Question

If X∼Binom(n,p), E[X] = np. Calculate V[X] = np(1−p) by:

a) First show E[X(X−1)] + E[X] − (E[X])^2 = V[X] (Hint: Use propertie of E[·] and V[·]).

b) Show E[X(X−1)] = n(n−1)p^2

c) Use E[X] = np, a) and b) to discuss, V[X] = np(1−p).

Answer #1

a)

Now,

b)

c) Variance of X =

Show that if X ∈ N(µ, σ2 ), then E(X) = µ, and V ar(X) = σ 2

Suppose X is binomial random variable with n = 18 and p = 0.5.
Since np ≥ 5 and n(1−p) ≥ 5, please use binomial distribution to
find the exact probabilities and their normal approximations. In
case you don’t remember the formula, for a binomial random variable
X ∼ Binomial(n, p), P(X = x) = n! x!(n−x)!p x (1 − p) n−x . (a) P(X
= 14). (b) P(X ≥ 1).

The normal approximation of the binomial distribution is
appropriate when:
A. np 10
B. n(1–p) 10
C. np ≤ 10
D. np(1–p) ≤ 10
E. np 10 and n(1–p) 10

Suppose that x has a binomial distribution with n
= 202 and p = 0.47. (Round np and n(1-p) answers
to 2 decimal places. Round your answers to 4 decimal places. Round
z values to 2 decimal places. Round the intermediate value (σ) to 4
decimal places.)
(a) Show that the normal approximation to the
binomial can appropriately be used to calculate probabilities about
x
np
n(1 – p)
Both np and n(1 – p) (Click to select)≥≤
5
(b)...

The normal approximation of the binomial distribution is
appropriate when
np ≥ 5.
n(1 − p) ≥ 5.
np ≤ 5.
n(1 −
p) ≤ 5 and np ≤ 5.
np ≥ 5 and n(1 − p) ≥ 5.

Compute E(N^2), for N∼Geom(p), by conditioning on the outcome of
the first trial. Use this to calculate Var(N)

Please show steps for solving all parts. im really confused,
thank you!
Calculate E, v, & r for Z=1 (Hydrogen), n=1 & 2; and E,
v, & r for Z=2(Helium), n=1

E(X)=3, V(X)=2, E(Y)=5, and V(Y)=1. If Z=3X-5Y, calculate its
expected value and variance. Please give in depth work for part
b.
a) E(Z)=-16
b)V(Z)=

Show that if x ∈ P, then x^(-1) ∈ P. Hint: show that a
contradiction will follow if one assumes that x ∈ P and x ∉
P.

Suppose that V is finite-dimensional and P ∈ L(V ) satisfies P^2
= P. Show that E(0, P) =null P and E(1, P) = range P. Deduce that P
is diagonalizable.

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