
= 0.49
Answer to the question 10
P (X=1) =0.6
P (Y=1) = 0.3
P (XY=1) = P (X=1) P(Y=1)
= 0.6*0.3
= 0.18
Answer to the question 11
There are two value in the set 0 and 1.
The probability of success is ½ means 0.5 and failure is also 0.5.
Hence the variance is p*q= 1*0.5= 0.5
Answer to the question 12
FW (w) = P (W ≤ w)
Now P (W= 1.92) = 1- F (2w-1-w2/2)

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PROBABILITY AND STATISTICS
= 1- (2*1.92-1-((1.92)2/2))
= 1-1
=0
Answer to the question 13
N= 4
Let X denote the total number of tails.
X = 0, 1, 2 , 3, 4
Let Y denote the number of head in the last flip.
P =1/4= probability of success
P (X=1 ∩ Y= 0) = P(X=1) P(Y=0)
= ¼ *1/4
= 0.0625
Answer to the question 14
PX , Y (X,Y) = 1/25
(X,Y) takes 25 pairs. X= 1,2,…
PX(X) = ∑
y
P X , Y (X , Y )
= ∑
y
p ( 1,1 ) + p ( 1,2 )+ p (1,3) +……….+ p(5, 5)
=25 (1/25)
= 1

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PROBABILITY AND STATISTICS
E (X) = ∑
x , y
X* P X ,Y (X , Y )
= 1/5 ( 1+2+3+4+5)
= 3
Answer to the question 15
f (X) =1 0 ≤ X ≤1
P (X< 0.77) = ∫
0
0.77
f ( x ) dx
= (0.77-0)
= 0.77
Bibliography
Li, N., Liu, X., Xie, W., Wu, J., & Zhang, P. (2013). The return period analysis of natural
disasters with statistical modeling of bivariate joint probability distribution. Risk
Analysis: An International Journal, 33(1), 134-145.
Man'ko, M. A. (2013). Joint probability distributions and conditional probabilities in the
tomographic representation of quantum states. Physica Scripta, 2013(T153), 014045.

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PROBABILITY AND STATISTICS