DAV Class 7 Maths Chapter 12 Worksheet 1
Data Handling Worksheet 1
1. The attendance in a school for a week is as follows: \( \color{black}425, 430, 400, 408, 410, 423 \). Calculate the mean of the attendance.
Solution
\[ \begin{align*} \boxed{\color{green}\text{Mean} = \frac{\text{Sum of all observations}}{\text{Number of observations}}} \end{align*} \] \[ \begin{align*} &= \frac{425 + 430 + 400 + 408 + 410 + 423}{6} \\ \\ &= \frac{\cancelto{416}{2496}}{\cancelto{1}{6}} \\ \\ &= 416 \\ \\ \end{align*} \]
Answer Mean of the attendance \( = \boxed{\color{red}416} \)
2. The heights of ten boys were measured in centimetres and the results were as folows: \( \color{blue}143, 132, 149, 148, 151, 146, 135, 128, 139, 150 \)
(i) What is the height of the tallest boy?
Answer \( {\color{red}151 \text{ cm}} \)
(ii) What is the height of the shortest boy?
Answer \( {\color{red}128 \text{ cm}} \)
(iii) What is the range of the data?
Solution
\[ \begin{align*} \boxed{\color{green}\text{Range} = \text{Highest observation} - \text{Lowest observation}} \end{align*} \] \[ \begin{align*} &= 151 - 128 \\ &= 23 \end{align*} \]
Answer Range \( = {\color{red}23} \)
(iv) What is the range of the data?
Solution
\[ \begin{align*} \boxed{\color{green}\text{Mean} = \frac{\text{Sum of all observations}}{\text{Number of observations}}} \end{align*} \] \[ \begin{align*} &= \frac{143 + 132 + 149 + 148 + 151 + 146 + 135 + 128 + 139 + 150}{10} \\ \\ &= \frac{1421}{10} \\ \\ &= 142.1 \end{align*} \]
Answer Mean height \( = {\color{red}142.1 \text{ cm}} \)
(v) How many boys are there whose heights are less than the mean height?
Answer \( {\color{red}4 \text{ boys}} \)
3. A group of students was given a special test. The test was completed by various students in the following duration of time (in minutes):
\( \color{black} 17, 19, 20, 22, 24, 24, 28, 30, 30, 36 \)(i) Find the mean time taken by the students to complete the test.
Solution
\[ \begin{align*} \boxed{\color{green}\text{Mean} = \frac{\text{Sum of all observations}}{\text{Number of observations}}} \end{align*} \] \[ \begin{align*} &= \frac{ 17 + 19 + 20 + 22 + 24 + 24 + 28 + 30 + 30 + 36 }{10} \\ \\ &= \frac{250}{10} \\ \\ &= 25 \text{ mins}\\ \\ \end{align*} \]
Answer Mean time \( = \boxed{\color{red}25 \text{ mins}} \)
(ii) Find the new mean, if another student who took 14 minutes is also included.
Solution
\[ \begin{align*} \text{Sum of observations} &= 250 + 14 \\ &= 264 \\ \text{No. of observations} &= 10 + 1 \\ &= 11 \\ \\ \text{New Mean} &= \frac{264}{11} \\ \\ &= 24 \text{ mins} \end{align*} \]
Answer New Mean \( = \boxed{\color{red}24 \text{ mins}} \)
4. The enrolment of a school during seven consecutive years was as follows:
\( \color{black} 1580, 2020, 2580, 3200, 3550, 3710, 4010 \)(i) Find the mean enrolment of the school for this period.
Solution
\[ \begin{align*} \boxed{\color{green}\text{Mean} = \frac{\text{Sum of all observations}}{\text{Number of observations}}} \end{align*} \] \[ \begin{align*} &= \frac{ 1580 + 2020 + 2580 + 3200 + 3550 + 3710 + 4010 }{7} \\ \\ &= \frac{20650}{7} \\ \\ &= 2950 \\ \\ \end{align*} \]
Answer Mean enrolment \( = \boxed{\color{red}2950} \)
(ii) Find the new mean, if the enrolment of the current year, that is, 4,270 is also included.
Solution
\[ \begin{align*} \text{Sum of observations} &= 20650 + 4270 \\ &= 24920 \\ \text{No. of observations} &= 7 + 1 \\ &= 8 \\ \\ \text{New Mean} &= \frac{24920}{8} \\ \\ &= 3115 \end{align*} \]
Answer New Mean \( = \boxed{\color{red}3115} \)
5. In a factory the total number of workers are 25. Out of them, the average monthly salary of 15 workers is Rs 8,500 whereas the other 10 workers have an average salary of Rs 9,800. Find the monthly average salary of all the 25 workers.
Solution
\[ \begin{align*} \text{Average salary of 15 workers} &= 8500 \\ \text{Sum salary of 15 workers} &= 8500 \times 15 \\ & = 127500 \\ \\ \text{Average salary of 10 workers} &= 9800 \\ \text{Sum salary of 10 workers} &= 9800 \times 10 \\ & = 98000 \\ \\ \text{Total salary of 25 workers} &= 127500 + 98000 \\ &= 225500 \\ \\ \color{green}\text{Average salary of all workers} &= \color{green}\frac{ \text{Total salary}}{\text{No. of workers}} \\ \\ \text{Average salary of 25 workers}&= \frac{225500}{25} \\ \\ & = \text{Rs } 9020 \end{align*} \]
Answer Average salary of all the 25 workers \( = \boxed{\color{red} \text{Rs } 9020} \)
6. The average salary of 19 workers in a factory is Rs 12,000 per month. If the salary of the manager is Rs 42,000 per month, find the average monthly salary paid to all the employees.
Solution
\[ \begin{align*} \text{Average salary of 19 workers} &= 12000 \\ \text{Sum of the salary of 19 workers} &= 12000 \times 19 \\ & = 228000 \\ \text{Salary of the manager} &= 42000 \\ \text{Total salary } &= 228000 + 42000 \\ &= 270000 \\ \\ \color{green}\text{Average salary } &= \color{green}\frac{ \text{Total salary}}{\text{No. of employees}} \\ \\ \text{Average salary of all employees}&= \frac{270000}{20} \\ \\ & = \text{Rs } 13500 \end{align*} \]
Answer Average monthly salary of all the employees \( = \boxed{\color{red} \text{Rs } 13500} \)
7. The mean of nine observations was found to be 35. Later on, it was discovered that an observation 81 was misread as 18. Find the correct mean of the observations.
Solution
\[ \begin{align*} \text{No. of observation} &= 9 \\ \text{Mean of 9 observation} &= 35 \\ \text{Sum of observation} &= 35 \times 9 \\ &= 315 \\ \text{Correct sum of observations } &= 315+81-18 \\ &= 315+63 \\ &= 378 \\ \\ \color{green}\text{Correct mean}&= \frac{378}{9} \\ \\ & = 42 \end{align*} \]
Answer Correct mean of the observations \( = \boxed{\color{red} 42} \)
8. The mean of five numbers is 27. If one number is excluded, their mean is 25. Find the excluded observation.
Solution
\[ \begin{align*} \text{Mean of 5 numbers} &= 27 \\ \text{Sum of 5 numbers} &= 27 \times 5 \\ &= 135 \\ \\ \text{Mean of 4 numbers} &= 25 \\ \text{Sum of 4 numbers} &= 25 \times 4 \\ &= 100 \\ \\\color{green}\text{Excluded number} &=\color{green} \text{Sum of 5 numbers - Sum of 4 numbers } \\ &= 135 - 100 \\ &= 35 \end{align*} \]
Answer Excluded observation \( = \boxed{\color{red} 35} \)
9. The mean of eight numbers is 27. If one more number is included, then the mean is 26. Find the included number.
Solution
\[ \begin{align*} \text{Mean of 8 numbers} &= 27 \\ \text{Sum of 8 numbers} &= 27 \times 8 \\ &= 216 \\ \\ \text{Mean of 9 numbers} &= 26 \\ \text{Sum of 9 numbers} &= 26 \times 9 \\ &= 234 \\ \\\color{green}\text{Included number} &=\color{green} \text{Sum of 9 numbers - Sum of 8 numbers } \\ &= 234 - 216 \\ &= 18 \end{align*} \]
Answer Included number \( = \boxed{\color{red} 18} \)
10. In a certain hospital, the mean birth rate of a week was 35. If the mean birth rate from Monday to Thursday was 32 and that of Thursday ot Sunday was 36, find the number of births on Thursday.
Solution
\[ \begin{align*} \text{Mean birth of 7 days} &= 35 \\ \text{Sum birth of 7 days} &= 35 \times 7 \\ &= 245 \\ \\\text{Mean birth (Monday to Thursday)} &= 32 \\ \text{Sum of birth of first 4 days} &= 32 \times 4 \\ &= 128 \\ \\\text{Mean birth (Thursday to Sunday)} &= 36 \\ \text{Sum of birth of last 4 days} &= 36 \times 4 \\ &= 144 \\ \\ \text{Total sum birth of 8 days} &= 128 + 144 \\ &= 272 \\ \\ \text{No. of birth on Thursday} &= 272 - 245 \\ &= 27 \end{align*} \]
Answer Births on Thursday \( = \boxed{\color{red} 27} \)