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2 years ago

Here is a uniform probability distribution.

Mathematics

1 answer:

Darya [45]2 years ago

4 0

a) $P(4\le X\le6)=\dfrac25$ because it is equal to the area of the shaded region between X=4 and X=6, and the probability that X falls within some interval is given by the area under the PDF.

b) $a=3$ because the shaded region is a rectangle of height 1/5 (by virtue of X following a uniform distribution over the interval [2, 7], which has length 5).

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A brownie recipe calls for 1/2 cup of flour, 1/8 cup of cocoa powder, and 3/8 cup of sugar. How many cups of dry ingredients are

natka813 [3]

1/2 x 2= 1

1/8 x 2= 1/4

3/8 x 2= 3/4

3/4 x 1/4 = 3/16

3/16 x 1 = 3/16

YOUR ANSWER IS 3/16 dry ingredients

7 0

2 years ago

Find the slope of (4, -1), (0, -3). Reduce all fractional answers to lowest terms.

Harman [31]

Slope= (y2-y1)/(x2-x1)= (-3-(-1))/(0-4)= -2/-4 =2/4 =1/2
Answer: slope =1/2.

3 0

2 years ago

Read 2 more answers

A flare is designed to follow the path modeled by the function h(t) = –16t2 + 100t, where t is the time elapsed, in seconds, and

gogolik [260]

Answer: The answer is D. P(t) = - 16t^2 + 100t / 3.2808

Step-by-step explanation:

4 0

2 years ago

14. Sharon earns $25 per item she sells plus a base salary of $100 per week. Write and solve an inequality to find how many item

kozerog [31]

Amount earned by Sharon per item sold = $25
Base salary of Sharon per week = $100
Amount that needs to be earned by Sharon per week = $700
Let us assume the number of items sold by Sharon per week = x
Then
100 + 25x = 700
25x = 700 - 100
25x = 600
x = 600/25
= 24
So Sharon needs to sell 24 items to earn a total of $700 per week. I hope you have understood the method of solving such problems.

8 0

3 years ago

The joint probability density function of X and Y is given by fX,Y (x, y) = ( 6 7 x 2 + xy 2 if 0 < x < 1, 0 < y < 2

fredd [130]

I'm going to assume the joint density function is

$f_{X,Y}(x,y)=\begin{cases}\frac67(x^2+\frac{xy}2\right)&\text{for }0$

a. In order for $f_{X,Y}$ to be a proper probability density function, the integral over its support must be 1.

$\displaystyle\int_0^2\int_0^1\frac67\left(x^2+\frac{xy}2\right)\,\mathrm dx\,\mathrm dy=\frac67\int_0^2\left(\frac13+\frac y4\right)\,\mathrm dy=1$

b. You get the marginal density $f_X$ by integrating the joint density over all possible values of $Y$ :

$f_X(x)=\displaystyle\int_0^2f_{X,Y}(x,y)\,\mathrm dy=\boxed{\begin{cases}\frac67(2x^2+x)&\text{for }0$

c. We have

$P(X>Y)=\displaystyle\int_0^1\int_0^xf_{X,Y}(x,y)\,\mathrm dy\,\mathrm dx=\int_0^1\frac{15}{14}x^3\,\mathrm dx=\boxed{\frac{15}{56}}$

d. We have

$\displaystyle P\left(X$

and by definition of conditional probability,

$P\left(Y>\dfrac12\mid X\frac12\text{ and }X$

$\displaystyle=\dfrac{28}5\int_{1/2}^2\int_0^{1/2}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\boxed{\frac{69}{80}}$

e. We can find the expectation of $X$ using the marginal distribution found earlier.

$E[X]=\displaystyle\int_0^1xf_X(x)\,\mathrm dx=\frac67\int_0^1(2x^2+x)\,\mathrm dx=\boxed{\frac57}$

f. This part is cut off, but if you're supposed to find the expectation of $Y$ , there are several ways to do so.

Compute the marginal density of $Y$ , then directly compute the expected value.

$f_Y(y)=\displaystyle\int_0^1f_{X,Y}(x,y)\,\mathrm dx=\begin{cases}\frac1{14}(4+3y)&\text{for }0$

$\implies E[Y]=\displaystyle\int_0^2yf_Y(y)\,\mathrm dy=\frac87$

Compute the conditional density of $Y$ given $X=x$ , then use the law of total expectation.

$f_{Y\mid X}(y\mid x)=\dfrac{f_{X,Y}(x,y)}{f_X(x)}=\begin{cases}\frac{2x+y}{4x+2}&\text{for }0$

The law of total expectation says

$E[Y]=E[E[Y\mid X]]$

We have

$E[Y\mid X=x]=\displaystyle\int_0^2yf_{Y\mid X}(y\mid x)\,\mathrm dy=\frac{6x+4}{6x+3}=1+\frac1{6x+3}$

$\implies E[Y\mid X]=1+\dfrac1{6X+3}$

This random variable is undefined only when $X=-\frac12$ which is outside the support of $f_X$ , so we have

$E[Y]=E\left[1+\dfrac1{6X+3}\right]=\displaystyle\int_0^1\left(1+\frac1{6x+3}\right)f_X(x)\,\mathrm dx=\frac87$

5 0

3 years ago