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dsp73
2 years ago
7

Find o% of 176: Find 100% of 176: Find 50% of 176: Find 25% of 176: Find 75% of 176:

Mathematics
1 answer:
DaniilM [7]2 years ago
3 0
0
176
88
44
132
Hope this helps
You might be interested in
A random variable X has a gamma density function with parameters α= 8 and β = 2.
DerKrebs [107]

I know you said "without making any assumptions," but this one is pretty important. Assuming you mean \alpha,\beta are shape/rate parameters (as opposed to shape/scale), the PDF of X is

f_X(x) = \dfrac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha - 1} e^{-\beta x} = \dfrac{2^8}{\Gamma(8)} x^7 e^{-2x}

if x>0, and 0 otherwise.

The MGF of X is given by

\displaystyle M_X(t) = \Bbb E\left[e^{tX}\right] = \int_{-\infty}^\infty e^{tx} f_X(x) \, dx = \frac{2^8}{\Gamma(8)} \int_0^\infty x^7 e^{(t-2) x} \, dx

Note that the integral converges only when t.

Define

I_n = \displaystyle \int_0^\infty x^n e^{(t-2)x} \, dx

Integrate by parts, with

u = x^n \implies du = nx^{n-1} \, dx

dv = e^{(t-2)x} \, dx \implies v = \dfrac1{t-2} e^{(t-2)x}

so that

\displaystyle I_n = uv\bigg|_{x=0}^{x\to\infty} - \int_0^\infty v\,du = -\frac n{t-2} \int_0^\infty x^{n-1} e^{(t-2)x} \, dx = -\frac n{t-2} I_{n-1}

Note that

I_0 = \displaystyle \int_0^\infty e^{(t-2)}x \, dx = \frac1{t-2} e^{(t-2)x} \bigg|_{x=0}^{x\to\infty} = -\frac1{t-2}

By substitution, we have

I_n = -\dfrac n{t-2} I_{n-1} = (-1)^2 \dfrac{n(n-1)}{(t-2)^2} I_{n-2} = (-1)^3 \dfrac{n(n-1)(n-2)}{(t-2)^3} I_{n-3}

and so on, down to

I_n = (-1)^n \dfrac{n!}{(t-2)^n} I_0 = (-1)^{n+1} \dfrac{n!}{(t-2)^{n+1}}

The integral of interest then evaluates to

\displaystyle I_7 = \int_0^\infty x^7 e^{(t-2) x} \, dx = (-1)^8 \frac{7!}{(t-2)^8} = \dfrac{\Gamma(8)}{(t-2)^8}

so the MGF is

\displaystyle M_X(t) = \frac{2^8}{\Gamma(8)} I_7 = \dfrac{2^8}{(t-2)^8} = \left(\dfrac2{t-2}\right)^8 = \boxed{\dfrac1{\left(1-\frac t2\right)^8}}

The first moment/expectation is given by the first derivative of M_X(t) at t=0.

\Bbb E[X] = M_x'(0) = \dfrac{8\times\frac12}{\left(1-\frac t2\right)^9}\bigg|_{t=0} = \boxed{4}

Variance is defined by

\Bbb V[X] = \Bbb E\left[(X - \Bbb E[X])^2\right] = \Bbb E[X^2] - \Bbb E[X]^2

The second moment is given by the second derivative of the MGF at t=0.

\Bbb E[X^2] = M_x''(0) = \dfrac{8\times9\times\frac1{2^2}}{\left(1-\frac t2\right)^{10}} = 18

Then the variance is

\Bbb V[X] = 18 - 4^2 = \boxed{2}

Note that the power series expansion of the MGF is rather easy to find. Its Maclaurin series is

M_X(t) = \displaystyle \sum_{k=0}^\infty \dfrac{M_X^{(k)}(0)}{k!} t^k

where M_X^{(k)}(0) is the k-derivative of the MGF evaluated at t=0. This is also the k-th moment of X.

Recall that for |t|,

\displaystyle \frac1{1-t} = \sum_{k=0}^\infty t^k

By differentiating both sides 7 times, we get

\displaystyle \frac{7!}{(1-t)^8} = \sum_{k=0}^\infty (k+1)(k+2)\cdots(k+7) t^k \implies \displaystyle \frac1{\left(1-\frac t2\right)^8} = \sum_{k=0}^\infty \frac{(k+7)!}{k!\,7!\,2^k} t^k

Then the k-th moment of X is

M_X^{(k)}(0) = \dfrac{(k+7)!}{7!\,2^k}

and we obtain the same results as before,

\Bbb E[X] = \dfrac{(k+7)!}{7!\,2^k}\bigg|_{k=1} = 4

\Bbb E[X^2] = \dfrac{(k+7)!}{7!\,2^k}\bigg|_{k=2} = 18

and the same variance follows.

6 0
2 years ago
Please don't answer if you don't know, please!!
Snezhnost [94]

Answer:

<u>S</u><u>e</u><u>e</u><u> </u><u>T</u><u>h</u><u>e</u><u> </u><u>A</u><u>t</u><u>t</u><u>a</u><u>c</u><u>h</u><u>m</u><u>e</u><u>n</u><u>t</u>

<u></u>

7 0
2 years ago
Read 2 more answers
In the number 1934 is a value of 9 in 100 place 10 times the value of the 3 in the tens place
dybincka [34]

no because mathematically 30X10=300 not 900 its would have to be 30X30 to get the product of 900 so therefore no your statement is incorrect

8 0
3 years ago
List multiples to find LMC of 27 and 45
PtichkaEL [24]

Answer:

135

Step-by-step explanation:

I think you mean the LCM instead of LMC. LCM stands for Least Common Multiple. Anyway, some of the multiples of 27 are 27, 54, 81, 108, and 135. Some multiples for 45 are 45, 90, 135. We can see here that 27 and 45 share the mutiple of 135 which is the first number they have in common.

6 0
3 years ago
P divided by 28 = -26
juin [17]
The answer is p= -728
5 0
3 years ago
Read 2 more answers
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