Answer:
μ₁`= 1/6
μ₂= 5/36
Step-by-step explanation:
The rolling of a fair die is described by the binomial distribution, as the
- the probability of success remains constant for all trials, p.
- the successive trials are all independent
- the experiment is repeated a fixed number of times
- there are two outcomes success, p, and failure ,q.
The moment generating function of the binomial distribution is derived as below
M₀(t) = E (e^tx)
= ∑ (e^tx) (nCx)pˣ (q^n-x)
= ∑ (e^tx) (nCx)(pe^t)ˣ (q^n-x)
= (q+pe^t)^n
the expansion of the binomial is purely algebraic and needs not to be interpreted in terms of probabilities.
We get the moments by differentiating the M₀(t) once, twice with respect to t and putting t= 0
μ₁`= E (x) = [ d/dt (q+pe^t)^n] t= 0
= np
μ₂`= E (x)² =[ d²/dt² (q+pe^t)^n] t= 0
= np +n(n-1)p²
μ₂=μ₂`-μ₁` =npq
in similar way the higher moments are obtained.
μ₁`=1(1/6)= 1/6
μ₂= 1(1/6)5/6
= 5/36
9(9n-5)
9=0 9n-5=0
9n=5
9/5 is the answer and 0 or 9(9n-5)
The answer is .48 because 6/15 is 4. which means .48 is bigger
