Question

Question 3 [20 marks] A measure of skewness is defined by Y such that: Note that when a distribution is symmetrical about the mean, the skewness is equal to zero. If it is skewed to the right, the measure of skewness will be positive; if it is skewed to the left, the measure of skewness will be negative. Let X be random variable, and a function f(x) is defined such that 21, 2, 3, 4, 5, 6, f(x)- (a) Create a table for f(x) for each value of x. Does f(x) represent a probability distribution function for X? Give reasons for your answer. (2 marks) (b) Draw a bar chart for f(x) against a and comment on its shape. This may be done meatly by hand or in any software package of your choosing). (2 marks) (i) Comment on the shape of the distribution. (I mark) (ii) Given your answer to (b)(0), would you expect the measure of skewness to be positive or negative or zero? Justify your answer. (1 mark) (c) Create a table for the cumulative distribution function, F(z), and plot it. Hence, median of X, denoted by to.s. [This may be done neatly by hand or in any software package of your choosing] (3 marks) (d) Determine the mean u - E(X). Show all your working. Leave your answer as a fraction and as a decimal correct to 4 decimal places. (2 marks) (e) Determine E(X2), leaving your answer as a fraction in simplest form. (2 marks) (f) Calculate the variance of X and hence the standard deviation of X. Leave your swers as decimals correct to 4 decimal places. (2 marks) (g) Determine an expression for EL(X )] in terms of E(X3), E(X2) and (3 marks (h) Hence calculate the numeric value of the measure of skewness, Y. Comment on its value with reference to your answer to (b) (ii). (2 marks)

Answer 1

Given the pdf of the discrete random variable ,

$f\left ( x \right )=\left\{\begin{matrix} \frac{2^{6-x}}{64} ,& x=1,2,3,4,5,6\\ \frac{1}{64}, & x=7 \end{matrix}\right.$

a) A table of discrete probability values corresponding to each RV is created as follows,

X	1	2	3	4	5	6	7
f(x)	$\frac{1}{2}$	$\frac{1}{4}$	$\frac{1}{8}$	$\frac{1}{16}$	$\frac{1}{32}$	$\frac{1}{64}$	$\frac{1}{64}$

We can see that

$\sum_{x=1}^{7}f\left ( x \right )=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{64}=1$ .

Hence the given $f\left ( x \right )$ is probability density function.

b) A barchart of $f\left ( x \right )$ is plotted using R.

The commands are

> p <- c(0.5,0.25,0.125,0.0625,0.03125,0.015625,0.015625)
> barplot(p,xlab = "X",ylab = "probability")

The plot is

i) The chart is skewed to right.

ii) Since the chart skewed to the right, the skewness is positive.

c) The cumulative density function is table is given below.

X	1	2	3	4	5	6	7
F(x)	$\frac{1}{2}$	$\frac{3}{4}$	$\frac{7}{8}$	$\frac{15}{16}$	$\frac{31}{32}$	$\frac{63}{64}$

The CDF is plotted as given below.

R commands are

> cp <- c(1/2,3/4,7/8,15/16,31/64,63/64,1)
> barplot(cp,xlab = "X",ylab = "cumulative probability")

From the graph $F\left ( x \right )=0.5$ , when $x=1$ . Hence median is ${\color{Blue} x_{0.5}=1}$ .

d) The expected value is

$\mu =E\left ( X \right )\\ \mu =\sum_{x=1}^{7}xf\left ( x \right )\\ \mu =\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\frac{6}{64}+\frac{7}{64}\\ {\color{Blue} \mu =\frac{127}{64}=1.9844}$

e) The second moment is

$E\left ( X^2 \right ) =\sum_{x=1}^{7}x^2f\left ( x \right )\\ E\left ( X^2 \right ) =\frac{1}{2}+\frac{2^2}{4}+\frac{3^2}{8}+\frac{4^2}{16}+\frac{5^2}{32}+\frac{6^2}{64}+\frac{7^2}{64}\\ {\color{Blue}E\left ( X^2 \right ) =\frac{367}{64}}$