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Colt1911 [192]
3 years ago
13

I like to help in this problem: in the course of Juan 2/9 of the students are children of 13 years, 1/4 have 14 years and the re

st 15 years What kind of students do they have?
Mathematics
1 answer:
taurus [48]3 years ago
3 0
Rest are 19/36

it is 1 -(1/4) - (2/9)
= 1-9/36 - 8/36
= (36-9-8)/36
= 19/36.
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Please consider the following values for the variables X and Y. Treat each row as a pair of scores for the variables X and Y (wi
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Answer:

The Pearson's coefficient of correlation between the is 0.700.

Step-by-step explanation:

The correlation coefficient is a statistical degree that computes the strength of the linear relationship amid the relative movements of the two variables (i.e. dependent and independent).It ranges from -1 to +1.

The formula to compute correlation between two variables <em>X</em> and <em>Y</em> is:

r(X, Y)=\frac{Cov(X, Y)}{\sqrt{V(X)\cdot V(Y)}}

The formula to compute covariance is:

Cov(X, Y)=n\cdot \sum XY-\sum X \cdot\sum Y

The formula to compute the variances are:

V(X)=n\cdot\sum X^{2}-(\sum X)^{2}\\V(Y)=n\cdot\sum Y^{2}-(\sum Y)^{2}

Consider the table attached below.

Compute the covariance as follows:

Cov(X, Y)=n\cdot \sum XY-\sum X \cdot\sum Y

                 =(5\times 165)-(30\times 25)\\=75

Thus, the covariance is 75.

Compute the variance of X and Y as follows:

V(X)=n\cdot\sum X^{2}-(\sum X)^{2}\\=(5\times 226)-(30)^{2}\\=230\\\\V(Y)=n\cdot\sum Y^{2}-(\sum Y)^{2}\\=(5\times 135)-(25)^{2}\\=50

Compute the correlation coefficient as follows:

r(X, Y)=\frac{Cov(X, Y)}{\sqrt{V(X)\cdot V(Y)}}

            =\frac{75}{\sqrt{230\times 50}}

            =0.69937\\\approx0.70

Thus, the Pearson's coefficient of correlation between the is 0.700.

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