Question

1. Discrete distribution for X' is given by the following table: Probabilities p Values A 0.2 20 0.2 40 0.3 60 0.2 80 0.1 100 Find distribution function fax) and median Me(X). Calculate expectation value MX), variance (dispersion) DA), standard error σ(X), asymmetry coefficient As(X) and excess Ex(X). 2. Calculate multiplier k. Find distribution function foc), mode Moc), median Me(x), expectation value M(x), variance (dispersion) D(x), standard error σ(x), asymmetry coefficient As(x) and excess Ex(x) for continuous distributions with the given probability densities a) b) 0 r<-2 -2 r>6 0sxs8 ρ(x) = k(4x-x2 +12) 6 x 1r+1 8s10 10 x>10 Calculate probability that X E 1-28 Calculate probability that x e -3:2]

This is a statistics question, unfortunately I have difficulties understanding problems with statistics and I need this question solved with full workings and also in an understandable way to revise. Your help will be highly appreciated, *I will give positive rating to the fullest*. Thank you!!

Answer 1

1) f(X)

X	f(X= x)
20	0.2
40	0.2
60	0.3
80	0.2
100	0.1
	1

Median

P(X <median) = 0.5

X	f(X= x)	cumulative probability
20	0.2	0.2
40	0.2	0.4
60	0.3	0.7
80	0.2	0.9
100	0.1	1

note that  

P(X< = 40) = 0.4 and P(X<= 60) = 0.7

as 0.5 is between 0.4 and 0.7

60 is median

$M(X) = \sum p(X= x)X = 20*0.2 &plus; 40 * 0.2 &plus; 60 * 0.3 &plus; 80 *0.2 &plus;100 *0.1$

= 56

D(X)= E(X^2) - (E(X))^2

X	f(X= x)	x*p	x^2*p
20	0.2	4	80
40	0.2	8	320
60	0.3	18	1080
80	0.2	16	1280
100	0.1	10	1000
	1	56	3760

E(X^2) = 3760

hence

D(X) = 3760 - 56^2

= 624

standard error = sd(X) = sqrt(D(X)) = sqrt(624) = 24.979999

he most frequently employed measure of the asymmetry of a distribution, defined by the relationship

11 = S/2

where and are the second and third central moments of the distribution, respectively

X	f(X= x)	x*p	x^2*p	(x - mu)^3
20	0.2	4	80	-46656
40	0.2	8	320	-4096
60	0.3	18	1080	64
80	0.2	16	1280	13824
100	0.1	10	1000	85184
	1	56	3760	48320

As(X) =   48320/(3760)^(3/2)

= 0.2095777

A scalar characteristic of the pointedness of the graph of the probability density of a unimodal distribution. It is used as a certain measure of the deviation of the distribution in question from the normal one. The excess is defined by the formula

where is the second Pearson coefficient (cf. Pearson distribution), and and are the second and fourth central moments of the probability distribution

X	f(X= x)	x*p	x^2*p	(x - mu)^3	(x- mu)^4
20	0.2	4	80	-46656	1679616
40	0.2	8	320	-4096	65536
60	0.3	18	1080	64	256
80	0.2	16	1280	13824	331776
100	0.1	10	1000	85184	3748096
	1	56	3760	48320	5825280

Ex(X) = 5825280/3760^2 - 3

= -2.58796