
How do you find the sample variance in $ 89,57,104,73,26,121,81 $ ?
Answer
530.4k+ views
Hint: In the given question, we have to find the sample variance for the given set of seven numbers provided to us. So, we can find the variance of the seven numbers by using the formula of variance. So, Variance $ = \,\dfrac{1}{{n - 1}}\left( {\sum {{x^2} - \dfrac{{{{\left( {\sum x } \right)}^2}}}{n}} } \right) $ . So, we will need to compute the squares of all the numbers or observations for the calculation of variance. Then, we substitute the values of the sum of squares of the observations and sum of the observations to find the variance of the numbers.
Complete step by step solution:
In the given problem, we need to find the variance of the set of seven numbers provided to us.
So, the seven numbers given to us are: $ 89,57,104,73,26,121,81 $ .
Now, the sum of these numbers $ = 89 + 57 + 104 + 73 + 26 + 121 + 81 = 551 $
Hence, the mean of the seven numbers $ = \dfrac{{551}}{7} = 78.7143 $
Also, we need to compute the squares of all the seven numbers so as to find the mean of the squares and substitute the value in the formula for calculation of variance.
Square of $ 89 $ is $ {\left( {89} \right)^2} = 7921 $
Square of $ 57 $ is $ {\left( {57} \right)^2} = 3249 $
Square of $ 104 $ is $ {\left( {104} \right)^2} = 10816 $
Square of $ 73 $ is $ {\left( {73} \right)^2} = 5329 $
Square of $ 26 $ is $ {\left( {26} \right)^2} = 676 $
Square of $ 121 $ is $ {\left( {121} \right)^2} = 14641 $
Square of $ 81 $ is $ {\left( {81} \right)^2} = 6561 $
So, the sum of squares $ = 7921 + 3249 + 10861 + 5329 + 676 + 14641 + 6561 $
$ = 49193 $
So, substituting all the values in the formula of variance, we get,
Variance $ = \,\dfrac{1}{{n - 1}}\left( {\sum {{x^2} - \dfrac{{{{\left( {\sum x } \right)}^2}}}{n}} } \right) $
\[ = \,\dfrac{1}{{7 - 1}}\left( {49193 - \dfrac{{{{\left( {551} \right)}^2}}}{7}} \right)\]
Simplifying the calculations further, we get,
\[ = \,\dfrac{1}{6}\left( {49193 - \dfrac{{303601}}{7}} \right)\]
\[ = \,970.238\]
So, the sample variance in the seven numbers given to us $ 89,57,104,73,26,121,81 $ is \[970.238\].
So, the correct answer is “\[970.238\].”.
Note: Be careful while calculating all the squares and doing the addition of the same. Learn the formula of the variance carefully. While doing the basic calculations carefully. Don’t use calculators because in the exam you have to do the calculations yourself.
Complete step by step solution:
In the given problem, we need to find the variance of the set of seven numbers provided to us.
So, the seven numbers given to us are: $ 89,57,104,73,26,121,81 $ .
Now, the sum of these numbers $ = 89 + 57 + 104 + 73 + 26 + 121 + 81 = 551 $
Hence, the mean of the seven numbers $ = \dfrac{{551}}{7} = 78.7143 $
Also, we need to compute the squares of all the seven numbers so as to find the mean of the squares and substitute the value in the formula for calculation of variance.
Square of $ 89 $ is $ {\left( {89} \right)^2} = 7921 $
Square of $ 57 $ is $ {\left( {57} \right)^2} = 3249 $
Square of $ 104 $ is $ {\left( {104} \right)^2} = 10816 $
Square of $ 73 $ is $ {\left( {73} \right)^2} = 5329 $
Square of $ 26 $ is $ {\left( {26} \right)^2} = 676 $
Square of $ 121 $ is $ {\left( {121} \right)^2} = 14641 $
Square of $ 81 $ is $ {\left( {81} \right)^2} = 6561 $
So, the sum of squares $ = 7921 + 3249 + 10861 + 5329 + 676 + 14641 + 6561 $
$ = 49193 $
So, substituting all the values in the formula of variance, we get,
Variance $ = \,\dfrac{1}{{n - 1}}\left( {\sum {{x^2} - \dfrac{{{{\left( {\sum x } \right)}^2}}}{n}} } \right) $
\[ = \,\dfrac{1}{{7 - 1}}\left( {49193 - \dfrac{{{{\left( {551} \right)}^2}}}{7}} \right)\]
Simplifying the calculations further, we get,
\[ = \,\dfrac{1}{6}\left( {49193 - \dfrac{{303601}}{7}} \right)\]
\[ = \,970.238\]
So, the sample variance in the seven numbers given to us $ 89,57,104,73,26,121,81 $ is \[970.238\].
So, the correct answer is “\[970.238\].”.
Note: Be careful while calculating all the squares and doing the addition of the same. Learn the formula of the variance carefully. While doing the basic calculations carefully. Don’t use calculators because in the exam you have to do the calculations yourself.
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