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Goryan [66]
3 years ago
10

a survey asked 72 randomly chosen students if they were going to attend the school play twelve said yes if there are 210 tickets

sold for the play predict the number of students who attend the school
Mathematics
2 answers:
stellarik [79]3 years ago
4 0

Answer:

1260

Step-by-step explanation:

Since we have to predict, we can use ratios to help us solve this.

12 said yes out of 72 randomly chosen, SO,

if 210 would say yes (goes to play), how many is total??

12 is to 72  AS  210 is to HOW MUCH (let it be x)??

<em>We can setup ratio, cross multiply, and solve for x:</em>

<em>\frac{12}{72}=\frac{210}{x}\\12x=210*72\\12x=15,120\\x=\frac{15,120}{12}\\x=1260</em>

<em />

<em>around 1260 students attend the school</em>

Ymorist [56]3 years ago
4 0

Answer:

1260 is the answer

Step-by-step explanation:

12 = 210

----   -----

72       x

= 12x=  15120

divide 15120 by 12 and you get:

x= 1260

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

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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

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