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svetoff [14.1K]
3 years ago
8

What is the relationship between the two variables represented in the table?

Mathematics
2 answers:
dlinn [17]3 years ago
6 0

Answer:

d positive linear association with one deviation

Step-by-step explanation:

Given is a table showing different heights of people and their show size

Height of People (cm) Shoe Size

170                               8.5

172                                9

174                                9.5

176                            10

178                                  11

We find that for every 2 cm increase in height there is 0.5increase in show size

from 170 to 176

Only for 176-178 we have 1 increase for 2 cm increase in height

Thus we find that both increase together hence positive relationship

But there is one deviation from 176 to 178

Hence answer is option

d positive linear association with one deviation

Zina [86]3 years ago
5 0
I think that it's - positive linear association with one deviation
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Suppose X, Y, and Z are random variables with the joint density function f(x, y, z) = Ce−(0.5x + 0.2y + 0.1z) if x ≥ 0, y ≥ 0, z
kompoz [17]

a.

f_{X,Y,Z}(x,y,z)=\begin{cases}Ce^{-(0.5x+0.2y+0.1z)}&\text{for }x\ge0,y\ge0,z\ge0\\0&\text{otherwise}\end{cases}

is a proper joint density function if, over its support, f is non-negative and the integral of f is 1. The first condition is easily met as long as C\ge0. To meet the second condition, we require

\displaystyle\int_0^\infty\int_0^\infty\int_0^\infty f_{X,Y,Z}(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz=100C=1\implies \boxed{C=0.01}

b. Find the marginal joint density of X and Y by integrating the joint density with respect to z:

f_{X,Y}(x,y)=\displaystyle\int_0^\infty f_{X,Y,Z}(x,y,z)\,\mathrm dz=0.01e^{-(0.5x+0.2y)}\int_0^\infty e^{-0.1z}\,\mathrm dz

\implies f_{X,Y}(x,y)=\begin{cases}0.1e^{-(0.5x+0.2y)}&\text{for }x\ge0,y\ge0\\0&\text{otherwise}\end{cases}

Then

\displaystyle P(X\le1.375,Y\le1.5)=\int_0^{1.5}\int_0^{1.375}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy

\approx\boxed{0.12886}

c. This probability can be found by simply integrating the joint density:

\displaystyle P(X\le1.375,Y\le1.5,Z\le1)=\int_0^1\int_0^{1.5}\int_0^{1.375}f_{X,Y,Z}(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz

\approx\boxed{0.012262}

7 0
3 years ago
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