Question

Please don’t answer me by hand written.. Would be better if you use your PC to answer so it’s clear for me to read . Thanks !

Question 1 A continuous random variable X which represents the amount of sugar (in kg) used by a family per week, has the probability density function c(x-1(2-xsxs2 ; otherwise f(x) (i) Determine the value of c ii) Obtain cumulative distribution function (iii) Find P(X<1.2). Question 2 Consider the following cumulative distribution function for X 0.3 0.6 0.8 0.9 1.0 (i) Determine the probability distribution. ii) Find P(X<1). iii Find P(O <Xs5). Consider the following pdf ,f(x) = 2k ; 1<x<2 0otherwise (G) Determine the value of k. i) Find POX 0.3. iii) Find (0.1 < X1.5)

Answer 1

As HOMEWORKLIB answering guideline i have answered only the first question

1)

i) by definition of PDF we get

$\int_{1}^{2}f(x)dx=1$

So, here,

$\int_{1}^{2}c(x-1)(2-x)dx=1$

now, (x-1)(2-x)=2x-x²-2+x=3x-x²-2

So we have

$\int_{1}^{2}c(3x-x^{2}-2)dx=1$

or,

$c(\frac{3}{2}x^{2}-\frac{1}{3}x^{3}-2x)=1$ at x is ranged from 1 to 2

or,

$c\left \{(\frac{3}{2}2^{2}-\frac{1}{3}2^{3}-2*2)-(\frac{3}{2}1^{2}-\frac{1}{3}1^{3}-2*1) \right \}=1$

or, c=6

--------------------------------------------------------------------------------------------

ii)

The CDF=F(x)

$F(x)=\int_{1}^{x}f(t)dt$

$\int_{1}^{x}6(t-1)(2-t)dt=\int_{1}^{x}6(3t-t^{2}-2)dt=6\left \{ \frac{3}{2}t^{2} -\frac{1}{3}t^{3}-2t\right \}$ at t from 1 to x

Hence,

$6\left \{( \frac{3}{2}x^{2}-\frac{1}{3}x^{3}-2x)-(-\frac{5}{6}) \right \}=6( \frac{3}{2}x^{2}-\frac{1}{3}x^{3}-2x&plus;\frac{5}{6})$

$F(x)=6((\frac{3}{2}x^{2}-\frac{1}{3}x^{3}-2x)-(-\frac{5}{6}))=6( \frac{3}{2}x^{2}-\frac{1}{3}x^{3}-2x&plus;\frac{5}{6});1\leq x\leq 2$

And F(x)=0;x<1

And F(x0=1;x>2

-----------------------------------------------------------------------------------------------------------------

iii)

$P(x<1.2)=\int_{1}^{1.2}f(x)dx=\int_{1}^{1.2}6(3x-x^{2}-2)dx=6(\frac{3}{2}x^{2}-\frac{1}{3}x^{3}-2x)$

at x =1 to 1.2

So,

$P(x<1.2)=6((\frac{3}{2}1.2^{2}-\frac{1}{3}1.2^{3}-2*1.2)-(\frac{3}{2}1^{2}-\frac{1}{3}1^{3}-2*1))=0.104$

----------------------------------------------------------------------------------------------------------------------------

PLEASE UPVOTE IF YOU LIKE MY ANSWER

THANK YOU