Question

Find the variance of the first 10 multiples of 3.
A.\[72.65\]
B.\[74.05\]
C.\[74.25\]
D.\[73.85\]

Answer 1

Hint: First, we will use the formula of sum, \[S = \dfrac{n}{2}\left( {a + l} \right)\], where \[l\] is the last term and \[a\] is the first term and formula to compute the variance is \[\dfrac{{\sum {\left( {{x_i}^2} \right)} }}{n} - {\mu ^2}\], where \[\mu \] is the mean. Then substituting the values in these formulas to find the required value.

Complete step-by-step answer:
We are given that the first 10 multiples of 3.
Using the given conditions, we have that 3, 6, 9, …, 30.
The given terms form an A.P where first term=3 and common difference=3.
We know that the use the formula of sum, \[S = \dfrac{n}{2}\left( {a + l} \right)\], where \[l\] is the last term and \[a\] is the first term.
Finding the value of \[n\], \[a\], and \[l\] from the given information, we get
\[n = 10\]
\[a = 3\]
\[l = 30\]
Substituting the values of \[a\], \[n\] and \[l\] to find the sum from the above formula, we get
\[
   \Rightarrow {\text{Sum}} = \dfrac{{10}}{2}\left( {3 + 30} \right) \\
   \Rightarrow {\text{Sum}} = \dfrac{{10}}{2}\left( {33} \right) \\
   \Rightarrow {\text{Sum}} = 5\left( {33} \right) \\
   \Rightarrow {\text{Sum}} = 165 \\
 \]
Dividing the above sum by 10 to find the value of mean, we get
\[ \Rightarrow {\text{Mean}} = \dfrac{{165}}{{10}}\]
We know the formula to compute the variance is \[\dfrac{{\sum {\left( {{x_i}^2} \right)} }}{n} - {\mu ^2}\], where \[\mu \] is the mean.
Substituting the above values in the formula of variance, we get
\[ \Rightarrow {\text{Variance}} = \dfrac{{{3^2} + {6^2} + ... + {{30}^2}}}{{10}} - {16.5^2}\]
Taking 9 common from the numerator of the above equation, we get
\[
   \Rightarrow {\text{Variance}} = \dfrac{{{3^2} \times \left( {{1^2} + {2^2} + ... + {{10}^2}} \right)}}{{10}} - {16.5^2} \\
   \Rightarrow {\text{Variance}} = \dfrac{{9 \times \left( {{1^2} + {2^2} + ... + {{10}^2}} \right)}}{{10}} - 272.25 \\
 \]

Using the formula of sum, \[{1^2} + {2^2} + ... + {n^2} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\] in the above equation, we get
\[
   \Rightarrow {\text{Variance}} = \dfrac{{{3^2} \times 10\left( {10 + 1} \right)\left( {2\left( {10} \right) + 1} \right)}}{{6 \times 10}} - 272.25 \\
   \Rightarrow {\text{Variance}} = \dfrac{{9 \times 10\left( {10 + 1} \right)\left( {20 + 1} \right)}}{{6 \times 10}} - 272.25 \\
   \Rightarrow {\text{Variance}} = \dfrac{{9 \times 10\left( {11} \right)\left( {30} \right)}}{{6 \times 10}} - 272.25 \\
   \Rightarrow {\text{Variance}} = 346.5 - 272.25 \\
   \Rightarrow {\text{Variance}} = 74.25 \\
 \]
Hence, the variance of the given data is \[74.5\].
Therefore, option C is correct.

Note: In solving these types of questions, students should know the formulae of mean, median and mode. The question is really simple, students should note down the values from the problem really carefully, else the answer can be wrong. One should take care while finding the mean, variance and avoid calculation mistakes. In solving these types of questions, you should be familiar with the formula of sum of the arithmetic progression and their sums.