STA301 Assignment No. 3 solution
Wednesday, June 15, 2011 Posted In STA Edit ThisAssignment No.3 (Course STA301)
Spring 2011 (Total Marks 30)
Deadline
Your Assignment must be uploaded/ submitted before or on 15th June, 2011
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Objective(s) of this Assignment:
The assignment is being uploaded to strengthen
- The students’ BASIC concepts regarding the Probability and Probability Distributions.
- To learn the applications of the Probability Distributions.
- To learn the use of Bivariate distribution/function for finding the probabilities and correlations.
Assignment 3 (Lessons 23-30)
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.
b) A random variable ‘x’ is said to be uniformly distributed such that 
(i) Write down the density function of ‘x’
(ii) Prove that total area between the given limits is unity.
c) What is the probability that a poker hand of 5 cards contain exactly 1 ace?
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?
Question 2: Marks:3+3+4+5=15
a) The Probability distribution of X is given below:
Compute,
i) P(x=2)
ii) E(X)
X | 0 1 2 3 4 |
P(X) |      |
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).
c) The given information is obtained from a joint distribution of X and Y. Calculate Correlation coefficient () between X and Y.
d) Find the value of A, such that the function f(x, y) is a density function.
Solution:
QUESTION 1; Part B
here is the Q:1 part B solution. Also include it in your file..
b) If Z is a standard normal random variable with mean 0 and variance 1, then find the Lower quartile.
Solution
Mean , Āµ = 0
Variance Š±2 = 1
Q1 = Āµ - 0.6745 Š±
Q1 = 0 - .6745 (1)
Q1 = -.6745
here is the Q:1 part B solution. Also include it in your file..
b) If Z is a standard normal random variable with mean 0 and variance 1, then find the Lower quartile.
Solution
Mean , Āµ = 0
Variance Š±2 = 1
Q1 = Āµ - 0.6745 Š±
Q1 = 0 - .6745 (1)
Q1 = -.6745
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:
0.29947... which is just under 30%.
QUESTION 1; Part C
100% correct but its not my subject.... Plz do make changes in it.
ALI Also include this one as well in your file. i have solved it for urself...
c) What is the probability that a poker hand of 5 cards contain exactly 1 ace?
Solution:
Since it is "at least" one ace, you have to account for drawing 1, 2, 3, or 4 aces. It is easier to calculate the following: Probability of drawing 0 aces. The probability of drawing
"at least" 1 ace is:
Pr[drawing at least 1 ace] = 1 - Pr[drawing 0 aces]
The probability of drawing 0 aces is:
C(48,5)/C(52,5)
The numerator is the ways we can draw 5-card hands out of 48 cards
(all the cards except for the aces). The denominator is the ways we
can draw any 5-card hands. So the ratio is the number of ways of
drawing no aces. If we subtract it from 1, we get the probability of
drawing "at least" 1 ace.
Pr[drawing at least 1 ace] = 1 - Pr[drawing 0 aces]
= 1 - (C(48,5)/C(52,5))
= 1 – (1712304/ 2598960)
= 1 - 0.658842
= 0.34 Ans
STA301 Assignment No. 3 solution
QUESTION 1; Part B
here is the Q:1 part B solution. Also include it in your file..
b) If Z is a standard normal random variable with mean 0 and variance 1, then find the Lower quartile.
Solution
Mean , Āµ = 0
Variance Š±2 = 1
Q1 = Āµ - 0.6745 Š±
Q1 = 0 - .6745 (1)
Q1 = -.6745
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:
0.29947... which is just under 30%.
QUESTION 1; Part C
100% correct but its not my subject.... Plz do make changes in it.
ALI Also include this one as well in your file. i have solved it for urself...
c) What is the probability that a poker hand of 5 cards contain exactly 1 ace?
Solution:
Since it is "at least" one ace, you have to account for drawing 1, 2, 3, or 4 aces. It is easier to calculate the following: Probability of drawing 0 aces. The probability of drawing
"at least" 1 ace is:
Pr[drawing at least 1 ace] = 1 - Pr[drawing 0 aces]
The probability of drawing 0 aces is:
C(48,5)/C(52,5)
The numerator is the ways we can draw 5-card hands out of 48 cards
(all the cards except for the aces). The denominator is the ways we
can draw any 5-card hands. So the ratio is the number of ways of
drawing no aces. If we subtract it from 1, we get the probability of
drawing "at least" 1 ace.
Pr[drawing at least 1 ace] = 1 - Pr[drawing 0 aces]
= 1 - (C(48,5)/C(52,5))
= 1 – (1712304/ 2598960)
= 1 - 0.658842
= 0.34 Ans
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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.
ANS
Mean,��= 0
Variance 62= 1
Q 1 =��-0.67456
Q1=0-.6745(1)
Q1= -.6745 ANS
b) A random variable ‘x’ is said to be uniformly distributed such that
(i) Write down the density function of ‘x’
(ii) Prove that total area between the given limits is unity.
ANSWER
DEFINTION
THE PROBIBLITY DENSITY OF A CONTINOUS RANDOM VARIABLE IS A FUNCTION WHICH CAN BE INTEGRATED TO ONTAIN THE PROBIBLITY THAT THE RANDOM VARIABLE TAKE A VALUE IN A GIVEN INTERVAL.
PART B
PROVE THAT TOTAL AREA BETWEEEN THE GIVEN LIMIT IS UNITY
ANS
THE TOTAL PROBIBLITY FOR ALL POSSIBLE VALUE OF THE CONTINOUS RANDOM VARIABLE X IS 1:
∫F(x) dx= 1
b) What is the probability that a poker hand of 5 cards contain exactly 1 ace?
Answer
First find the probility that the first card is an Ace and the 2nd to 5th are not:
(4/25) *(48/51)*(47/50)*(46/49)*(45/48).
Next to find the probability that he second card is an Ace and the first ,3rd, 4th and 5th are.
(48/52)*(4/51)*(47/50)*(46/49)*(45/48).
Next , the 3rd card is an Ace; the other is not:
(48/52)*(47/51)*(4/50)*(46/49)*(45/48).
Next , 4rd card is an Ace ; not the other:
(45/52)*(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 this answer is,
0.29947…which is just under 30%
c) 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
Z=2.3-3/0.5 = -1.4→p(x<2.3)=p( z < -1.4)=0.0808.
d) 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
X 0 1 2 3 4
P(X)
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
⌠ (2x-1)dx=[2x-x]=1
c) The given information is obtained from a joint distribution of X and Y. Calculate Correlation coefficient ( ) between X and Y.
Answer
Pxy=Cov(X,Y)/√ver(x) ver(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,
Var(x)=E(x)² -[E(x)]² =0.24
Var(y)=E(y)²-[E(y)]²=0.6
Pxy=0.096/√0.24 x 0.6= 0.25
d) Find the value of A, such that the function f(x, y) is a density function.
∫∫ ay (1-x)dxdy=1
a∫y(1-x)dx)dy=1
a∫(∫(y-yx)dx)dy=1
a∫[yx-y x²/2 dy]=1
after addining limit we get
A∫[y²/2-y²/4]=1
A[2-1/2]=1
A[1/4]=1
A=4
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Correction in Assignment no.3 Dated: Jun 20, 11
Dear Students,
It is to inform you that there was a mistake in the Question 2 (b), Assignment No. 3.
The given (incorrect) pdf was f(x) = (2x-1), 1<x < 2. This violates the basic property of a probability density function (pdf); that its total area under the curve MUST be equal to one.
The correct pdf is: f(x) = 2(x-1), 1<x < 2.
This ambiguity created difficulty for student in solving the problem. So, ALL the students will be awarded FULL marks in Q. No. 2 (b)
STA department sincerely apologize for the inconvenience!
