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attashe74 [19]
3 years ago
9

F(x) = (128/127)(1/2)x, x = 1,2,3,...7. determine the requested values: round your answers to three decimal places (e.g. 98.765)

. (a)p(x ≤ 1) (b)p(x > 1) (c) mean (d) variance
Mathematics
1 answer:
Marrrta [24]3 years ago
6 0
A.
\mathbb P(X\le 1)=\mathbb P(X=1)=\dfrac{128}{127}\left(\dfrac12\right)^1=\dfrac{64}{127}

b.
\mathbb P(X>1)=1-\mathbb P(X\le1)=1-\dfrac{64}{127}=\dfrac{63}{127}

c.
\mathbb E(X)=\displaystyle\sum_{x=1}^7 x\,f_X(x)=\frac{64}{127}\sum_{x=1}^7 x\left(\frac12\right)^{x-1}

Suppose f(y)=\displaystyle\sum_{x=0}^7 y^x. Then f'(y)=\displaystyle\sum_{x=1}^7 xy^{x-1}. So if we can find a closed form for f(y), in terms of y, we can find \mathbb E(X) by evaluating the derivative of f(y) at y=\dfrac12.

f(y)=\displaystyle\sum_{x=0}^7 y^x=y^0+y^1+y^2+\cdots+y^6+y^7
y\,f(y)=y^1+y^2+y^3+\cdots+y^7+y^8
f(y)-y\,f(y)=y^0-y^8
(1-y)f(y)=1-y^8
f(y)=\dfrac{1-y^8}{1-y}
\implies f'(y)=\dfrac{7y^8-8y^7+1}{(1-y)^2}
\implies\mathbb E(X)=\dfrac{64}{127}f'\left(\dfrac12\right)=\dfrac{64}{127}\times\dfrac{247}{64}=\dfrac{247}{127}

d.
\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2

We find \mathbb E(X^2) in a similar manner as in (c).

\mathbb E(X^2)=\displaystyle\sum_{x=1}^7 x^2\,f_X(x)=\frac{32}{127}\sum_{x=1}^7x^2\left(\frac12\right)^{x-2}

Now,

f(y)=\displaystyle\sum_{x=0}^7y^x
\implies f'(y)=\displaystyle\sum_{x=1}^7xy^{x-1}
\implies f''(y)=\displaystyle\sum_{x=2}^7x(x-1)y^{x-2}

We know that

f''(y)=-\dfrac{42y^8-96y^7+56y^6-2}{(1-y)^3}
\implies f''\left(\dfrac12\right)=\dfrac{219}{16}

We also have

f''(y)=\displaystyle\sum_{x=2}^7x(x-1)y^{x-2}
f''(y)=\displaystyle\sum_{x=2}^7x^2y^{x-2}-\sum_{x=2}^7xy^{x-2}
f''(y)=\displaystyle\frac1{y^2}\left(\sum_{x=2}^7x^2y^x-\sum_{x=2}^7xy^x\right)
f''(y)=\displaystyle\frac1{y^2}\left(\bigg(\sum_{x=1}^7x^2y^x-y\bigg)-\bigg(\sum_{x=1}^7xy^x-y\bigg)\right)
f''(y)=\displaystyle\frac1{y^2}\left(\sum_{x=1}^7x^2y^x-\sum_{x=1}^7xy^x\right)

so that when y=\dfrac12, we get

\dfrac{219}{16}=4\left(\dfrac{127}{128}\mathbb E(X^2)-\dfrac{127}{128}\mathbb E(X)\right)\implies\mathbb E(X^2)=\dfrac{685}{127}

Then

\mathbb V(X)=\dfrac{685}{127}-\left(\dfrac{247}{127}\right)^2=\dfrac{25,986}{16,129}
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