DAV Class 7 Maths Chapter 12 Worksheet 2
Data Handling Worksheet 2
1. Find the median of the following data. \( \color{black} 11, 39, 43, 45, 25, 46, 43, 42, 37 \).
Solution
\[ \begin{align*} \text{Ascending order} &= 11, 25, 37, 39, 42, 43, 43, 45, 46 \\ \\ \text{Number of }& \text{ observation} = 9 (odd) \\ \\ \text{Median} &= \text{Middle most observation} \\ \\ & = \left(\frac{n+1}{2}\right)^{th} \text{ observation} \\ \\ & = \left(\frac{9+1}{2}\right)^{th} \text{ observation} \\ \\ & = \left(\frac{10}{2}\right)^{th} \text{ observation} \\ \\ & = 5^{th} \text{ observation} \\ \\ 11, 25, 37, 39, &\boxed{\color{green}42}, 43, 43, 45, 46 \\ \\\end{align*} \]
Answer Median \( = \boxed{\color{red}42} \)
2. Find the median of:
(i) \( 85, 42, 56, 69, 51, 81, 72, 92, 67 \)
Solution
\[ \begin{align*} \text{Ascending order:} & \ 42, 51, 56, 67, 69, 72, 81, 85, 92 \\ \\ \text{Number of observations:} & \ 9 \text{ (odd)} \\ \\ \text{Median} & = \text{Middle observation} \\ \\ & = \left(\frac{n+1}{2}\right)^{th} \text{ observation} \\ \\ & = \left(\frac{9+1}{2}\right)^{th} \text{ observation} \\ \\ & = \left(\frac{10}{2}\right)^{th} \text{ observation} \\ \\ & = 5^{th} \text{ observation} \\ \\ 42, 51, 56, 67, &\boxed{\color{green}69}, 72, 81, 85, 92 \\ \\ \end{align*} \]
Answer Median \( = \boxed{\color{red}69} \)
(ii) \( 15, 47, 48, 81, 17, 27, 9, 3, 10, 75 \).
Solution
\[ \begin{align*} \text{Ascending order:} & \ 3, 9, 10, 15, 17, 27, 47, 48, 75, 81 \\ \\ \text{Number of observations:} & \ 10 \text{ (even)} \\ \\ \text{Median} & = \left( \frac{\text{Sum of two middle most observation}}{2}\right) \\ \\ &= 3, 9, 10, 15, \boxed{\color{green} 17, 27}, 47, 48, 75, 81 \\ \\& = \left(\frac{5^{th} \text{ observation} + 6^{th} \text{ observation}}{2}\right) \\ \\ & = \frac{17 + 27}{2} \\ \\ & = \frac{44}{2} \\ \\ & = 22 \\ \\ \end{align*} \]
Answer Median \( = \boxed{\color{red}22} \)
3. The following number of goals were scored by a team in a series of ten matches: \( \color{black} 2, 3, 5, 4, 0, 1, 3, 3, 3, 4 \). Find the median of these scores.
Solution
\[ \begin{align*} \text{Ascending order:} & \ 0, 1, 2, 3, 3, 3, 3, 4, 4, 5 \\ \\ \text{Number of observations:} & \ 10 \text{ (even)} \\ \\ \text{Median} & = \left( \frac{\text{Sum of two middle most observations}}{2}\right) \\ \\ &= 0, 1, 2, 3, \boxed{\color{green} 3, 3}, 3, 4, 4, 5 \\ \\ & = \left(\frac{5^{th} \text{ observation} + 6^{th} \text{ observation}}{2}\right) \\ \\ & = \frac{3 + 3}{2} \\ \\ & = \frac{6}{2} \\ \\ & = 3 \\ \\ \end{align*} \]
Answer Median \( = \boxed{\color{red}3} \)
4. Find the median of the weights (in kg) of 11 students of Class-VII in April 2015: \( \color{black} 42, 52, 48, 52, 55, 58, 53, 54, 46, 49, 57 \). In July 2015, one student of weight 58 kg left and a new student of weight 68 kg joined. Find the new median.
Solution
\[ \begin{align*} \text{Ascending order:} & \ 42, 46, 48, 49, 52, 52, 53, 54, 55, 57, 58 \\ \\ \text{Number of students:} & \ 11 \text{ (odd)} \\ \\ \text{Median} & = \text{Middle most observation}\\ \\ &= \left(\frac{n+1}{2}\right)^{th} \text{ observation} \\ \\ &= \left(\frac{11+1}{2}\right)^{th} \text{ observation} \\ \\ &= \left(\frac{12}{2}\right)^{th} \text{ observation} \\ \\ &= 6^{th} \text{ observation} \\ \\ & = \boxed{52} \\ \\\text{Updated weights:} & \ 42, 46, 48, 49, 52, 52, 53, 54, 55, 57, \cancel{58 } \color{green}68 \\ \\ \text{Ascending order:} & \ 42, 46, 48, 49, 52, 52, 53, 54, 55, 57, 68 \\ \\ \text{Median} & = \text{Middle most observation} \\ \\ &= 6^{th} \text{ observation} \\ \\ & = \boxed{52} \\ \\ \end{align*} \]
Answer New median weight in July \( = \boxed{\color{red}52 \text{ kg}} \)