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Elena-2011 [213]
3 years ago
15

CHAPTER 1,2/CLO 1,2

Mathematics
1 answer:
hram777 [196]3 years ago
8 0

a. In order for f to be a proper density function, its integral over its domain must evaluate to 1:

\displaystyle\int_{-\infty}^\infty f(x)\,\mathrm dx=k\int_0^1(1-x)\,\mathrm dx=\frac k2=1\implies k=2

f also must be non-negative over its support, which is the case here.

I assume the instruction regarding the distribution function doesn't apply to parts (b) and (c).

(b) Integrate the density over the interval [0.1, 0.2]:

P(0.1

(c) Integrate the density over the interval [0.5, 1]:

P(X>0.5)=\displaystyle\int_{0.5}^12(1-x)\,\mathrm dx=0.25

The distribution function is obtained by integrating the density:

F(x)=\displaystyle\int_{-\infty}^xf(t)\,\mathrm dt=\begin{cases}0&\text{for }x

(d) Using the distribution function, we have

P(X

(e) Using F again, we get

P(0.4

(f) The mean is

E[X]=\displaystyle\int_{-\infty}^\infty x\,f(x)\,\mathrm dx=\int_0^12x(1-x)\,\mathrm dx=\frac13

The variance is

V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2

where

E[X^2]=\displaystyle\int_{-\infty}^\infty x^2\,f(x)\,\mathrm dx=\int_0^12x^2(1-x)\,\mathrm dx=\frac16

so that the variance is

V[X]=\dfrac16-\left(\dfrac13\right)^2=\dfrac1{18}

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Answer:

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