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4 years ago

What is 32+4 X (16 X 1/2) -2 [PLEASE SHOW YOUR WORK}

Mathematics

1 answer:

aivan3 [116]4 years ago

8 0

your answer would be 62.

-- first, in the parentheses, 16 times half is just 16 divided by 2. so your answer would be 8.

-- second, multiple 4 times the 8. you would get 32.

-- lastly, do 32 plus 32 and you would get 64. after that, its simple because 64 minus "-2" would be 62.

thats how I got my answer. i hope this helps! please mark brainliest x

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Let Y1, Y2, . . . , Yn be independent, uniformly distributed random variables over the interval [0, θ]. Let Y(n) = max{Y1, Y2, .

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Answer:

a) $F(y) = 0, y$

$F(y) = \frac{y}{\theta} , 0 \leq y \leq \theta$

$F(y)= 1, y>1$

b) $f_{Y_{(n)}} = \frac{d}{dy} (\frac{y}{\theta})^n = n \frac{y^{n-1}}{\theta^n}, 0 \leq y \leq \theta$

$f_{Y_{(n)}} =0$ for other case

c) $E(Y_{(n)}) = \frac{n}{\theta^n} \frac{\theta^{n+1}}{n+1}= \theta [\frac{n}{n+1}]$

$Var(Y_{(n)}) =\theta^2 [\frac{n}{(n+1)(n+2)}]$

Step-by-step explanation:

We have a sample of $Y_1, Y_2,...,Y_n$ iid uniform on the interval $[0,\theta]$ and we want to find the cumulative distribution function.

Part a

For this case we can define the CDF for $Y_i$ , $i =1,2.,,,n$ like this:

$F(y) = 0, y$

$F(y) = \frac{y}{\theta} , 0 \leq y \leq \theta$

$F(y)= 1, y>1$

Part b

For this case we know that:

$F_{Y_{(n)}} (y) = P(Y_{(n)} \leq y) = P(Y_1 \leq y,....,Y_n \leq y)$

And since are independent we have:

$F_{Y_{(n)}} (y) = P(Y_1 \leq y) * ....P(Y_n \leq y) = (\frac{y}{\theta})^n$

And then we can find the density function calculating the derivate from the last expression and we got:

$f_{Y_{(n)}} = \frac{d}{dy} (\frac{y}{\theta})^n = n \frac{y^{n-1}}{\theta^n}, 0 \leq y \leq \theta$

$f_{Y_{(n)}} =0$ for other case

Part c

For this case we can find the mean with the following integral:

$E(Y_{(n)}) = \frac{n}{\theta^n} \int_{0}^{\theta} y y^{n-1} dy$

$E(Y_{(n)}) = \frac{n}{\theta^n} \int_{0}^{\theta} y^n dy$

$E(Y_{(n)}) = \frac{n}{\theta^n} \frac{y^{n+1}}{n+1} \Big|_0^{\theta}$

And after evaluate we got:

$E(Y_{(n)}) = \frac{n}{\theta^n} \frac{\theta^{n+1}}{n+1}= \theta [\frac{n}{n+1}]$

For the variance first we need to find the second moment like this:

$E(Y^2_{(n)}) = \frac{n}{\theta^n} \int_{0}^{\theta} y^2 y^{n-1} dy$

$E(Y^2_{(n)}) = \frac{n}{\theta^n} \int_{0}^{\theta} y^{n+1} dy$

$E(Y^2_{(n)}) = \frac{n}{\theta^n} \frac{y^{n+2}}{n+2} \Big|_0^{\theta}$

And after evaluate we got:

$E(Y^2_{(n)}) = \frac{n}{\theta^n} \frac{\theta^{n+2}}{n+2}= \theta^2 [\frac{n}{n+2}]$

And the variance is given by:

$Var(Y_{(n)}) = E(Y^2_{(n)}) - [E(Y_{(n)})]^2$

And if we replace we got:

$Var(Y_{(n)}) =\theta^2 [\frac{n}{n+2}] -\theta^2 [\frac{n}{n+1}]^2$

$Var(Y_{(n)}) =\theta^2 [\frac{n}{n+2} -(\frac{n}{n+1})^2]$

And after do some algebra we got:

$Var(Y_{(n)}) =\theta^2 [\frac{n}{(n+1)(n+2)}]$

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