Connecting the Dots... Class 7 Mathematics NCERT Solutions

Q1Figure it Out (Section 5.2)

Shreyas is playing with a bat and a ballbut not cricket. He counts the number of times he can bounce the ball on the bat before it falls to the ground. The data for 8 attempts is 6, 2, 9, 5, 4, 6, 3, 5. Calculate the average number of bounces of the ball that Shreyas is able to make with his bat.

Solution

Given: The data for 8 attempts is: 6, 2, 9, 5, 4, 6, 3, 5.

To Find: The average number of bounces.

Formula:

\text{Average (Mean)} = \frac{\text{Sum of all the values in the data}}{\text{Number of values in the data}}

Solution: First, we find the sum of the number of bounces in all attempts. Sum of bounces =

6 + 2 + 9 + 5 + 4 + 6 + 3 + 5 = 40

Next, we count the number of attempts. Number of attempts = 8

Now, we calculate the average. Average =

\frac{40}{8} = 5

Final Answer: The average number of bounces Shreyas is able to make is 5.

Q2Figure it Out (Section 5.2)

Try the activity above on your own. Collect data for 7 or more attempts and find the average.

Solution

This is an activity-based question. The answer will vary for each individual. To solve this:

Perform the activity: Take a bat and a ball. Count how many times you can bounce the ball on the bat in one attempt. Repeat this for at least 7 attempts.
Record the data: Write down the number of bounces for each attempt. For example, your data for 7 attempts might be: 8, 12, 7, 10, 9, 11, 13.
Calculate the average:
- Sum the values: $8 + 12 + 7 + 10 + 9 + 11 + 13 = 70$
- Divide by the number of attempts: Average = $\frac{70}{7} = 10$

In this example, the average number of bounces is 10. Your own result will depend on your collected data.

Q3Figure it Out (Section 5.2)

Identify a flowering plant in your neighbourhood. Track the number of flowers that bloom every day over a week during its flowering season. What is the average number of flowers that bloomed per day?

Solution

This is an activity-based question. The answer will depend on the plant observed and the data collected.

Procedure:

Choose a plant: Select a flowering plant near your home.
Collect data: For 7 consecutive days (a week), count the number of new flowers that have bloomed each day.
Record the data: Create a table. For example:
- Day 1: 3 flowers
- Day 2: 5 flowers
- Day 3: 4 flowers
- Day 4: 6 flowers
- Day 5: 2 flowers
- Day 6: 5 flowers
- Day 7: 3 flowers
Calculate the average:
- Sum the values: $3 + 5 + 4 + 6 + 2 + 5 + 3 = 28$
- Divide by the number of days: Average = $\frac{28}{7} = 4$

In this example, the average number of flowers that bloomed per day is 4. Your result will be based on your own observations.

Q4Figure it Out (Section 5.2)

Two friends are training to run a 100 m race. Their running times over the past week are given in seconds— Nikhil: 17, 18, 17, 16, 19, 17,18 ; Sunil: 20, 18, 18, 17, 16, 16, 17. Who on average ran quicker?

Solution

Given: Nikhil's running times (in seconds): 17, 18, 17, 16, 19, 17, 18 Sunil's running times (in seconds): 20, 18, 18, 17, 16, 16, 17

To Find: Who ran quicker on average. A quicker runner will have a lower average time.

Formula:

\text{Average (Mean)} = \frac{\text{Sum of all values}}{\text{Number of values}}

Solution:

For Nikhil: Number of races = 7 Sum of times =

17 + 18 + 17 + 16 + 19 + 17 + 18 = 122

seconds Average time for Nikhil =

\frac{122}{7} \approx 17.43

seconds

For Sunil: Number of races = 7 Sum of times =

20 + 18 + 18 + 17 + 16 + 16 + 17 = 122

seconds Average time for Sunil =

\frac{122}{7} \approx 17.43

seconds

Comparison: Both Nikhil and Sunil have the same average running time of approximately 17.43 seconds. Therefore, on average, they ran equally quick.

Final Answer: On average, both Nikhil and Sunil ran equally quick, as their average times are the same.

Q5Figure it Out (Section 5.2)

The enrolment in a school during six consecutive years was as follows: 1555, 1670, 1750, 2013, 2040, 2126. Find the mean enrolment in the school during this period.

Solution

Given: School enrolment for six consecutive years: 1555, 1670, 1750, 2013, 2040, 2126.

To Find: The mean enrolment during this period.

Formula:

\text{Mean} = \frac{\text{Sum of all values}}{\text{Number of values}}

Solution: First, we find the sum of the enrolments for all six years. Sum of enrolments =

1555 + 1670 + 1750 + 2013 + 2040 + 2126 = 11154

Number of years = 6

Now, we calculate the mean enrolment. Mean enrolment =

\frac{11154}{6} = 1859

Final Answer: The mean enrolment in the school during this period was 1859 students.

Q1Figure it Out (Section 5.2, Part 2)

Find the median of onion prices in Yahapur and Wahapur.

Solution

Given: Monthly onion prices for Yahapur and Wahapur. Yahapur: 25, 24, 26, 28, 30, 35, 39, 43, 44, 49, 56, 59 Wahapur: 19, 17, 23, 30, 35, 38, 42, 39, 53, 60, 52, 42

To Find: The median price for each town.

Solution: To find the median, we must first sort the data for each town in ascending order.

For Yahapur: Number of data points (N) = 12. Sorted data: 24, 25, 26, 28, 30, 35, 39, 43, 44, 49, 56, 59 Since N is even, the median is the average of the two middle values, which are the 6th and 7th values. Median for Yahapur =

\frac{35 + 39}{2} = \frac{74}{2} = 37

For Wahapur: Number of data points (N) = 12. Original data: 19, 17, 23, 30, 35, 38, 42, 39, 53, 60, 52, 42 Sorted data: 17, 19, 23, 30, 35, 38, 39, 42, 42, 52, 53, 60 Since N is even, the median is the average of the two middle values, the 6th and 7th values. Median for Wahapur =

\frac{38 + 39}{2} = \frac{77}{2} = 38.5

Final Answer: The median price of onions in Yahapur is ₹37. The median price of onions in Wahapur is ₹38.5.

Q2Figure it Out (Section 5.2, Part 2)

Sanskruti asked her class how many domestic animals and pets each had at home. Some of the students were absent. The data values are 0, 1, 0, 4, 8, 0, 0, 2, 1, 1, 5, 3, 4, 0, 0, -, 10, 25, 2, -, 2, 4. Find the mean and median. How would you describe this data?

Solution

Given: The data values are 0, 1, 0, 4, 8, 0, 0, 2, 1, 1, 5, 3, 4, 0, 0, 10, 25, 2, 2, 4. The '-' indicates absent students, so we ignore them.

To Find: The mean and median of the data, and a description of the data.

Solution: First, let's list the valid data points and count them. Data: 0, 1, 0, 4, 8, 0, 0, 2, 1, 1, 5, 3, 4, 0, 0, 10, 25, 2, 2, 4 Number of students who responded (N) = 20.

Calculating the Mean: Sum of values =

0+1+0+4+8+0+0+2+1+1+5+3+4+0+0+10+25+2+2+4 = 72

Mean =

\frac{\text{Sum of values}}{\text{Number of values}} = \frac{72}{20} = 3.6

Calculating the Median: First, sort the data in ascending order: 0, 0, 0, 0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 4, 4, 4, 5, 8, 10, 25 There are 20 data points (an even number). The median is the average of the two middle values, the 10th and 11th values. 10th value = 2 11th value = 2 Median =

\frac{2 + 2}{2} = 2

Description of the Data: The data represents the number of pets/domestic animals owned by students.

The data is skewed to the right. Most students have a small number of pets (0 to 4), as shown by the median of 2.
There are a few students with a significantly larger number of animals (8, 10, and especially 25). The value 25 is a clear outlier.
This outlier pulls the mean (3.6) to be much higher than the median (2).
In this case, the median (2) is a better representation of the 'typical' number of pets for a student in this class than the mean.

Final Answer: The mean number of pets is 3.6. The median number of pets is 2. The data is right-skewed with an outlier at 25, which makes the mean higher than the median.

Q3Figure it Out (Section 5.2, Part 2)

Rintu takes care of a date-palm tree farm in Habra. The heights of the trees (in feet) in his farm are given as: 50, 45, 43, 52, 61, 63, 46, 55, 60, 55, 59, 56, 56, 49, 54, 65, 66, 51, 44, 58, 60, 54, 52, 57, 61, 62, 60, 60, 67. Fill the dot plot, and mark the mean and median. How would you describe the heights of these palm trees? Can you think of quicker ways to find the mean? How many trees are shorter than the average height?

Solution

Given: Heights of 29 date-palm trees (in feet): 50, 45, 43, 52, 61, 63, 46, 55, 60, 55, 59, 56, 56, 49, 54, 65, 66, 51, 44, 58, 60, 54, 52, 57, 61, 62, 60, 60, 67. Number of trees (N) = 29.

To Find: Mean, median, description of data, quicker way to find mean, and number of trees shorter than average.

Solution:

1. Calculating the Mean: Sum of heights =

50+45+43+52+61+63+46+55+60+55+59+56+56+49+54+65+66+51+44+58+60+54+52+57+61+62+60+60+67 = 1611

Mean =

\frac{1611}{29} \approx 55.55

feet.

2. Calculating the Median: First, sort the data: 43, 44, 45, 46, 49, 50, 51, 52, 52, 54, 54, 55, 55, 56, 56, 57, 58, 59, 60, 60, 60, 60, 61, 61, 62, 63, 65, 66, 67 Since N=29 (odd), the median is the middle value, which is the

(\frac{29+1}{2})^{th} = 15^{th}

value. The 15th value is 56. Median = 56 feet.

3. Dot Plot and Marking:

A dot plot would be drawn with a horizontal axis from about 40 to 70. Dots would be placed above each height value corresponding to its frequency (e.g., four dots above 60).
The mean (~55.55) would be marked with a symbol between 55 and 56.
The median (56) would be marked with another symbol at 56.

4. Description of Heights: The heights of the palm trees range from 43 feet to 67 feet. The data is clustered around the mid-50s and 60. The mean (55.55 ft) and median (56 ft) are very close, which suggests the data is roughly symmetric with no significant outliers. The most frequent height (mode) is 60 feet.

5. Quicker ways to find the mean (Assumed Mean Method): Instead of summing all the large numbers, we can assume a mean (e.g., A = 55) and find the average of the deviations from this assumed mean.

Calculate deviations (d = x - A) for each value: e.g., for 50, d= -5; for 61, d= +6.
Sum all deviations: $\sum d$ .
Calculate the actual mean: Mean = $A + \frac{\sum d}{N}$ . This method simplifies calculations if done manually.

6. Number of trees shorter than the average height: The average height is ~55.55 feet. We need to count how many trees have a height less than 55.55. From the sorted list: 43, 44, 45, 46, 49, 50, 51, 52, 52, 54, 54, 55, 55. There are 13 trees shorter than 55.55 feet.

Final Answer:

Mean: ~55.55 feet
Median: 56 feet
Description: The tree heights are fairly symmetrically distributed around a central value in the mid-50s, ranging from 43 to 67 feet.
Quicker mean calculation: Assumed Mean Method.
Trees shorter than average: 13 trees.

Q4Figure it Out (Section 5.2, Part 2)

The daily water usage from a tap was measured. The usage in liters for the first few days are: 5.6, 8, 3.09, 12.9, 6.5, 12.1, 11.3, 20.5, 7.4.

(a)

Can the mean or median daily usage lie between 25 and 30? Justify your claim using the meaning of mean and median.

(b)

Can the mean or median be lesser than the minimum value or greater than the maximum value in a data?

Solution

Given: Data on daily water usage (in liters): 5.6, 8, 3.09, 12.9, 6.5, 12.1, 11.3, 20.5, 7.4.

(a) Can the mean or median daily usage lie between 25 and 30?

Justification: First, let's identify the minimum and maximum values in the data set. Minimum value = 3.09 liters. Maximum value = 20.5 liters.

Mean: The mean (average) is calculated by summing all values and dividing by the count. Since all data values are between 3.09 and 20.5, their sum, when divided by the number of values, must also result in a number within this range. The mean is a measure of central tendency and will always lie between the minimum and maximum values of the data set.
Median: The median is the middle value of a sorted data set. When we sort the data, the median will be one of the values from the set (if N is odd) or the average of two values from the set (if N is even). In either case, the median must also lie between the minimum and maximum values.

Since the maximum value in the data is 20.5, neither the mean nor the median can be greater than 20.5. Therefore, they cannot lie in the range of 25 to 30.

Answer (a): No, the mean or median cannot lie between 25 and 30 because both measures of central tendency must fall within the range of the data, and the maximum value is 20.5.

(b) Can the mean or median be lesser than the minimum value or greater than the maximum value in a data?

Justification: Based on the reasoning in part (a):

The mean is the 'balancing point' of the data and is influenced by all data points. It cannot be smaller than the smallest value or larger than the largest value.
The median is the physical center of the sorted data. By definition, it cannot be outside the range of the data.

Answer (b): No, the mean and median can never be lesser than the minimum value or greater than the maximum value in a data set. They are always bounded by the range of the data.

Final Answer: (a) No. Both mean and median must lie within the range of the data [3.09, 20.5], so they cannot be between 25 and 30. (b) No. The mean and median are always greater than or equal to the minimum value and less than or equal to the maximum value of the data set.

Q2Figure it Out (Section 5.3)

Preyashi asked her students 'If you were to get a super power to become aquatic (water-borne), aerial (air-borne), or spaceborne which one would you choose?'. The responses are shown below. Some chose none. Draw a double-bar graph comparing how both grades chose each option. Choose an appropriate scale. Grade 5: w, a, a, a, w, n, s, a, n, w, a, a, a, a, a, w, w, s, a, a, n, w, a, a, n Grade 9: n, w, s, a, s, w, s, s, a, a, w, s, s, a, s, a, n, w, s, s, a, w, a, w, a

Solution

Given: Responses from Grade 5 and Grade 9 students. w = aquatic, a = aerial, s = spaceborne, n = none.

To Do: Draw a double-bar graph.

Step 1: Tally the data for each grade.

For Grade 5 (Total 25 students):

Aquatic (w): 6
Aerial (a): 12
Spaceborne (s): 2
None (n): 5 (Check: $6+12+2+5 = 25$ )

For Grade 9 (Total 25 students):

Aquatic (w): 6
Aerial (a): 7
Spaceborne (s): 9
None (n): 3 (Check: $6+7+9+3 = 25$ )

Step 2: Plan the double-bar graph.

Title: Superpower Choices of Grade 5 and Grade 9 Students.
Horizontal Axis (X-axis): Categories of Superpowers (Aquatic, Aerial, Spaceborne, None).
Vertical Axis (Y-axis): Number of Students.
Scale: The maximum frequency is 12. A suitable scale for the Y-axis would be 1 unit = 2 students. The markings would be 0, 2, 4, 6, 8, 10, 12, 14.
Legend: We need two different colors or patterns for the bars. For example:
- Blue bar = Grade 5
- Green bar = Grade 9

Step 3: Describe the bars to be drawn.

For the 'Aquatic' category:
- Draw a blue bar up to the height of 6 for Grade 5.
- Draw a green bar next to it, also up to the height of 6, for Grade 9.
For the 'Aerial' category:
- Draw a blue bar up to the height of 12 for Grade 5.
- Draw a green bar next to it up to the height of 7 for Grade 9.
For the 'Spaceborne' category:
- Draw a blue bar up to the height of 2 for Grade 5.
- Draw a green bar next to it up to the height of 9 for Grade 9.
For the 'None' category:
- Draw a blue bar up to the height of 5 for Grade 5.
- Draw a green bar next to it up to the height of 3 for Grade 9.

This description outlines all the necessary components and values to create the required double-bar graph.

Q3Figure it Out (Section 5.3)

The temperature variation over two days in different months in Jodhpur, Rajasthan, is given below. Draw a double-bar graph. Use the scale 1 unit = $4^{\circ} \mathrm{C}$ . Can you guess which two months these days might belong to?
12 am 3 am 6 am 9 am 12 pm 3 pm 6 pm 9 pm
Day 1 $20^{\circ} \mathrm{C}$ $18^{\circ} \mathrm{C}$ $16^{\circ} \mathrm{C}$ $20^{\circ} \mathrm{C}$ $26^{\circ} \mathrm{C}$ $34^{\circ} \mathrm{C}$ $30^{\circ} \mathrm{C}$ $24^{\circ} \mathrm{C}$
Day 2 $37^{\circ} \mathrm{C}$ $34^{\circ} \mathrm{C}$ $30^{\circ} \mathrm{C}$ $33^{\circ} \mathrm{C}$ $37^{\circ} \mathrm{C}$ $43^{\circ} \mathrm{C}$ $42^{\circ} \mathrm{C}$ $39^{\circ} \mathrm{C}$

	12 am	3 am	6 am	9 am	12 pm	3 pm	6 pm	9 pm
Day 1	$20^{\circ} \mathrm{C}$	$18^{\circ} \mathrm{C}$	$16^{\circ} \mathrm{C}$	$20^{\circ} \mathrm{C}$	$26^{\circ} \mathrm{C}$	$34^{\circ} \mathrm{C}$	$30^{\circ} \mathrm{C}$	$24^{\circ} \mathrm{C}$
Day 2	$37^{\circ} \mathrm{C}$	$34^{\circ} \mathrm{C}$	$30^{\circ} \mathrm{C}$	$33^{\circ} \mathrm{C}$	$37^{\circ} \mathrm{C}$	$43^{\circ} \mathrm{C}$	$42^{\circ} \mathrm{C}$	$39^{\circ} \mathrm{C}$

Solution

Given: Temperature data for two different days at various times.

To Do:

Draw a double-bar graph with the specified scale.
Guess the months for Day 1 and Day 2.

Part 1: Drawing the Double-Bar Graph

Title: Temperature Variation in Jodhpur on Two Different Days.
Horizontal Axis (X-axis): Time of Day (12 am, 3 am, 6 am, 9 am, 12 pm, 3 pm, 6 pm, 9 pm).
Vertical Axis (Y-axis): Temperature in degrees Celsius ( $^{\circ} \mathrm{C}$ ).
Scale: 1 unit = $4^{\circ} \mathrm{C}$ . The Y-axis markings will be 0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44.
Legend:
- Bar 1 (e.g., Blue) = Day 1
- Bar 2 (e.g., Red) = Day 2

Description of the bars to be drawn: For each time slot on the X-axis, two bars will be drawn side-by-side.

At 12 am: Blue bar to height 20, Red bar to height 37.
At 3 am: Blue bar to height 18, Red bar to height 34.
At 6 am: Blue bar to height 16, Red bar to height 30.
At 9 am: Blue bar to height 20, Red bar to height 33.
At 12 pm: Blue bar to height 26, Red bar to height 37.
At 3 pm: Blue bar to height 34, Red bar to height 43.
At 6 pm: Blue bar to height 30, Red bar to height 42.
At 9 pm: Blue bar to height 24, Red bar to height 39.

Part 2: Guessing the Months

Analysis of Day 1: The temperatures range from a low of $16^{\circ} \mathrm{C}$ to a high of $34^{\circ} \mathrm{C}$ . This represents a pleasant to warm day. In a desert climate like Jodhpur, these temperatures are typical of spring (March/April) or autumn (October/November).
Analysis of Day 2: The temperatures are much higher, ranging from a low of $30^{\circ} \mathrm{C}$ to a scorching high of $43^{\circ} \mathrm{C}$ . This is indicative of the peak summer season in Rajasthan.

Guess:

Day 1 could be a day in October or March.
Day 2 could be a day in May or June, which are the hottest months.

Final Answer: A double-bar graph should be drawn as described above. Day 1 likely belongs to a month like October, while Day 2 likely belongs to a hot summer month like May or June.

Q1Figure it Out (Section 5.4)

The dot plots below show the distribution of the number of pockets on clothing for a group of boys and for a group of girls. Based on the dot plots, which of the following statements are true?

(a)

The data varies more for the boys than for the girls.

(b)

The median number of pockets for the boys is more than that for the girls.

(c)

The mean number of pockets for the girls is more than that for the boys.

(d)

The maximum number of pockets for boys is greater than that for the girls.

Solution

Analysis of the Dot Plots (as described in the text):

For Boys:

The data points are spread out from 0 to 12 pockets.
The range is $12 - 0 = 12$ .

For Girls:

The data points are clustered at the lower end, from 0 to 6 pockets.
The range is $6 - 0 = 6$ .

Let's evaluate each statement:

(a) The data varies more for the boys than for the girls.

The range for boys (12) is larger than the range for girls (6). The data for boys is more spread out.
This statement is TRUE.

(b) The median number of pockets for the boys is more than that for the girls.

For girls, the data is heavily clustered at the lower numbers. The median will be a low number (likely around 2 or 3).
For boys, the data is more spread out towards higher numbers. The median will be higher than the girls' median (likely around 4 or 5).
This statement is TRUE.

(c) The mean number of pockets for the girls is more than that for the boys.

The girls' data has lower values overall compared to the boys' data. Therefore, the sum of values for girls will be proportionally lower, leading to a lower mean.
The boys' data has higher values, which will result in a higher mean.
This statement is FALSE.

(d) The maximum number of pockets for boys is greater than that for the girls.

The maximum number of pockets for boys is 12.
The maximum number of pockets for girls is 6.
12 is greater than 6.
This statement is TRUE.

Final Answer: Statements (a), (b), and (d) are true.

Q2Figure it Out (Section 5.4)

The following table shows the points scored by each player in four games: | Player | Game 1 | Game 2 | Game 3 | Game 4 | |---|---|---|---|---| | A | 14 | 16 | 10 | 10 | | B | 0 | 8 | 6 | 4 | | C | 8 | 11 | Did not play | 13 | Now answer the following questions:

(a)

Find the average number of points scored per game by A.

(b)

To find the mean number of points scored per game by C, would you divide the total points by 3 or by 4? Why? What about B?

(c)

Who is the best performer?

Solution

Given: A table of points scored by three players in four games.

(a) Find the average number of points scored per game by A.

Player A's scores: 14, 16, 10, 10
Number of games played: 4
Total points: $14 + 16 + 10 + 10 = 50$
Average for A: $\frac{50}{4} = 12.5$ points per game.

(b) To find the mean number of points scored per game by C, would you divide the total points by 3 or by 4? Why? What about B?

For Player C: Player C played in Game 1, Game 2, and Game 4. He did not play in Game 3. To find the average points per game played, we should only consider the games he participated in. Therefore, we should divide his total points by 3.
- Reason: The average should reflect his performance in the games he actually played.
For Player B: Player B played in all four games. He scored 0 points in Game 1, but he did play. Therefore, to find his average, we must include this game in the count. We would divide his total points by 4.
- Reason: Scoring 0 is part of the performance data for the games played.

(c) Who is the best performer?

To determine the best performer, we should compare their average scores per game played.

Average for A: 12.5 points per game.
Average for B:
- Total points: $0 + 8 + 6 + 4 = 18$
- Games played: 4
- Average: $\frac{18}{4} = 4.5$ points per game.
Average for C:
- Total points: $8 + 11 + 13 = 32$
- Games played: 3
- Average: $\frac{32}{3} \approx 10.67$ points per game.

Comparison:

Player A's average: 12.5
Player C's average: ~10.67
Player B's average: 4.5

Player A has the highest average number of points per game.

Final Answer: (a) The average for player A is 12.5 points per game. (b) For C, you would divide by 3 because he only played 3 games. For B, you would divide by 4 because he played all 4 games. (c) Player A is the best performer based on the average score.

Q5Figure it Out (Section 5.4)

Consider a group of 17 students with the following heights (in cm): 106, 110, 123, 125, 117, 120, 112, 115, 110, 120, 115, 102, 115, 115, 109, 115, 101. The sports teacher wants to divide the class into two groups so that each group has an equal number of students: one group has students with height less than a particular height and the other group has students with heights greater than the particular height. Suggest a way to do this. Can you guess the age of these students based on the tabular data in the 'Telling Tall Tales' section?

Solution

Given: Heights of 17 students (in cm): 106, 110, 123, 125, 117, 120, 112, 115, 110, 120, 115, 102, 115, 115, 109, 115, 101.

Part 1: Suggest a way to divide the class into two equal groups.

To divide the students into two groups of equal size based on height, the best measure to use is the median. The median is the central value that separates the higher half of a data set from the lower half.

Step 1: Sort the data. 101, 102, 106, 109, 110, 110, 112, 115, 115, 115, 115, 115, 117, 120, 120, 123, 125

Step 2: Find the median. There are 17 students (an odd number). The median is the middle value, which is the

(\frac{17+1}{2})^{th} = 9^{th}

value. The 9th value in the sorted list is 115 cm.

Step 3: Form the groups. The teacher can use the median height of 115 cm to divide the class. However, there are multiple students with the height of 115 cm. To create two groups of equal size (8 students each), with one student left in the middle, the teacher can do the following:

Group 1 (Shorter): The 8 students with heights less than or equal to the 8th value (115 cm). Heights: 101, 102, 106, 109, 110, 110, 112, 115.
Group 2 (Taller): The 8 students with heights greater than or equal to the 10th value (115 cm). Heights: 115, 115, 115, 117, 120, 120, 123, 125.
The student with the median height (the 9th student, 115 cm) can be a group leader or be assigned to either group to balance activities.

A simpler way:

Group 1: Students with heights less than 115 cm. (7 students)
Group 2: Students with heights greater than 115 cm. (5 students) This does not create equal groups. The best way is to use the median's position. The 8 students below the median position and the 8 students above the median position form the two groups.

Part 2: Guess the age of the students.

Let's find the average height of this group. Sum of heights = 1925 cm Mean height =

\frac{1925}{17} \approx 113.2

cm. The median height is 115 cm.

Referring to the 'Telling Tall Tales' table for the year 2019:

Age 5: ~107 cm
Age 6: ~113 cm
Age 7: ~118 cm

The average height of the students (~113.2 cm) is very close to the average height of 6-year-old children in India in 2019 (113.1 cm for boys, 112.9 cm for girls).

Final Answer:

The teacher should find the median height, which is 115 cm. The 8 students with heights below this median position form one group, and the 8 students with heights above it form the other group.
Based on the average height of approximately 113.2 cm, the students are likely around 6 years old.

Q1In-text Questions (Know Your Onions!)

Find the average price of onions at Yahapur and Wahapur.

Solution

Given: Monthly onion prices (in rupees per kg) for Yahapur and Wahapur.

Yahapur prices: 25, 24, 26, 28, 30, 35, 39, 43, 49, 56, 59, 44 Wahapur prices: 19, 17, 23, 30, 38, 35, 42, 39, 53, 60, 52, 42

To Find: The average price of onions at each location.

Formula:

\text{Average (Mean)} = \frac{\text{Sum of all values}}{\text{Number of values}}

Solution:

For Yahapur: Number of months = 12 Sum of prices =

25 + 24 + 26 + 28 + 30 + 35 + 39 + 43 + 49 + 56 + 59 + 44 = 458

Average price at Yahapur =

\frac{458}{12} \approx 38.17

For Wahapur: Number of months = 12 Sum of prices =

19 + 17 + 23 + 30 + 38 + 35 + 42 + 39 + 53 + 60 + 52 + 42 = 450

Average price at Wahapur =

\frac{450}{12} = 37.5

Final Answer: The average price of onions at Yahapur is approximately ₹38.17 per kg, and at Wahapur it is ₹37.50 per kg.

Q1In-text Questions (Outliers and Medians)

Find the average height of each family. Can we say that Yaangba's family is taller than Poovizhi's family?

Solution

Given: Heights of Yaangba's family (6 members): 169 cm, 173 cm, 155 cm, 165 cm, 160 cm, 164 cm. Heights of Poovizhi's family (5 members): 170 cm, 173 cm, 165 cm, 118 cm, 175 cm.

To Find: The average height of each family and compare them.

Solution:

Average height of Yaangba's family: Sum of heights =

169 + 173 + 155 + 165 + 160 + 164 = 986

cm Number of members = 6 Average height =

\frac{986}{6} \approx 164.33

cm

Average height of Poovizhi's family: Sum of heights =

170 + 173 + 165 + 118 + 175 = 801

cm Number of members = 5 Average height =

\frac{801}{5} = 160.2

cm

Comparison: The average height of Yaangba's family (164.33 cm) is greater than the average height of Poovizhi's family (160.2 cm). Based on the average, we can say that Yaangba's family is taller than Poovizhi's family. However, it is important to note that the average for Poovizhi's family is skewed by one very low value (118 cm), which is an outlier.

Final Answer: The average height of Yaangba's family is approximately 164.33 cm. The average height of Poovizhi's family is 160.2 cm. Based on the average, Yaangba's family is taller.

Q2In-text Questions (Outliers and Medians)

Find the mean and median in Poovizhi's data without the outlier value 118. What change do you notice?

Solution

Given: Original data for Poovizhi's family: 170 cm, 173 cm, 165 cm, 118 cm, 175 cm. Outlier value: 118 cm.

To Find: The mean and median of the data without the outlier and describe the change.

Solution:

Data without the outlier: 170, 173, 165, 175. Number of values = 4.

Calculating the new mean: Sum of heights =

170 + 173 + 165 + 175 = 683

cm New mean =

\frac{683}{4} = 170.75

cm

Calculating the new median: First, sort the data: 165, 170, 173, 175. Since there are an even number of values (4), the median is the average of the two middle values (170 and 173). New median =

\frac{170 + 173}{2} = \frac{343}{2} = 171.5

cm

Comparison and Change:

Original Mean: 160.2 cm
New Mean: 170.75 cm
Original Median: 170 cm (sorted data: 118, 165, 170, 173, 175)
New Median: 171.5 cm

Noticeable Changes:

The mean increased significantly from 160.2 cm to 170.75 cm. This shows that the mean is very sensitive to outliers.
The median changed only slightly from 170 cm to 171.5 cm. This shows that the median is much more resistant to the effect of outliers.

Final Answer: Without the outlier, the new mean is 170.75 cm and the new median is 171.5 cm. The mean increased substantially, while the median changed very little, demonstrating the median's robustness against outliers.

Q3In-text Questions (Outliers and Medians)

After the summer vacation, a class teacher asked his class how many short stories they had read. Each student answered the number of stories read on a piece of paper, as shown below. Find the mean and median number of short stories read. Before calculating them, can you guess whether the mean will be less than or greater than the median?

Solution

Given: The data for the number of short stories read is given in the image. Let's list the values from the image: 2, 3, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 8, 8, 9, 10, 12, 40. Total number of students = 24.

To Find: The mean and median of the data. Also, to guess if the mean will be less than or greater than the median.

Guess: The data includes a very high value, 40, which is significantly different from the rest of the data. This is an outlier. An outlier on the higher end will pull the mean towards it. The median, being the middle value, will be less affected. Therefore, it is expected that the mean will be greater than the median.

Solution:

Calculating the Mean: Sum of all values =

2 + 3(2) + 4(3) + 5(4) + 6(5) + 7(3) + 8(2) + 9 + 10 + 12 + 40

Sum =

2 + 6 + 12 + 20 + 30 + 21 + 16 + 9 + 10 + 12 + 40 = 178

Number of students = 24 Mean =

\frac{178}{24} \approx 7.42

Calculating the Median: The data is already sorted. There are 24 values (an even number). The median is the average of the 12th and 13th values. The 12th value is 6. The 13th value is 6. Median =

\frac{6 + 6}{2} = 6

Comparison: The mean is approximately 7.42, and the median is 6. As guessed, the mean is greater than the median due to the high outlier (40).

Final Answer: The mean number of short stories read is approximately 7.42. The median number of short stories read is 6. The guess was correct; the mean is greater than the median.

Q4In-text Questions (Outliers and Medians)

Which of the values would you consider an outlier?

Solution

Given: The data for the number of short stories read: 2, 3, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 8, 8, 9, 10, 12, 40.

Analysis: An outlier is a value that significantly deviates from the rest of the values in the data set. Most of the data points are clustered between 2 and 12. The value 40 is much larger than all the other values. The next highest value is 12, and the difference between 40 and 12 is 28, which is a very large gap compared to the spread of the rest of the data.

Final Answer: The value 40 would be considered an outlier.

Q5In-text Questions (Outliers and Medians)

Find the mean and median in the absence of the outlier. What change do you notice?

Solution

Given: The data for short stories read, with the outlier 40 removed. New data set: 2, 3, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 8, 8, 9, 10, 12. Number of values = 23.

To Find: The new mean and median, and to describe the change.

Solution:

Calculating the new mean: Original sum = 178. Outlier = 40. New sum =

178 - 40 = 138

New number of values = 23 New mean =

\frac{138}{23} = 6

Calculating the new median: There are now 23 values (an odd number). The median is the middle value, which is the

(\frac{23+1}{2})^{th} = 12^{th}

value. The 12th value in the sorted list is 6. New median = 6

Comparison and Change:

Original Mean: ~7.42. New Mean: 6.
Original Median: 6. New Median: 6.

Noticeable Changes:

The mean decreased significantly from ~7.42 to 6. Removing the high outlier brought the mean down to be more representative of the central cluster of the data.
The median did not change at all. It remained 6. This again shows that the median is not affected by outliers at the extremes of the data.

Final Answer: Without the outlier, the new mean is 6 and the new median is 6. The mean decreased to become equal to the median, which is now a better representation of the center of the data.

Q6In-text Questions (Outliers and Medians)

Mark the data, the mean, and the median on the dot plot below.

Solution

This question requires marking values on a dot plot. Since I cannot draw, I will state the values to be marked.

Data: The original data for short stories read is: 2, 3, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 8, 8, 9, 10, 12, 40.

Values Calculated:

Mean: ~7.42
Median: 6

Instructions for Marking the Dot Plot:

Mark the data points: For each number on the horizontal axis, place a dot vertically for every time that number appears in the data set.
- One dot above 2.
- Two dots above 3.
- Three dots above 4.
- Four dots above 5.
- Five dots above 6.
- Three dots above 7.
- Two dots above 8.
- One dot above 9.
- One dot above 10.
- One dot above 12.
- One dot above 40.
Mark the Median: The median is 6. Draw a vertical line or a special marker (like a triangle or an arrow) on the horizontal axis at the position for 6.
Mark the Mean: The mean is approximately 7.42. Draw another vertical line or a different marker on the horizontal axis at the position for 7.42 (which would be between 7 and 8, closer to 7).

Q1In-text Questions (Section 5.1)

Which of the following are statistical questions?

(a)

What is the price of a tennis ball in India?

(b)

How old are the dogs that live on this street?

(c)

What fraction of the students in your class like walking up a hill?

(d)

Do you like reading?

(e) Approximately how many bricks are in this wall?

(f) Who was the best bowler in the match yesterday?

(g) What was the rainfall pattern in Barmer last year?

Solution

A statistical question is one that can be answered by collecting data that is expected to vary.

(a) Not a statistical question. The price of a tennis ball can vary from shop to shop and brand to brand. If the question was phrased as "What is the typical price of a tennis ball in India?", it would be a statistical question requiring data collection from various sources. As written, it could be interpreted as asking for a single price.

(b) Statistical question. To answer this, one would need to collect the age of each dog on the street. The ages will likely vary, so we would be collecting data to find a summary (like an average age or a range of ages).

(c) Statistical question. This requires collecting data from every student in the class (do they like it or not?). The answers will vary, and the result will be a fraction or percentage, which is a statistical summary.

(d) Not a statistical question. This question is directed at a single person ("you") and expects a single answer (yes or no). It does not require collecting data that varies.

(e) Not a statistical question. Although it involves a number, it refers to a single, fixed quantity. One would estimate or count it, but there is no data variability to analyze. It is an estimation question.

(f) Not a statistical question. The term "best" is subjective unless defined by a specific statistic (e.g., most wickets taken, best economy rate). If a specific metric is used, it becomes a question of fact from the match records, not a statistical analysis of variable data.

(g) Statistical question. This requires collecting rainfall data for Barmer over the last year. The data will vary daily or monthly, and analyzing this data would reveal a pattern (e.g., average rainfall, seasonal trends).

Connecting the Dots...Class 7 Mathematics NCERT Solutions

Shreyas is playing with a bat and a ballbut not cricket. He counts the number of times he can bounce the ball on the bat before it falls to the ground. The data for 8 attempts is 6, 2, 9, 5, 4, 6, 3, 5. Calculate the average number of bounces of the ball that Shreyas is able to make with his bat.

Solution

Try the activity above on your own. Collect data for 7 or more attempts and find the average.

Solution

Identify a flowering plant in your neighbourhood. Track the number of flowers that bloom every day over a week during its flowering season. What is the average number of flowers that bloomed per day?

Solution

Two friends are training to run a 100 m race. Their running times over the past week are given in seconds— Nikhil: 17, 18, 17, 16, 19, 17,18 ; Sunil: 20, 18, 18, 17, 16, 16, 17. Who on average ran quicker?

Solution

The enrolment in a school during six consecutive years was as follows: 1555, 1670, 1750, 2013, 2040, 2126. Find the mean enrolment in the school during this period.

Solution

Find the median of onion prices in Yahapur and Wahapur.

Solution

Sanskruti asked her class how many domestic animals and pets each had at home. Some of the students were absent. The data values are 0, 1, 0, 4, 8, 0, 0, 2, 1, 1, 5, 3, 4, 0, 0, -, 10, 25, 2, -, 2, 4. Find the mean and median. How would you describe this data?

Solution

Solution

The daily water usage from a tap was measured. The usage in liters for the first few days are: 5.6, 8, 3.09, 12.9, 6.5, 12.1, 11.3, 20.5, 7.4.

Solution

Solution

Solution

The dot plots below show the distribution of the number of pockets on clothing for a group of boys and for a group of girls. Based on the dot plots, which of the following statements are true?

Solution

The following table shows the points scored by each player in four games: | Player | Game 1 | Game 2 | Game 3 | Game 4 | |---|---|---|---|---| | A | 14 | 16 | 10 | 10 | | B | 0 | 8 | 6 | 4 | | C | 8 | 11 | Did not play | 13 | Now answer the following questions:

Solution

Solution

Find the average price of onions at Yahapur and Wahapur.

Solution

Find the average height of each family. Can we say that Yaangba's family is taller than Poovizhi's family?

Solution

Find the mean and median in Poovizhi's data without the outlier value 118. What change do you notice?

Solution

Solution

Which of the values would you consider an outlier?

Solution

Find the mean and median in the absence of the outlier. What change do you notice?

Solution

Mark the data, the mean, and the median on the dot plot below.

Solution

Which of the following are statistical questions?

Solution