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Assignment No.3 (Course STA301)
Fall 2012 (Total Marks 15)
Your Assignment must be uploaded/ submitted before or on 23:59 January 16th, 2013
STUDENTS ARE STRICTLY DIRECTED TO SUBMIT THEIR ASSIGNMENT BEFORE OR BY DUE DATE. NO ASSIGNMNENT AFTER DUE DATE WILL BE ACCEPTED VIA E.MAIL).
Rules for Marking
It should be clear that your Assignment will not get any credit IF:
 The Assignment submitted, via email, after due date.
 The submitted Assignment is not found as MS Word document file.
 There will be unnecessary, extra or irrelevant material.  The Statistical notations/symbols are not well-written i.e., without using MathType software.
 The Assignment will be copied from handouts, internet or from any other student’s file. Copied material (from handouts, any book or by any website) will be awarded ZERO MARKS. It is PLAGIARISM and an Academic Crime.
 The medium of the course is English. Assignment in Urdu or Roman languages will not be accepted.
 Assignment means Comprehensive yet precise accurate details about the given topic quoting different sources (books/articles/websites etc.). Do not rely only on handouts. You can take data/information from different authentic sources (like books, magazines, website etc) BUT express/organize all the collected material in YOUR OWN WORDS. Only then you will get good marks.
Objective(s) of this Assignment:
 Properties of probability distributions.
 Application of probability distribution in real life.
 Mathematical expectation and its properties.
Assignment No: 3 (Lessons 23-30)
Question 1: Marks: 4+4=8
a) Given the function below
f (x)  c(1 x) 0  x 1
Obtain the value of “c”, so that f (x) is a density function.
b) A random variable “X” following binomial distribution has its mean and variance,
18 and 3.52 respectively. Calculate the value of “n” and “p”.
Question 2: Marks: 5+2=7
a) During a laboratory experiment the number of radio active particles passing
through a counter in one milli-second is 4.
I. What is the probability that 6 particles enter the counter in a given millisecond?
II. What is the probability that at most two particles enter the counter in a given
milli-second?
b) If E(X)  5 , then calculate the value of E(Y) . Where Y is defined as
Y  3X 5
-END

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Helo*-+*Ning aj extand day ha plz ab or na formula niklo phla he final solution submit krwe he kitne bar chnge kr k STA301 ASSIGNMNT 3 SOLUTION.  STA301 VU Current Assignment No. 3 Fall 2012 SolutionPosted on January 16, 2013 by JUST AN IDEA SOLUTIONS..Assignment No.3 ( STA301)Fall 2012 (Total Marks 15)DeadlineYour Assignment must be uploaded/ submitted before or on 23:59 January 16th, 2013Assignment N 3 (Lessons 23-30)Question 1: Marks: 4+4=8a) Given the function belowf (x)  c(1 x) 0  x 1Obtain the value of “c”, so that f (x) is a density function.b) A random variable “X” following binomial distribution has its mean and variance, 18 and 3.52 respectively. Calculate the value of “n” and “p”.Question 2: Marks: 5+2=7a) During a laboratory experiment the number of radio active particles passing through a counter in one milli-second is 4.I. What is the probability that 6 particles enter the counter in a given millisecond?II. What is the probability that at most two particles enter the counter in a given milli-second?b) If E(X)  5 , then calculate the value of E(Y) . Where Y is defined as Y  3X 5 -ENDSOLUTION:Question 2:a) During a laboratory experiment the number of radio active particles passing through a counter in one milli-second is 4.I. What is the probability that 6 particles enter the counter in a given milli-second?x=6,μ=4,e=0.01832Solution :P(X=6) = 0.01832 x 466!= 75.0208720P(X=6) = 0.1042II. What is the probability that at most two particles enter the counter in a given milli-second?x= 2,μ= 4,Soution:= 0.01832 x 40 + 0.01832 x 41 + 0.01832 x 420! 1! 2!= 0.01832 x 40 + 0.01832 x 41 + 0.01832 x 421 1 2= 0.01832 + 0.07328 + 0.14656= 0.23816b) If , then calculate the value of . Where Y is defined asY = 3X + 5E(Y) = 3 E (x) + 5= 3 (5) + 5= 15 + 5Y = 20……………………………….Q1 b)b) A random variable “X” following binomial distribution has its mean and variance, 18 and 3.52 respectively. Calculate the value of “n” and “p”.solutionmean=np=18variance=npq=3.52npq/np=0.195=qp=1-q=1-.195=.80np=18n(.80)=18n=22.50……………………..Lecture no 27 is related with the Assignment no.3For complete understanding first readout the lecture……………………..Question 1 me a part or question 2 me a part discuss kro in dono me koi help kr dey gaquestion 2 b part= 3 E(x)+ 5= 3 (5) + 5= 15+5= 20……………………..Q1 part b me integral lena he 0 to 1.densty function ki wja se ye integration 1 ki equal ho ga.integral 0 to 1(c(1-x)dx=1thenc(x-x2/2)value 0 to 1=1ans will be 2……………………..E(X)=5 then E(Y)=?Y=3X+5Y=3*5+5=20orY=20what is differenec bt Y and E(Y) ??then how to get E(Y)……………………..integral 0 to 1(c(1-x)dx=1thenc(x-x2/2)value 0 to 1=12Ans.b) A random variable “X” following binomial distribution has its mean and variance, 18 and 3.52 respectively. Calculate the value of “n” and “p”.Ans:Solutionmean=np=18variance=npq=3.52npq/np=0.195=qp=1-q=1-.195=.80np=18n(.80)=18n=22.50Question 2: Marks: 5+2=7a) During a laboratory experiment the number of radioactive particles passing through a counter in one milli-second is 4.1. 1. I. What is the probability that 6 particles enter the counter in a given milli-second?2. 2. II. What is the probability that at most two particles enter the counter in a given milli-second?Ans:This question is from 29 lec….!(a)(i)Formula: P(X=x)= e−μμx / x!where x=6, μ=4, e=2.71828Answer will be P(X=6)=0.1042(ii)Formula: P(X=x)= e−μμx / x!where x=2, μ=4, e=2.71828Answer will be P(X=2)= ?? (Calculate yourself)Probability formula:b) If, then calculate the value of. Where Y is defined asAns:= 3 E(x)+ 5= 3 (5) + 5= 15+5= 20……………………..QUESTION 2 (a)(i)Formula: P(X=x)= e−μμx / x!where x=6, μ=4, e=2.71828Answer will be P(X=6)=0.1042(ii)Formula: P(X=x)= e−μμx / x!where x=2, μ=4, e=2.71828Answer will be P(X=2)= 0.14652

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