Question

If \[r = 0.5,\,\,\sum {xy = 120} ,\,\,\sum {{x^2}} = 90,\,\,{\sigma _y} = 8\], then \[n = \]
A. \[100\]
B. \[10\]
C. \[15\]
D. \[50\]

Answer 1

Hint: In this question, we need to find the value of \[n\]. For that, we apply the formula of standard deviation \[{\sigma _y} = \sqrt {\dfrac{{\sum {{y^2}} }}{n}} \] then substitute all the given values in it and simplifies it to get the required value which is n.

Formula used:
We have been using the following formulas of standard deviation:
1. \[{\sigma _y} = \sqrt {\dfrac{{\sum {{y^2}} }}{n}} \] when \[y = Y - \overline Y \]
Where y is the difference between each value in the data set and the mean value of the data set and n is the number of items

Complete step-by-step solution:
We are given that
\[
  r = 0.5 \\
  \sum {xy} = 120 \\
  \sum {{x^2}} = 90 \\
  {\sigma _y} = 8
 \]
Now we know the formula of standard deviation
\[{\sigma _y} = \sqrt {\dfrac{{\sum {{y^2}} }}{n}} \]when \[y = Y - \overline Y \]
Now we substitute the given values in the formula, and we get
\[8 = \sqrt {\dfrac{{\sum {{y^2}} }}{n}} \]
Now by squaring on both sides and simplifying it, we get
\[
  {\left( 8 \right)^2} = {\left( {\sqrt {\dfrac{{\sum {{y^2}} }}{n}} } \right)^2} \\
  64 = \dfrac{{\sum {{y^2}} }}{n} \\
  64n = \sum {{y^2}}
 \]
Now by rearranging the above equation, we get
\[\sum {{y^2}} = 64n\]
Now, we know that \[r = \dfrac{{\sum {xy} }}{{\sqrt {\sum {{x^2} \times \sum {{y^2}} } } }}\] where \[r\] is the correlation coefficient, \[xy\] is the sum of the product of \[x\] and \[y\] values, \[{x^2}\] is the sum of squares of \[x - \] values and \[{y^2}\] is sum of squares of \[y - \]values.

By substituting given values in above formula, we will get
\[0.5 = \dfrac{{120}}{{\sqrt {90 \times 64n} }}\]
By squaring on both sides, we will get
\[
  {\left( {0.5} \right)^2} = \dfrac{{{{\left( {120} \right)}^2}}}{{{{\left( {\sqrt {90 \times 64n} } \right)}^2}}} \\
  0.25 = \dfrac{{14,400}}{{90 \times 64n}}
 \]
Now by cross multiply both the terms, we will get
\[
  0.25 \times 90 \times 64n = 14,400 \\
  22.5 \times 64n = 14,400 \\
  64n = \dfrac{{14,400}}{{22.5}} \\
  64n = 640
 \]
On further simplification, we will get
\[
  n = \dfrac{{640}}{{64}} \\
  n = 10
 \]
Therefore, if \[r = 0.5,\,\,\sum {xy = 120} ,\,\,\sum {{x^2}} = 90,\,\,{\sigma _y} = 8\], then \[n = 10\]
Hence, option (B) is correct option

Note: To solve these types of questions, students must know the concept of standard deviation. The standard deviation formula used in the above question is \[{\sigma _y} = \sqrt {\dfrac{{\sum {{y^2}} }}{n}} \] to find the value of n. Students must be careful while calculating the square of the summation, do the square of each number and not the square of the summation of x as it gives a different answer.