Question

For the data 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, …., 9, 9, the product of the mean and mode equals
(a) 9
(b) 45
(c) 57
(d) 285

Answer 1

Hint: To solve this question, we will first define mean and mode. The formula of mean is given by \[\text{Mean}=\dfrac{\text{Sum of observations}}{\text{Number of observations}}\] and mode is the number that occurs out of the most number of terms. The sum of n terms is given by \[\dfrac{n\left( n+1 \right)}{2}\] and that of the square of n terms is given by \[\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\] to get the result. To solve this question, let us first define the mean of the terms and mode of the terms.

Complete step by step answer:
Let us first define mean and its formula. Mean is also called the Arithmetic Mean n the average of the numbers also called a central value of a set of numbers. To calculate mean, we will first add all the numbers and then divide by the number of terms. The formula is given by
\[\text{Mean}=\dfrac{\text{Sum of observations}}{\text{Number of observations}}\]
And let us define mode now. The mode is the value that appears most frequently in a data set. A set of data may have one mode or more than one mode. We are given the terms as
\[1,2,2,3,3,3,4,4,4,4,......,9,9\]
Observing this, we see that 1 occurs 1 time, 2 occurs 2 times, 3 occurs 3 times and so on. We can say that the digits occur according to their value. Then the sum of the observations will be
\[1+2+2+3+3+3+.....+9+9\]
\[\Rightarrow 1+4+9+16.....+81\]
\[\Rightarrow 1+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}.....+{{9}^{2}}\]
So, the sum of observations is \[1+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}.....+{{9}^{2}}.\] And also the number of observations is also 1 + 2 + 3 + 4 + ….. + 9, as 1 occurs once, 2 occurs twice and similarly for all others.
Therefore, the number of observations is \[1+2+3+4......+9.\]
The formula of the sum of n terms is given by \[\dfrac{n\left( n+1 \right)}{2}\] and the formula of the square of the sum of the n terms is given by
\[{{1}^{2}}+{{2}^{2}}+....+{{n}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\]
Here, we have, n = 9.
Therefore, the number of observations will be
\[\Rightarrow \dfrac{n\left( n+1 \right)}{2}\]
\[\Rightarrow \dfrac{9\left( 10 \right)}{2}\]
\[\Rightarrow 45\]
And the sum of the observations will be
\[\Rightarrow \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\]
\[\Rightarrow \dfrac{9\left( 10 \right)\left( 18+1 \right)}{6}\]
\[\Rightarrow \dfrac{3\times 10\times 19}{2}\]
\[\Rightarrow 15\times 19\]
\[\Rightarrow 285\]
Therefore, we will get the mean as
\[\text{Mean}=\dfrac{285}{45}\]
\[\Rightarrow \text{Mean}=\dfrac{19}{3}\]
So, the mean is given by \[\dfrac{19}{3}.\]
And the mode is the value which occurs the maximum number of times. Because 9 is occurring 9 times, the mode will be 9. Therefore, the product of mean and mode will be
\[\Rightarrow \dfrac{19}{3}\times 9=57\]
Hence, the product of the mean and mode of the given terms is given by 57.

So, the correct answer is “Option C”.

Note: Another method to calculate mean can be directly substituting \[\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\] and \[\dfrac{n\left( n+1 \right)}{2}\] in the numerator and denominator respectively and then solving.
\[\Rightarrow \text{Mean}=\dfrac{\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}}{\dfrac{n\left( n+1 \right)}{2}}\]
\[\Rightarrow \text{Mean}=\dfrac{2n+1}{3}\]
Here, n = 9. Therefore, we get the mean as
\[\Rightarrow \text{Mean}=\dfrac{18+1}{3}=\dfrac{19}{3}\]
Therefore, the answer of the mean is the same.