DAV Class 7 Maths Chapter 12 Worksheet 3
Data Handling Worksheet 3
1. Find the mode of the following observations.
(i) \( 25, 14, 28, 17, 18, 14, 25, 14, 17, 14 \)
Solution
\[ \begin{align*} \text{Ascending order} & =\, \boxed{\color{green} 14, \, 14, \, 14, \, 14}, \, 17, \, 17, \, 18, \, 25, \, 25, \, 28 \\ \text{Mode} & = 14 \end{align*} \]
Answer Mode \( = \boxed{\color{red}14} \)
(ii) \( 0, 6, 1, 5, 6, 2, 0, 4, 6, 3, 4, 6, 3, 1 \)
Solution
\[ \begin{align*} \text{Ascending order} &= 0, \, 0, \, 1, \, 1, \, 2, \, 3, \, 3, \, 4, \, 4, \, 5, \, \boxed{\color{green} 6, \, 6, \, 6, \, 6 }\\ \text{Mode} & = 6 \end{align*} \]
Answer Mode \( = \boxed{\color{red}6} \)
2. The following table shows the weight of 15 students. Find the mode. \[ \begin{array}{|c|c|} \hline \text{Weight (in kg)} & \text{No. of students} \\ \hline 46 & 3 \\ 51 & 1 \\ 53 & 5 \\ 56 & 2 \\ 60 & 4 \\ \hline \end{array} \]
Solution
\[ \begin{align*} & \text{Here the maximum frequency is 5.} \\ & \text{The weight of maximum number of students is 53 kg.} \\ & \text{Hence mode is } \boxed{53}. \end{align*} \]
Answer Mode \( = \boxed{\color{red}53} \)
3. The following table shows the sale of shirts having different sizes from a certain shop in a month. Find the mode.
\[ \begin{array}{|c|c|c|c|c|c|} \hline \text{Size} & 38 & 39 & 40 & 42 & 44 \\ \hline \text{No. of shirts} & 24 & 31 & 23 & 14 & 7 \\ \hline \end{array} \]
Solution
\[ \begin{align*} & \text{Here the maximum frequency is 31.} \\ & \text{Size 39 has the maximum frequency of shirts.} \\ & \text{Hence mode is } \boxed{39}. \end{align*} \]
Answer Mode \( = \boxed{\color{red}39} \)
4. In January 2015, the number of children in 10 families of a locality are: \(\color{black} 4, 3, 4, 0, 2, 2, 5, 2, 1, 3 \). Find the mean, median and mode. At the end of the year, wto families having children 0and 1vacated the house. As a result, wt o more families having children 2and 5got the vacant accommodation. Fnid the new mean, median and mode.
Solution
\[ \begin{align*} \text{Ascending order} & = \, 0, 1, 2, 2, 2, 3, 3, 4, 4, 5 \\ \\\text{Mean} &= \frac{0 + 1 + 2 + 2 + 2 + 3 + 3 + 4 + 4 + 5}{10} \\ &= \frac{26}{10} \\ \\ \text{Mean} &= \boxed{\color{green}2.6} \\ \\& \text{Number of observations} = \ 10 \text{ (even)} \\ \\\text{Median} & = \left( \frac{\text{Sum of two middle most observations}}{2}\right) \\ \\ &= 0, 1, 2, 2, \boxed{\color{green}2, 3}, 3, 4, 4, 5 \\ \\ & = \left(\frac{5^{th} \text{ observation} + 6^{th} \text{ observation}}{2}\right) \\ \\ & = \frac{2 + 3}{2} \\ \\ \text{Median} &= \boxed{\color{green}2.5} \\ \end{align*} \]
\[ \begin{align*} \text{Mode} & = \text{The mode is the observation with the maximum frequency.} \\ \text{Ascending order} & = \, 0, 1, \boxed{\color{green}2, 2, 2}, 3, 3, 4, 4, 5 \\ & \text{Here, 2 occurs 3 times, which is the maximum frequency.} \\ & \text{Hence mode is } \boxed{2}. \end{align*} \]\[ \begin{align*} \color{red}\text{Mean} &= \color{red} 2.6 \\ \color{red}\text{Median} &= \color{red} 2.5 \\ \color{red}\text{Mode} &= \color{red} 2 \\ \end{align*} \]
\[ \begin{align*}& \boxed{\color{green}\text{New data}} \\ \\ \text{Updated observation} & = \, 4, 3, 4, 2, 2, 2, 5, 2, 3, 5 \\ \text{Ascending order} & = \, 2, 2, 2, 2, 3, 3, 4, 4, 5, 5 \\ \end{align*} \]
\[ \begin{align*} \text{Mean} & = \frac{2 + 2 + 2 + 2 + 3 + 3 + 4 + 4 + 5 + 5}{10} \\ \\ & = \frac{32}{10} \\ \\ \text{Mean} &= \boxed{\color{green}3.2} \\ \\ \end{align*} \]
\[ \begin{align*} & \text{Number of observations} = 10 \, (\text{even}) \\ \\\text{Median} & = \text{Mean of two middle most data} \\ \\ &= 2, 2, 2, 2, \boxed{\color{green}3, 3}, 4, 4, 5, 5 \\ \\ & = \frac{3 + 3}{2} \\ \\ \text{Median} &= \boxed{\color{green}3} \\ \end{align*} \]
\[ \begin{align*} & \text{Mode:} \\ & \text{The mode is the observation with the highest frequency.} \\ & \text{Here, 2 occurs 4 times, which is the highest frequency.} \\ & \text{Mode } = \boxed{\color{red}2} \end{align*} \]
Answer New Mean \( = \boxed{\color{red}3.2} \), New Median \( = \boxed{\color{red}3} \), New Mode \( = \boxed{\color{red}2} \)