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barxatty [35]
2 years ago
15

A set of n = 25 pairs of scores (x and y values) has a pearson correlation of r = 0. 80. how much of the variance for the y scor

es is predicted by its relationship with x?
Mathematics
1 answer:
Leno4ka [110]2 years ago
7 0

The amount of variance for y that is predicted by its relationship with x is 64%.

<h3>How much variance is predicted?</h3>

Variance measures the rate of dispersion of a data point around the dataset. It can be calculated by finding the square of the standard deviation of a dataset. Variance measures the variation of a data set.

Correlation is a statistical measure used to measure the linear relationship that exists between two variables. The greater the correlation coefficient is closer to one, the greater the linear relationship that exists between the two variables. A positive correlation occurs when the two variables move in the same direction.

Variance = (correlation coefficient²) x 100

(0.80²) x 100

0.64 x 100 = 64%

To learn more about correlation, please check: brainly.com/question/27246345

#SPJ1

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Rectangle ABCD has vertices A(3, 5), B(5,5), C(5, 1) and D(3, 1). Drag and drop the coordinates of each vertex when rectangle AB
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Answer:

Given the coordinates of rectangle ABCD are;

A = (3, 5)

B = (5,5)

C = (5, 1) and

D = (3, 1)

The rule of rotation 90 degree counter-clockwise around the origin is given by;

(x, y) \rightarrow (-y, x)

Now, the coordinates of each vertex when rectangle ABCD is rotated 90 degree counter-clockwise around the origin are;

A(3, 5) \rightarrow A'(-5, 3)

B(5, 5) \rightarrow B'(-5, 5)

C(5, 1) \rightarrow C'(-1, 5)

D(3, 1) \rightarrow D'(-1, 3)

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A state vector X for a four-state Markov chain is such that the system is four times as likely to be in state 2 as in 4, is not
GREYUIT [131]
All the components in the state vector need to sum to 1. You're given that component corresponding to state 1 is 0.2, and that the component for state 3 is 0.

That leaves states 2 and 4, for which you're told that the component for state 2 is four times as large. If x_i is the component for state i, then you have

x_1+x_2+x_3+x_4=1\iff 0.2+4x_4+0+x_4=1\implies5x_4=0.8\implies x_4=0.16

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One 16 inch pizza serves 4 students how many pizzas are needed for 240 students
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Let X be a random variable with probability density function fX(x) = ( c(1 − x 2 ) if − 1 &lt; x &lt; 1 0 otherwise. a) What is
xeze [42]

Answer:

a)   c=3/4

b)  for -1<x<1

 F(x) = (3/4) ( x + (2-x^3)/3 )

c)

  E(x)=0

  Var(x) = 1/5

Step-by-step explanation:

Hi!

a)

In order to f(x) be a probability density function it must be normalized, which means that its integral must be equal to 1:

1 = \int\limits^{\infty}_{-\infty} {f(x)} \, dx

Since f(x) is zero for x>1 and x<-1

1 = \int\limits^1_{-1} {f(x)} \, dx = \int \limits^1_{-1} {c(1-x^2)}\, dx \\\\\\1 = c ((1-\frac{1}{3}) - (-1 - \frac{-1}{3})) = c(\frac{2}{3} - \frac{-2}{3} ) = c\frac{4}{3}

Therefore:

c=3/4

b)

The cumulative distribution fucntion F(x) can be obtained integrating f(x) from -∞ to x:

Since f(x) = 0 for x<-1      

        F(x) = 0 for x<-1

for -1<x<1:

F(x) = \int \limits^x_{-1} f(x')\, dx' = c(x'-\frac{x'^3}{3})^x_{-1} = c (x-\frac{x^3}{3} + \frac{2}{3})\\\\F(x) = \frac{3}{4}(x + \frac{2-x^3}{3})

for x>1

  F(x)=1

c)

The mean E(x) can be found integrating  xf(x)

E(x) = \int \limits_{-1}^1 {x f(x)}\, dx

We can easily infer that the mean must be zero, since f(x) is an even function and thus xf(x) is an odd function integrated in a simetric interval:

E(x) = 0

The variance of x, Var(x), can be evaluated integrating  (x-E(x))^2f(x), since E(x)=0:

Var(x) = \int \limits^{1}_{-1} {x^2f(x)}\, dx\\Var(x) = \int \limits^{1}_{-1} {c (x^2 - x^4 )dx}\\Var(x) = c \frac{4}{15}

Since c=3/4

Var(x) = 1/5

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