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STA301 Statistics and Probability Assignment 3 Solution Fall 2012

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.
Solution:

Obtain the value of “c”, so that is a density function.
density function ki value 1 hoti hai so hum nay is function ko limits k sath integrate karna hai 1 k equal rakh k so c ki value aa jaye gi

is mein 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=1
then
c(x-x2/2)value 0 to 1=1

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”.
solution
mean=np=18
variance=npq=3.52
npq/np=0.195=q
p=1-q
=1-.195=.80
np=18
n(.80)=18
n=22.50

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

Solution:

QUESTION 2 (a)

(i)

Formula: P(X=x)= e−μμx / x!

where x=6,  μ=4,  e=2.71828

Answer will be P(X=6)=0.1042

(ii)

Formula: P(X=x)= e−μμx / x!

where x=2,  μ=4,  e=2.71828

Answer will be P(X=2)=  0.14652

b) If, then calculate the value of. Where Y is defined as
Ans:
= 3 E(x)+ 5
= 3 (5) + 5
= 15+5
= 20

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