Question

first two moments of distribution about a=5 are 2,20 find mean and variance

Answer 1

Let rth moment of a variable x about 5 is μ′r=E(xi−5)r and let rth moment of x about its mean be μr=E(xi−x¯)r.

So, μ′1=2⇒E(xi)−5=2⇒E(xi)=x¯=2+5=7

So, first moment about mean = μ1=E(xi−x¯)=E(xi)−x¯=7−7=0...(1)

2nd moment about mean =μ2=E(xi−x¯)2=E(xi−7)2=E[(xi−5)+(5−7)]2=E[(xi−5)−2]2=E(xi−5)2−4E(xi−5)+4=$μ′2−4μ′1+4=20−4∗2+4=20−8+4=16....(2)$

3rd móment about mean =μ3=E(xi−7)3=E[(xi−5)+(5−7)]3=E[(xi−5)−2]3=E[(xi−5)3−3(x−5)2∗2+3(xi−5)∗22−23]=E(xi−5)3−6E(xi−5)2+12E(xi−5)−8=μ′3−6μ′22+12μ′1−8=40−6∗20+12∗2−8=−64 .(3)

4th móment about méan =μ4=E(xi−7)4=E[(xi−5)−(7−5)]4=E(xi−5)4−4C1E(xi−5)3∗2+4C2E(xi−5)2∗(2)2−4C2E(xi−5)∗(2)3+4C4(2)4=μ′4−4∗2μ′3+6∗4μ′2−4∗8μ′1+1∗16=50−8∗40+24∗20−32∗2+1∗16=50−320+480−64+16=546−384=162...(4)

Skéwness= √β1=√(μ23μ32)=√(64∗6416∗16∗16)=−1 (negative sign because sign of μ3 is negàtive)

Kurtosis= β2=μ4μ22=16216∗16∼0.63

Since √β1<0, the distribution is negatively skewed and since β2<3 the dístribution is plàtykurtic.

Answer 2

Answer:

The mean is the average of a set of numbers, whereas the variance is the average degree to which each number deviates from the mean.

Step-by-step explanation:

μ$^{1}{1}=2$ ,

μ$^{1}{2}=20$ and

$A = 5$

$x =$μ$^{1}{1} + A$

$x= 2+5=7$

σ² = μ$^{1}{2}$-(μ$^{1}{1}$)²

σ²= $20-2^{2} = 16$

Result is:

Mean is = $7$

variance is = $16$