Question

(4) For each of the following distributions, write the support for the random variable and calculate the second absolute moment.

a. X ~ normal (0,1)

b. Y ~ binomial (100,0.2)

c. Z ~ Poisson (1)

Answer 1

4.

given data,
X,Y,Z are support the random variables
a. X ~ normal (0,1)
b. Y ~ binomial (100,0.2)
c. Z ~ Poisson (1)
finding the second absolute moment,
it means generally in statistics,
the first moment is the mean and
the second moment about the mean is the sample variance. so that,

a.
normal distribution (0,1) = (mean,variance)
the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 0
standard Deviation ( sd )= 1
Answer:
sample variance = sqrt(standard deviation) = 1

b.
Y ~ binomial (100,0.2)
BIONOMIAL DISTRIBUTION
pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where   
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial
I.
mean = np
where
n = total number of repetitions experiment is executed
p = success probability
mean = 100 * 0.2
= 20
II.
variance = npq
where
n = total number of repetitions experiment is executed
p = success probability
q = failure probability
variance = 100 * 0.2 * 0.8
= 16
III.
standard deviation = sqrt( variance ) = sqrt(16)
=4
Answer:
second moment (sample variance) = 16

c.
c. Z ~ Poisson (1)
POSSION DISTRIBUTION
pmf of P.D is = f ( k ) = e-λ λx / x!
where   
λ = parameter of the distribution.
x = is the number of independent trials
I.
mean = λ
= 1
possion distribution, mean = variance = λ=1

answer:
sample variance (second moment) = 1