Question

First three raw moments of a distribution about 2 are 3, 12 and 17 respectively. Find mean and

standard deviation of the distribution.​

Answer 1

Answer:

denote the n-moment about the value 2 by M2 = E[(x - 2)"]

E[(X-2)]=1... (1) E[(X-2)] = 16... (2)

M3 = E(X-2)³] = -40 ... (3)

From (1), E[X]-E[2] =1=E[X]-2=1= Mean = H₁ = E[X] = 3

From (2), E[(x - 2)²] =E[(X-3 + 1)²] = E[(X-3)² + 2(x-3) + 1]

= E(X-3)²]+2E(X-3)+E(1) = 16

|Variance = μ₂ = E[(X-3)²] = 16-2E[X-3]-E[1] = 16-2(E[X]-3) -1

= 16-2(3-3) -1 = 15

From (3), E[(x - 2)³] = E((x − 3) + 1)²]

= E[(X-3)³] +3E[(X-3)2] +3E[X - 3] + E[1] = E[(x-3)³] + 3(15) + 3(0) + 1

<=-40

Skewness = μ3=E[(x-2)³]=-40-3(15)-3(0)-1 = -86

n-moment about the value 0 is M,,o= E[(X-0)"]=E[X"]

M₁.0=E[X]=H₂ = 3

M2.0=E[X²] = E[(X−2+2)²] = E[(x - 2)²] + 4E[X-2] + E[4] = 16+4(1) + 4 = 24

M3.0=E[X³] = E[(X−2+2)³] =E[(X-2)³] +6E[(X-2)²]+12E[X-2] + E[8] =-40+ 6(16) +12(1) +8=7

Answer 2

SOLUTION

GIVEN

First three raw moments of a distribution about 2 are 3, 12 and 17 respectively.

TO DETERMINE

The mean and standard deviation of the distribution.

EVALUATION

We know that the r th raw moments of a distribution about a is

$\displaystyle \sf{ \mu_r = \frac{ \sum {(x_i - a )}^{r} }{n} }$

Now by the given condition a = 2

$\displaystyle \sf{ \mu_r = \frac{ \sum {(x_i - 2 )}^{r} }{n} }$

Now

$\displaystyle \sf{ \mu_1= 3 }$

$\displaystyle \sf{ \implies \frac{ \sum {(x_i - 2 )}^{} }{n} = 3 }$

$\displaystyle \sf{ \implies \frac{ \sum {x_i}^{} }{n} - 2= 3 }$

$\displaystyle \sf{ \implies \frac{ \sum {x_i}^{} }{n}= 5 }$

$\displaystyle \sf{ \implies Mean = 5 }$

$\boxed{ \: \: \displaystyle \sf{ Mean = 5 } \: \: }$

Now

$\displaystyle \sf{ \mu_2 = 12 }$

$\displaystyle \sf{ \implies \: \frac{ \sum {(x_i - 2 )}^{2} }{n} = 12 }$

$\displaystyle \sf{ \implies \: \frac{ \sum {x_i}^{2} - 4 \sum \:x_i + 4n }{n} = 12 }$

$\displaystyle \sf{ \implies \: \frac{ \sum {x_i}^{2} }{n} -\frac{ 4 \sum \:x_i }{n} + 4 = 12 }$

$\displaystyle \sf{ \implies \: \frac{ \sum {x_i}^{2} }{n} - (4 \times 5) + 4= 12 }$

$\displaystyle \sf{ \implies \: \frac{ \sum {x_i}^{2} }{n} - 16= 12 }$

$\displaystyle \sf{ \implies \: \frac{ \sum {x_i}^{2} }{n} = 28 }$

Variance

$\displaystyle \sf{ = \frac{ \sum {x_i}^{2} }{n} - {( \bar{x})}^{2} }$

$\displaystyle \sf{ = 28 - {5}^{2} }$

$\displaystyle \sf{ = 28 - 25 }$

$\displaystyle \sf{ = 3 }$

$\boxed{ \sf{ \: \: Variance} = 3 \: \: \: }$

Now

$\sf{Standard \: Deviation = + \sqrt{Variance} }$

$\sf{ \implies \: Standard \: Deviation = 1.732}$

$\boxed{ \sf{ \: \: Standard \: Deviation = 1.732 \: \: \: }}$

━━━━━━━━━━━━━━━━

Learn more from Brainly :-

1. If the variance of the poissons distribution is 2,find the probability for r=1,2,3 and 4 from the recurrence relation of...

https://brainly.in/question/23494639

2. If random variables has Poisson distribution such that p(2) =p(3), then p(5) =

https://brainly.in/question/23859240