Answer:
The Ballwin Bears are taller on average, and the Aviston Aces have players whose heights are more consistent.
Step-by-step explanation:
for derby dragon;
mean height = 72 inches
standard deviation = 1.2 inches
for aviston aces:
mean height = 70.8
standard deviation = 0.7 inches
for balwin bears;
mean height = 73 inches
standard deviation = 1.0 inches
the mean height of aviston aces < mean height of derby < mean height of balwin bears
So on average the balwin bears are taller.
The standard deviation of derby dragon is > that of balwin bears > aviston aces.
So the more consistent height is that of aviston aces players.
306 i did the math to the best of my ability
Answer:
m= -9/2
Step-by-step explanation:
hope i helped :) !!!
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Answer:
Your answer is
A & E
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Step-by-step explanation:
Answer:
M₀ (t) = p / e^-t -q = p (e^-t -q) ^ -1
Step-by-step explanation:
Let the random variable Y have a geometric distribution g (y;p) = pq y-¹
The m.g.f of the geometric distribution is derived as below
By definition , M₀ (t) = E (e^ ty) = ∑ (e^ ty )( q ^ y-1)p ( for ∑ , y varies 1 to infinity)
= pe^t ∑(e^tq)^y-1
= pe^t/1- qe^t, where qe^t <1
In order to differentiate the m.g.f we write it as
M₀ (t) = p / e^-t -q = p (e^-t -q) ^ -1
M₀` (t) = pe^-t (e^-t -q) ^ -2 and
M₀^n(t) = 2pe^-2t (e^-t -q) ^ -3 - pe^-t (e^-t -q) ^ -2
Hence
E (y) = p (1-q)-² = 1/p
E (y²) =2 p (1-q)-³ - p (1-q)-²
= 2/p² - 1/p and
σ² = [E (y²) -E (y)]²
= 2/p² - 1/p - (1/p)²
= q/p²
