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प्रश्न
Given the following information, `sum"x"_"i"^2` = 90, `sum"x"_"i""y"_"i"` = 60, r = 0.8, `sigma_"y"` = 2.5, where xi and yi are the deviations from their respective means, find the number of items.
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उत्तर
Here, r = 0.8, `sum"x"_"i"^2` = 90, `sum"x"_"i""y"_"i"` = 60, `sigma_"y"` = 2.5,
Here, xi and yi are the deviations from their respective means.
∴ If Xi, Yi are elements of x and y series respectively, then `"X"_"i" - bar"x" = "x"_"i"` and `"Y"_"i" - bar"y" = "y"_"i"`
∴ `sum"x"_"i""y"_"i" = sum("X"_"i" - bar"x")("Y"_"i" - bar"y")` = 60, `sum"x"_"i"^2 = sum("X"_"i" - bar"x")^2 = 90`
Now, `sigma_"x"^2 = (sum("X"_"i" - bar"x")^2)/"n"`
∴ `sigma_"x"^2 = 90/"n"`
∴ `sigma_"x" = sqrt(90/"n")`
Also, Cov (X, Y) = `1/"n" sum("X"_"i" - bar"x")("Y"_"i" - bar"y")`
∴ Cov (X, Y) = `60/"n"`
r = `("Cov(X, Y)")/(sigma_"x" sigma_"y")`
∴ 0.8 = `(60/"n")/(sqrt(90/"n") xx 2.5)`
∴ `0.8 xx 2.5 xx sqrt(90/"n") = 60/"n"`
∴ `2 xx sqrt(90)/sqrt("n") = 60/"n"`
∴ `"n"/sqrt("n") = 60/(2 xx sqrt(90))`
∴ `(sqrt("n") xx sqrt("n"))/sqrt("n") = 30/sqrt(90) = (sqrt(30) xx sqrt(30))/(sqrt(3) sqrt(30)`
∴ `sqrt("n") = sqrt(10)`
∴ n = 10
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