CBSE Class 7 Mathematics Data Handling Worksheet Set 06

Question Q1. Define the terms-
a) data
b) frequency of an observation
c) mean of a data
d) median of a data
e) mode of a data
f) bar graph
g) double bar graph
Answer:
Here are the definitions for each of the terms:

a) **data:** A collection of facts, numbers, or measurements gathered to give information.

b) **frequency of an observation:** The number of times a particular value occurs in a dataset.

c) **mean of a data:** The average value. You find it by adding all values and dividing by the total count.

d) **median of a data:** The middle value when all numbers are arranged in order from smallest to largest.

e) **mode of a data:** The value that appears most often in a dataset.

f) **bar graph:** A chart that uses rectangular bars of different heights to compare different values.

g) **double bar graph:** A bar graph that shows two sets of data side-by-side to make comparisons easy.
In simple words: These terms help us organize, calculate, and visualize numbers to make sense of our records.

Exam Tip: Memorize these short definitions because they often show up as direct, one-mark questions on exams.

Question Q2. Prepare the frequency table for the following data:
5,2,1,3,4,4,5,6,2,2,4,5,5,6,2,2,4,5,5,1.
Answer:
To prepare the frequency table, we count how many times each number appears in the list:

In simple words: A frequency table counts how many times each number shows up in our list.

Exam Tip: Always add up your frequency column to verify that the total matches the count of numbers given in the question.

Question Q3. Find the mean of first five natural numbers.
Answer:
The first five natural numbers are 1, 2, 3, 4, and 5.
To find their mean, we add these numbers together:
\( \text{Sum} = 1 + 2 + 3 + 4 + 5 = 15 \)

Since there are 5 numbers, we divide this sum by 5:
\( \text{Mean} = \frac{15}{5} = 3 \)

Therefore, the mean of the first five natural numbers is 3.
In simple words: The average of the numbers 1, 2, 3, 4, and 5 is 3.

Exam Tip: Natural numbers start from 1, whereas whole numbers start from 0. Be sure not to mix them up!

Question Q4. Find the mean of first five prime numbers.
Answer:
The first five prime numbers are 2, 3, 5, 7, and 11.
To find their mean, we add these numbers together:
\( \text{Sum} = 2 + 3 + 5 + 7 + 11 = 28 \)

Since there are 5 numbers, we divide this sum by 5:
\( \text{Mean} = \frac{28}{5} = 5.6 \)

Therefore, the mean of the first five prime numbers is 5.6.
In simple words: The average of the first five prime numbers is 5.6.

Exam Tip: Note that 1 is not a prime number, and 2 is the only even prime number.

Question Q5. The ages (in years) of 50 players of a school are given below:
Age (yr) 14 15 16 17 18
No. of players 15 14 10 8 3
Find the mean age.
Answer:
To find the mean age of the players, we multiply each age by the number of players of that age and then sum them up:

Now, divide the total sum of ages by the total number of players:
\( \text{Mean Age} = \frac{\sum fx}{\sum f} = \frac{775}{50} = 15.5 \text{ years} \)

Therefore, the mean age of the players is 15.5 years.
In simple words: Multiply each age by its player count, add the results, and divide by 50 to get 15.5.

Exam Tip: Always construct a table with columns for \( x \), \( f \), and \( fx \) to systematically calculate grouped means.

Question Q6. The weights of 10 students (kg) are:
40,52,34,47,31,35,48,41,44,38,42.
Find the median weight.
Answer:
Although the question mentions 10 students, there are 11 weights listed. Let us arrange these 11 weights in ascending order:
31, 34, 35, 38, 40, 41, 42, 44, 47, 48, 52

Since \( n = 11 \) (which is an odd number), the median is the middle value at the \( \left(\frac{n+1}{2}\right) \)-th position:
\( \text{Median Position} = \frac{11+1}{2} = 6\text{-th term} \)

Looking at our sorted list, the 6th term is 41 kg.

Therefore, the median weight is 41 kg.
In simple words: Sort the weights from lightest to heaviest. The exact middle weight is 41 kg.

Exam Tip: Make sure to count the number of data points before and after sorting to verify no number was missed.

Question Q7. Calculate the median for the following data:
Marks 17 20 22 15 30 25
No. of students 5 9 4 3 10 6
Answer:
First, let us arrange the marks in ascending order and calculate their cumulative frequencies:

Since \( N = 37 \) (which is an odd number), the median is the value at the \( \left(\frac{N+1}{2}\right) \)-th position:
\( \text{Median Position} = \frac{37+1}{2} = 19\text{-th term} \)

Looking at the cumulative frequencies, the 19th term falls in the group corresponding to Marks 22 (since 17 < 19 ≤ 21).

Therefore, the median mark is 22.
In simple words: Put the marks in order, find the cumulative count, and look for the student right in the middle (the 19th student), who got 22 marks.

Exam Tip: For grouped data, always add a cumulative frequency column to find the exact median value.

Question Q8. Find the median of first 10 even numbers.
Answer:
The first 10 even numbers are:
2, 4, 6, 8, 10, 12, 14, 16, 18, 20

Since there is an even number of values (\( n = 10 \)), the median is the average of the two middle terms:
\( \text{Median} = \frac{\left(\frac{n}{2}\right)\text{-th term} + \left(\frac{n}{2} + 1\right)\text{-th term}}{2} \)
\( \text{Median} = \frac{5\text{-th term} + 6\text{-th term}}{2} \)

The 5th term is 10 and the 6th term is 12:
\( \text{Median} = \frac{10 + 12}{2} = 11 \)

Therefore, the median of the first 10 even numbers is 11.
In simple words: The middle numbers are 10 and 12, and their middle average is 11.

Exam Tip: If you have an even number of values, average the two middle values to find your median.

Question Q9. The ages (in years) of 11 cricket players are given below:
28,34,32,41,36,32,32,38,32,40,31.
Find the mode of ages.
Answer:
Let us count the frequency of each age in the list:
- 28 occurs 1 time
- 31 occurs 1 time
- 32 occurs 4 times
- 34 occurs 1 time
- 36 occurs 1 time
- 38 occurs 1 time
- 40 occurs 1 time
- 41 occurs 1 time

Since the age 32 occurs most frequently (4 times), the mode of the ages is 32 years.
In simple words: The age that appears the most in this list is 32.

Exam Tip: Mode is the value with the highest frequency, so simply count the occurrences to find it.

Question Q10. Daily wages of 45 workers in a factory are given below:
Daily wages (in Rs.) 100 125 150 175 200
No. of workers 6 8 9 12 10
Find the median, mean and mode.
Answer:
Let us prepare a combined table to find the Mean, Median, and Mode:

1. Mean:
\( \text{Mean} = \frac{\sum fx}{N} = \frac{7050}{45} = \text{Rs } 156.67 \)

2. Median:
Since \( N = 45 \) (odd), the median is at the \( \left(\frac{N+1}{2}\right) \)-th position:
\( \text{Median Position} = \frac{45+1}{2} = 23\text{-rd term} \)
Looking at the cumulative frequencies, the 23rd term corresponds to Rs 150.
\( \text{Median} = \text{Rs } 150 \)

3. Mode:
The highest frequency is 12, which corresponds to the wage of Rs 175.
\( \text{Mode} = \text{Rs } 175 \)
In simple words: The average wage is Rs 156.67, the middle wage is Rs 150, and the most common wage is Rs 175.

Exam Tip: Be precise with calculations of \( \sum fx \) to ensure your mean calculation is correct to two decimal places.

Question Q11. The following table shows the weight of 12 players:
Weight in kg 48 50 52 54 58
No. of players 4 3 2 2 1
Find the median and mean weights.
Using empirical formula, calculate its mode. [Note: Empirical formula for calculating mode is Mode = 3 Median - 2 Mean]
Answer:
Let us arrange the calculations in a table:

1. Mean:
\( \text{Mean} = \frac{\sum fx}{N} = \frac{612}{12} = 51 \text{ kg} \)

2. Median:
Since \( N = 12 \) (even), the median is the average of the 6th and 7th terms.
Looking at the cumulative frequencies, both the 6th and 7th terms fall in the weight category of 50 kg (since 4 < 6 and 7 ≤ 7).
\( \text{Median} = 50 \text{ kg} \)

3. Mode using Empirical Formula:
\( \text{Mode} = 3 \text{ Median} - 2 \text{ Mean} \)
\( \text{Mode} = 3(50) - 2(51) = 150 - 102 = 48 \text{ kg} \)
In simple words: The average weight is 51 kg, the middle weight is 50 kg, and the empirical mode is 48 kg.

Exam Tip: Always double check that you substitute Mean and Median in their correct positions in the empirical formula.

Question Q12. The following table shows the no. of scooters produced by a company during 6 consecutive years. Draw a bar graph to represent.
Year 2001 2002 2003 2004 2005 2006
No. of scooters 11000 14000 12500 17500 15000 24000
Answer:
Here is the bar graph showing the production of scooters over the years:

In simple words: The graph shows how scooter production changed each year, peaking at 24000 in 2006.

Exam Tip: Choose a clear, uniform scale for the Y-axis and label both axes clearly to score full presentation marks.

Question Q13. A coin is tossed 100 times and head is obtained 59 times. On tossing a coin at random, find the probability of getting :
a) A head.
b) A tail.
Answer:
Total number of trials = 100
Number of heads obtained = 59
Number of tails obtained = \( 100 - 59 = 41 \)

a) Probability of getting a head:
\( P(\text{Head}) = \frac{\text{Number of heads}}{\text{Total trials}} = \frac{59}{100} = 0.59 \)

b) Probability of getting a tail:
\( P(\text{Tail}) = \frac{\text{Number of tails}}{\text{Total trials}} = \frac{41}{100} = 0.41 \)
In simple words: The chance of getting a head is 59 out of 100, and the chance of getting a tail is 41 out of 100.

Exam Tip: Check that the sum of all probabilities in a single experiment equals exactly 1 (\( 0.59 + 0.41 = 1.00 \)).

Question Q14. A dice is thrown 200 times and the out comes are noted as shown below:
Outcome 1 2 3 4 5 6
Freqency 21 30 42 38 29 40
When a dice is thrown at random, find the probability of getting:
a) 5
b) 3
c) 4
d) 6.
Answer:
The total number of trials is 200.

a) Probability of getting 5:
\( P(5) = \frac{\text{Frequency of 5}}{\text{Total trials}} = \frac{29}{200} = 0.145 \)

b) Probability of getting 3:
\( P(3) = \frac{\text{Frequency of 3}}{\text{Total trials}} = \frac{42}{200} = 0.21 \)

c) Probability of getting 4:
\( P(4) = \frac{\text{Frequency of 4}}{\text{Total trials}} = \frac{38}{200} = 0.19 \)

d) Probability of getting 6:
\( P(6) = \frac{\text{Frequency of 6}}{\text{Total trials}} = \frac{40}{200} = 0.2 \)
In simple words: Divide the frequency of each outcome by 200 to find its chance of occurring.

Exam Tip: Express your probability answers as simplified fractions first, then write the decimal form.

Question Q15. 50 seeds were selected at random from each of 5 bags of seeds and were kept under standardised conditions favourable to germination. After 20 days the no. of seeds which had germinated in each collection were counted and recorded as follows:
Bag 1 2 3 4 5
No. of seeds germinated 40 48 42 39 41
What is the probability of germination of:
a) More than 40 seeds in a bag?
b) 49 seeds in a bag?
c) More than 35 seeds in a bag?
d) Atleast 40 seeds in a bag?
e) Atmost 40 seeds in a bag?
Answer:
The total number of bags evaluated is 5.

a) **More than 40 seeds in a bag (Bags with > 40 germinated seeds):**
Bags 2 (48), 3 (42), and 5 (41). (Total = 3 bags)
\( P(> 40) = \frac{3}{5} = 0.6 \)

b) **49 seeds in a bag:**
No bag has exactly 49 seeds germinated. (Total = 0 bags)
\( P(49) = \frac{0}{5} = 0 \)

c) **More than 35 seeds in a bag:**
All 5 bags have more than 35 germinated seeds. (Total = 5 bags)
\( P(> 35) = \frac{5}{5} = 1 \)

d) **At least 40 seeds in a bag (Bags with ≥ 40 germinated seeds):**
Bags 1 (40), 2 (48), 3 (42), and 5 (41). (Total = 4 bags)
\( P(\ge 40) = \frac{4}{5} = 0.8 \)

e) **At most 40 seeds in a bag (Bags with ≤ 40 germinated seeds):**
Bags 1 (40) and 4 (39). (Total = 2 bags)
\( P(\le 40) = \frac{2}{5} = 0.4 \)
In simple words: Look at the bags that match each condition, and divide that number of bags by 5.

Exam Tip: Be careful with keywords: "Atleast 40" includes 40, while "More than 40" does not.

Question Q16. The distance covered by 10 athletes in 1 minute in a sprint race of 400 m is :
299,248,315,320,286,370,388,379,295,300.
a) What is the maximum distance covered in 1 minute?
b) What is the minimum distance covered in 1 minute?
c) What is the range of the data?
d) What is the mean of the data?
e) How many athletes have covered distance more than the mean distance?
Answer:
Let us sort the data in ascending order first:
248, 286, 295, 299, 300, 315, 320, 370, 379, 388

a) **Maximum distance covered:** 388 m.
b) **Minimum distance covered:** 248 m.
c) **Range:** \( \text{Maximum} - \text{Minimum} = 388 - 248 = 140 \text{ m} \).
d) **Mean:**
\( \text{Sum of distances} = 248 + 286 + 295 + 299 + 300 + 315 + 320 + 370 + 379 + 388 = 3200\text{ m} \)
\( \text{Mean} = \frac{3200}{10} = 320 \text{ m} \).
e) **Athletes covering more than the mean distance (strictly > 320 m):**
The distances greater than 320 m are 370 m, 379 m, and 388 m.
So, exactly 3 athletes covered more than the mean distance.
In simple words: The longest distance is 388 m, the shortest is 248 m, the average is 320 m, and 3 runners beat that average.

Exam Tip: Note that an athlete covering exactly 320 m is not included in "more than 320 m".

Question Q17. The following table show the no. of boys and girls who have paricipated in various events conducted by a Rotary club of a city. Draw a double bar graph to represent the given data using an appropriate scale:
Event Dance Drama Sports Rangoli Singing
Boys 400 480 670 220 710
Girls 580 330 590 610 720
Answer:
Here is the double bar graph representing the participation data:

In simple words: The blue bars show boys and the pink bars show girls for each of the five events.

Exam Tip: Include a clear legend when drawing double bar graphs so the reader knows which bar stands for which group.

Question Q18. Given below is the marks (M.M-100) obtained by Samaira (class VII) in First and Second Terminal Exam 2014. Represent it using a double bar graph.
Subject English Hindi Maths Science So. Science Sanskrit
1st term 92 93 98 91 99 100
2nd term 98 97 100 100 99 100.
Answer:
Here is the double bar graph representing Samaira's terminal marks:

In simple words: The graph shows how Samaira's marks generally improved or stayed excellent between the first and second terms.

Exam Tip: On your exams, use different shading patterns or distinct colors to clearly separate the two terms.