DAV Class 7 Maths Chapter 12 Practice Worksheet
Data Handling Practice Worksheet
1. Find the mean of all factors of \( \color{black}{10} \).
Solution
\[ \begin{aligned} \text{Factors of } 10 &= 1, 2, 5, 10 \\[8pt] \color{green}\text{Mean} &= \color{green}\frac{\text{Sum of all observations}}{\text{Number of observations}} \\[8pt] \text{Mean} &= \frac{1 + 2 + 5 + 10}{4} \\[8pt] &= \frac{18}{4} \\[6pt] &= 4.5 \end{aligned} \]
Answer Mean of all factors of 10 \( = \color{red}{4.5} \)
2. The heights of five persons are \( \color{black}{140 \text{ cm}, 150 \text{ cm}, 152 \text{ cm}, 158 \text{ cm}, 161 \text{ cm}} \) respectively. Find the Range and the mean height of five persons.
Solution
\[ \begin{aligned} \color{green}\text{Range} &= \color{green}\text{Highest observation} - \text{Lowest observation} \\[8pt] \text{Highest} &= 161 \text{ cm} \\ \text{Lowest} &= 140 \text{ cm} \\[4pt] \text{Range} &= 161 - 140 \\[4pt] &= 21 \text{ cm} \end{aligned} \]
\[ \begin{aligned} \color{green}\text{Mean} &= \color{green} \frac{\text{Sum of all observations}}{\text{Number of observations}} \\[8pt] \text{Mean height} &= \frac{140 + 150 + 152 + 158 + 161}{5} \\[6pt] &= \frac{761}{5} \\[6pt] &= 152.2 \text{ cm} \end{aligned} \]
Answer Range \( = \color{red}{21 \text{ cm}} \), Mean height \( = \color{red}{152.2 \text{ cm}} \)
3. The mean of five numbers is \( \color{black}{27} \). If one number is excluded, their mean is \( \color{black}{25} \). Find the excluded number.
Solution
\[ \begin{aligned} \text{Mean of 5 numbers} &= 27 \\ \text{Sum of 5 numbers} &= 27 \times 5 \\ & \implies 135 \\[6pt] \text{Mean of 4 numbers} &= 25 \\ \text{Sum of 4 numbers} &= 25 \times 4 \\ & \implies 100 \\[8pt] \color{green}\text{Excluded number} &= \color{green}\text{Sum of 5 numbers} - \text{Sum of 4 numbers} \\[4pt] &= 135 - 100 \\[4pt] &= 35 \end{aligned} \]
Answer Excluded number \( = \color{red}{35} \)
4. Find the mean of the weights (in kg) of 10 new born babies in a hospital on a particular day: \( \color{black}{3.4, 3.6, 4.2, 4.5, 3.9, 4.1, 3.8, 4.5, 4.4, 3.6} \).
Solution
\[ \begin{aligned} \color{green}\text{Mean} = \frac{\text{Sum of all observations}}{\text{Number of observations}} \end{aligned} \] \[ \begin{aligned} \text{Mean weight} &= \frac{3.4 + 3.6 + 4.2 + 4.5 + 3.9 + 4.1 + 3.8 + 4.5 + 4.4 + 3.6}{10} \\[6pt] &= \frac{40}{10} \\[6pt] &= 4 \text{ kg} \end{aligned} \]
Answer Mean weight of babies \( = \color{red}{4 \text{ kg}} \)
5. The mean of 200 items was 50. Later on, it was discovered that the two items were misread as \(92\) and \(8\) instead of \(192\) and \(88\). Find the correct mean.
Solution
\[ \begin{aligned} \text{No. of items} &= 200 \\ \text{Mean} &= 50 \\[4pt] \text{Total sum} &= 200 \times 50 \\ & \implies 10000 \\[8pt] \text{Wrong items} &= 92, \; 8 \\ \text{Correct items} &= 192, \; 88 \\[4pt] \text{Correct total} &= 10000 - (92 + 8) + (192 + 88) \\[4pt] &= 10000 - 100 + 280 \\[4pt] &= 10180 \\[8pt] \color{green}\text{Correct mean} &= \color{green}\frac{10180}{200} \\[8pt] &= \frac{1018}{20} \\[8pt] &= 50.9 \end{aligned} \]
Answer Correct mean \( = \color{red}{50.9} \)
6.
The scores in Mathematics test (out of 25) of 15 students are as follows:
\( \color{black}{19, 25, 23, 20, 9, 20, 15, 10, 5, 16, 25, 20, 24, 12, 20} \).
Find the mode and median of this data.
Solution
Ascending order \( = 5, 9, 10, 12, 15, 16, 19, 20, 20, 20, 20, 23, 24, 25, 25 \)
\[ \begin{aligned} \text{No. of observations} &= 15 \; (\text{odd}) \\[4pt] \text{Median} &= \left(\frac{n+1}{2}\right)^{\text{th}} \text{ observation} \\[4pt] &= \left(\frac{15+1}{2}\right)^{\text{th}} \text{ observation} \\[4pt] &= 8^{\text{th}} \text{ observation} \\[4pt] Median &= \color{green}{20} \end{aligned} \]
\[ \begin{aligned} &\text{Mode is the observation with highest frequency.} \\ &\text{Here, } 20 \text{ occurs 4 times, which is the maximum.} \end{aligned} \]
Answer Median \( = \color{red}{20} \), Mode \( = \color{red}{20} \)
7. Consider the following data gathered from a survey of a colony:
\[ \begin{array}{|c|c|c|c|c|c|} \hline \text{Favourite Sport} & \text{Cricket} & \text{Basket Ball} & \text{Swimming} & \text{Hockey} & \text{Athletics} \\ \hline \text{Watching} & 1240 & 450 & 500 & 400 & 250 \\ \hline \text{Participating} & 620 & 350 & 300 & 250 & 100 \\ \hline \end{array} \]
Draw a double bar graph choosing an appropriate scale.
(a) Which sport is most popular?
Solution
y-axis: \(1\) unit \(= 100\) people
Answer \( \color{red}{{Cricket}} \)
(b) What is more preferred, watching or participating in sports?
Answer Watching sports is more preferred than participating.