The variance of the first 50 even natural numbers is______.
Answer
591.3k+ views
Hint:
Here we will first write the general form of the even numbers. Then we will write the actual numbers and find the value of the mean of the numbers. Then we will put the value of mean in the formula of the variance to get the variance of the first 50 even natural numbers. Even numbers are the numbers that are exactly divisible by 2.
Formula used:
Variance \[{\sigma ^2} = \dfrac{1}{n}\sum {{x_i}^2} - {\left( {\bar x} \right)^2}\], where \[{x_i}\] is the \[{i^{th}}\] value, \[n\] is the number of terms of the series and \[\bar x\] is the mean value of the series.
Complete Step by Step Solution:
General form of the even numbers is \[2r\] where, \[r\] varies from 1, 2, 3 to infinity.
It is given that the numbers are the first 50 even numbers. Therefore \[n = 50\]
So the given range of the numbers is 2, 4, 6, 8, 10, 12…., 100.
Firstly we will find the value of the mean of the numbers. Therefore, we get
Mean \[\bar x = \dfrac{{\sum {{x_i}} }}{n}\]
\[ \Rightarrow \bar x = \dfrac{{\sum\limits_{r = 1}^{r = 50} {2r} }}{n}\]
Now by putting the values in the formula of the mean, we get
\[ \Rightarrow \bar x = \dfrac{{2 + 4 + 6 + 8 + 10 + 12 + ...... + 100}}{{50}}\]
\[ \Rightarrow \bar x = \dfrac{{50 \times 51}}{{50}}\]
Simplifying the expression, we get
\[ \Rightarrow \bar x = 51\]
Now we will put the value of mean in the variance formula and by using that formula of variance we will calculate the value of the variance. Therefore, we get
\[{\sigma ^2} = \dfrac{{\sum\limits_{r = 1}^{r = 50} {4{r^2}} }}{n} - {\left( {51} \right)^2}\]
Now by solving the above equation of the variance, we get
Variance, \[{\sigma ^2} = 833\]
Hence the variance of the first 50 even natural numbers is 833.
Note:
we should know that the square root of the variance is equal to the standard deviation of the data.
\[Standard\,deviation = \sqrt {Variance} \]
Standard deviation is defined as the measure of the variation of the numbers within the given set of numbers. Standard deviation is used to check the range of deviation of the data from the mean of the data. Wider the range of the data from the mean in a set of numbers, then higher will be the standard deviation and lower the range of the data in a set of numbers then lower will be the standard deviation of the data.
Here we will first write the general form of the even numbers. Then we will write the actual numbers and find the value of the mean of the numbers. Then we will put the value of mean in the formula of the variance to get the variance of the first 50 even natural numbers. Even numbers are the numbers that are exactly divisible by 2.
Formula used:
Variance \[{\sigma ^2} = \dfrac{1}{n}\sum {{x_i}^2} - {\left( {\bar x} \right)^2}\], where \[{x_i}\] is the \[{i^{th}}\] value, \[n\] is the number of terms of the series and \[\bar x\] is the mean value of the series.
Complete Step by Step Solution:
General form of the even numbers is \[2r\] where, \[r\] varies from 1, 2, 3 to infinity.
It is given that the numbers are the first 50 even numbers. Therefore \[n = 50\]
So the given range of the numbers is 2, 4, 6, 8, 10, 12…., 100.
Firstly we will find the value of the mean of the numbers. Therefore, we get
Mean \[\bar x = \dfrac{{\sum {{x_i}} }}{n}\]
\[ \Rightarrow \bar x = \dfrac{{\sum\limits_{r = 1}^{r = 50} {2r} }}{n}\]
Now by putting the values in the formula of the mean, we get
\[ \Rightarrow \bar x = \dfrac{{2 + 4 + 6 + 8 + 10 + 12 + ...... + 100}}{{50}}\]
\[ \Rightarrow \bar x = \dfrac{{50 \times 51}}{{50}}\]
Simplifying the expression, we get
\[ \Rightarrow \bar x = 51\]
Now we will put the value of mean in the variance formula and by using that formula of variance we will calculate the value of the variance. Therefore, we get
\[{\sigma ^2} = \dfrac{{\sum\limits_{r = 1}^{r = 50} {4{r^2}} }}{n} - {\left( {51} \right)^2}\]
Now by solving the above equation of the variance, we get
Variance, \[{\sigma ^2} = 833\]
Hence the variance of the first 50 even natural numbers is 833.
Note:
we should know that the square root of the variance is equal to the standard deviation of the data.
\[Standard\,deviation = \sqrt {Variance} \]
Standard deviation is defined as the measure of the variation of the numbers within the given set of numbers. Standard deviation is used to check the range of deviation of the data from the mean of the data. Wider the range of the data from the mean in a set of numbers, then higher will be the standard deviation and lower the range of the data in a set of numbers then lower will be the standard deviation of the data.
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