Chapter 5 Connecting the Dots... (Class 7 - Latest Maths NCERT (Ganita Prakash II) Solutions)

Given:

Number of bounces recorded in $8$ attempts: $6, 2, 9, 5, 4, 6, 3, 5$

Total number of attempts = $8$

To Find:

The average number of bounces.

Solution:

To find the average, we use the formula for Arithmetic Mean:

First, we calculate the sum of the number of bounces:

$\text{Sum} = 6 + 2 + 9 + 5 + 4 + 6 + 3 + 5$

$\text{Sum} = 40$

Now, we divide the sum by the total number of attempts:

$\text{Average} = \frac{40}{8}$

$\text{Average} = \frac{\cancel{40}^{5}}{\cancel{8}_{1}}$

$\text{Average} = 5$

The average number of bounces Shreyas is able to make with his bat is $5$.

Activity: Perform the ball-bouncing activity and record the results for at least $7$ attempts.

Sample Data:

Let us assume the data collected for $7$ attempts is as follows:

Solution for Sample Data:

Given:

Observations: $12, 8, 15, 10, 7, 14, 11$

Number of attempts = $7$

To Find:

The average of the sample data.

Calculation of Sum:

$\text{Sum} = 12 + 8 + 15 + 10 + 7 + 14 + 11$

$\text{Sum} = 77$

Calculation of Average:

$\text{Average} = \frac{77}{7}$

$\text{Average} = \frac{\cancel{77}^{11}}{\cancel{7}_{1}}$

$\text{Average} = 11$

For the sample activity data, the average number of bounces is $11$.

Note: You should record your own actual bounces in a similar table and calculate your personal average using the same method.

This is a practical activity. Below is a sample observation of a Hibiscus plant in an Indian garden recorded over one week.

Given:

The number of flowers blooming each day for $7$ days is recorded in the table below:

To Find:

Average number of flowers bloomed per day.

Solution:

The average is calculated as the sum of all flowers divided by the total number of days.

Now, applying the average formula:

In this sample observation, the average number of flowers that bloomed per day is $4$.

Given:

Nikhil's times (in seconds): $17, 18, 17, 16, 19, 17, 18$

Sunil's times (in seconds): $20, 18, 18, 17, 16, 16, 17$

Total number of days for each = $7$

To Find:

Who is quicker on average (the person with the lower average time is quicker).

Solution:

Step 1: Calculate Nikhil's Average Time

$\text{Sum for Nikhil} = 17 + 18 + 17 + 16 + 19 + 17 + 18$

$\text{Sum for Nikhil} = 122 \text{ seconds}$

Step 2: Calculate Sunil's Average Time

$\text{Sum for Sunil} = 20 + 18 + 18 + 17 + 16 + 16 + 17$

$\text{Sum for Sunil} = 122 \text{ seconds}$

Comparison:

Comparing the average times of both friends:

Since both Nikhil and Sunil have the same average time of approximately $17.43$ seconds, on average, both ran with the same speed. Neither was quicker than the other on average during this week.

Given:

Enrolment data for $6$ years: $1555, 1670, 1750, 2013, 2040, 2126$

Total number of years = $6$

To Find:

Mean (Average) enrolment.

Solution:

We calculate the mean using the formula:

First, find the sum of all enrolments:

$\text{Sum} = 1555 + 1670 + 1750 + 2013 + 2040 + 2126$

Now, divide the total sum by the number of years:

The mean enrolment of the school during this six-year period is $1859$ students.

Given:

The monthly onion prices for two locations, Yahapur and Wahapur, for 12 months.

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 Median price for both Yahapur and Wahapur.

Solution:

To find the median, we must first arrange the data in ascending order. Since the number of observations ($n$) is $12$ (which is an even number), the median will be the average of the $\left(\frac{n}{2}\right)^{th}$ and $\left(\frac{n}{2} + 1\right)^{th}$ terms.

1. Calculating Median for Yahapur:

Arranging Yahapur prices in ascending order:

$24, 25, 26, 28, 30, \mathbf{35}, \mathbf{39}, 43, 44, 49, 56, 59$

Here, the $6^{th}$ term is $35$ and the $7^{th}$ term is $39$.

The median price in Yahapur is $\textsf{₹} 37$.

2. Calculating Median for Wahapur:

Arranging Wahapur prices in ascending order:

$17, 19, 23, 30, 35, \mathbf{38}, \mathbf{39}, 42, 42, 52, 53, 60$

Here, the $6^{th}$ term is $38$ and the $7^{th}$ term is $39$.

The median price in Wahapur is $\textsf{₹} 38.5$.

Final Answer:

The median price for Yahapur is $\textsf{₹} 37$ and the median price for Wahapur is $\textsf{₹} 38.5$.

Given:

Data values (excluding missing values "—"): $0, 1, 0, 4, 8, 0, 0, 2, 1, 1, 5, 3, 4, 0, 0, 10, 25, 2, 2, 4$

Total number of observations ($n$) = $20$

To Find:

1. Mean, 2. Median, 3. Description of data.

Solution:

1. Calculating the Mean:

$\text{Sum} = 0 + 1 + 0 + 4 + 8 + 0 + 0 + 2 + 1 + 1 + 5 + 3 + 4 + 0 + 0 $$ + 10 + 25 + 2 + 2 + 4$

$\text{Sum} = 72$

$\text{Mean} = \frac{72}{20}$

$\text{Mean} = \frac{\cancel{72}^{3.6}}{\cancel{20}_{1}}$

$\text{Mean} = 3.6$

2. Calculating the Median:

First, we arrange the $20$ observations in ascending order:

$0, 0, 0, 0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 4, 4, 4, 5, 8, 10, 25$

For $n = 20$ (even), the median is the average of the $10^{th}$ and $11^{th}$ terms.

$10^{th} \text{ term} = 2$

$11^{th} \text{ term} = 2$

$\text{Median} = \frac{2 + 2}{2} = 2$

Description:

The mean ($3.6$) is higher than the median ($2$) because of the outlier $25$. Most students have $0$ to $2$ pets, but a few high values pull the average up. The data is skewed to the right.

Given:

Heights (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$

Total number of trees = $29$

To Find:

1. Dot plot, 2. Mean, 3. Median, 4. Trees shorter than average.

Solution:

1. Dot Plot:

To fill the dot plot, place a dot for every occurrence of a height on the number line provided in the question. For example, place $4$ dots vertically above the number $60$.

2. Calculating the Mean:

$\text{Sum of heights} = 1621$

$\text{Mean} = \frac{1621}{29} \approx 55.89 \text{ feet}$

Quicker way: A quicker way to find the mean is to use an assumed mean (like $55$) and calculate deviations, or group the heights by frequency before multiplying.

3. Calculating the Median:

Arrange the $29$ heights in ascending order:

$43, 44, 45, 46, 49, 50, 51, 52, 52, 54, 54, 55, 55, 56, \mathbf{56}, 57, 58, 59, 60, $$ 60, 60, 60, 61, 61, 62, 63, 65, 66, 67$

The median is the $(\frac{29+1}{2})^{th} = 15^{th}$ term.

$\text{Median} = 56 \text{ feet}$

4. Trees shorter than the average height:

The average height is approximately $55.89$ feet. We count how many trees have heights less than $55.89$ feet:

Heights: $43, 44, 45, 46, 49, 50, 51, 52, 52, 54, 54, 55, 55$

Total count = $13$ trees.

Given:

Daily water usage (in liters): $5.6, 8, 3.09, 12.9, 6.5, 12.1, 11.3, 20.5, 7.4$

Number of days ($n$) = $9$

Solution (a):

To determine if the mean or median can lie between $25$ and $30$, let us look at the range of the data.

Minimum value = $3.09$

Maximum value = $20.5$

The Mean (average) represents the central value of the data. Since all observations are less than or equal to $20.5$, their average cannot be greater than the largest value. Similarly, the Median is the middle value when the data is arranged in order ($3.09, 5.6, 6.5, 7.4, 8, 11.3, 12.1, 12.9, 20.5$). The median here is $8$.

Therefore, No, the mean or median cannot lie between $25$ and $30$ because both must lie within the range of the actual data collected (i.e., between $3.09$ and $20.5$).

Solution (b):

No, the mean or median can never be lesser than the minimum value or greater than the maximum value of the data set.

The mean and median are measures of central tendency. They always fall somewhere within the boundaries of the smallest and largest values in the data set.

Given:

Weights of Boys ($6$ babies): $3.5, 4.1, 2.6, 3.2, 3.4, 3.8$

Weights of Girls ($6$ babies): $4.0, 3.1, 3.4, 3.7, 2.5, 3.4$

To Find:

1. Fill the dot plot, 2. Calculate Mean and Median for both groups, 3. Analyse and compare the data.

Solution:

Step 1: Analysis of Boys' Weights

Arranging weights of boys in ascending order: $2.6, 3.2, 3.4, 3.5, 3.8, 4.1$

$\text{Sum for boys} = 2.6 + 3.2 + 3.4 + 3.5 + 3.8 + 4.1 = 20.6 \text{ kg}$

$\text{Mean weight of boys} = \frac{20.6}{6} \approx 3.43 \text{ kg}$

Since the number of observations is $6$ (even), the median is the average of the $3^{rd}$ and $4^{th}$ terms ($3.4$ and $3.5$):

$\text{Median weight of boys} = \frac{3.4 + 3.5}{2} = 3.45 \text{ kg}$

Step 2: Analysis of Girls' Weights

Arranging weights of girls in ascending order: $2.5, 3.1, 3.4, 3.4, 3.7, 4.0$

$\text{Sum for girls} = 4.0 + 3.1 + 3.4 + 3.7 + 2.5 + 3.4 = 20.1 \text{ kg}$

$\text{Mean weight of girls} = \frac{20.1}{6} = 3.35 \text{ kg}$

The middle terms are both $3.4$ ($3^{rd}$ and $4^{th}$ terms):

$\text{Median weight of girls} = \frac{3.4 + 3.4}{2} = 3.4 \text{ kg}$

Step 3: Dot Plot and Comparison

The dot plot is filled by placing dots at the respective positions for each value. Note that $3.4$ will have three dots (one from boys and two from girls).

Comparison Table:

Conclusion:

Based on the analysis, we observe that the mean weight of boys ($3.43 \text{ kg}$) is slightly higher than the mean weight of girls ($3.35 \text{ kg}$). Similarly, the median weight for boys is also higher. This indicates that, in this specific sample, the newborn boys are generally slightly heavier than the newborn girls.

Given:

Whole class: Mean $= 141.21 \text{ cm}$, Median $= 142.5 \text{ cm}$

Boys: Mean $= 142.05 \text{ cm}$, Median $= 143 \text{ cm}$

Girls: Mean $= 140.14 \text{ cm}$, Median $= 140 \text{ cm}$

Observations:

1. Comparison of Boys and Girls: The average height of boys ($142.05 \text{ cm}$) is higher than that of girls ($140.14 \text{ cm}$). Similarly, the median height for boys ($143 \text{ cm}$) is greater than that of girls ($140 \text{ cm}$).

2. Distribution in Dot Plots: The dot plot for boys shows that most of the data points are clustered between $142 \text{ cm}$ and $146 \text{ cm}$. In contrast, the dot plot for girls shows a wider spread of heights, ranging from $126 \text{ cm}$ to $158 \text{ cm}$.

3. Central Tendency: For the whole class and the boys' group, the median is slightly higher than the mean, which suggests that the data is slightly negatively skewed (more students have heights on the higher side). For girls, the mean and median are very close, indicating a more symmetrical distribution.

Inference:

From the dot plots and central tendency measures, we can infer that boys in this section are generally taller than girls. While the girls have some individuals who are very tall ($158 \text{ cm}$) or very short ($126 \text{ cm}$), the boys' heights are more consistently grouped around the $142$–$146 \text{ cm}$ range. The whole class average is balanced by the combination of these two groups.

Given:

Weights of Sumo wrestlers: $295.2, 250.7, 234.1, 221.0, 200.9$ (in kg)

Weights of Ballet dancers: $40.3, 37.6, 38.8, 45.5, 44.1, 48.2$ (in kg)

To Find:

How many times heavier a sumo wrestler is compared to a ballet dancer on average.

Solution:

Step 1: Calculate the mean weight of Sumo wrestlers

Total sum of weights $= 295.2 + 250.7 + 234.1 + 221.0 + 200.9 $$ = 1201.9 \text{ kg}$

Number of wrestlers $= 5$

$\text{Mean weight of Sumo wrestlers} = \frac{1201.9}{5} = 240.38 \text{ kg}$

Step 2: Calculate the mean weight of Ballet dancers

Total sum of weights $= 40.3 + 37.6 + 38.8 + 45.5 + 44.1 + 48.2 $$ = 254.5 \text{ kg}$

Number of dancers $= 6$

$\text{Mean weight of Ballet dancers} = \frac{254.5}{6} \approx 42.42 \text{ kg}$

Step 3: Comparison

To find how many times heavier a sumo wrestler is, we divide the average weight of a sumo wrestler by the average weight of a ballet dancer:

$\text{Ratio} = \frac{240.38}{42.42} \approx 5.66$

Rounding to the nearest whole number, we find that a sumo wrestler is approximately $6$ times heavier than a ballet dancer.

Yes, this infographic can be classified as a horizontal bar graph. In this graph, the length of each colored bar represents the maximum speed of a specific animal, allowing for an easy visual comparison between different species and environments (air, land, and water).

(a) Scale:

Looking at the horizontal axis, the markings increase in equal intervals of $16$.

The scale used in the graph is: $1$ unit length $= 16 \text{ km/h (kph)}$.

(b) Observations:

One of the most interesting facts in this infographic is about the Australian tiger beetle. Even though its absolute speed is low ($8 \text{ kph}$), it moves so fast relative to its body size ($120$ lengths per second) that its brain cannot process visual information, making it "run blind." I would like to explore how other small insects, like ants or bees, compare in terms of body lengths per second.

(c) Identifying Pairs:

We need to find two animals where $Speed_{A} \approx 2 \times Speed_{B}$.

1. Flying fish ($56 \text{ kph}$) and Sailfish ($109 \text{ kph}$): Since $56 \times 2 = 112$, the Sailfish is approximately twice as fast as the Flying fish.

2. Pronghorn antelope ($88 \text{ kph}$) and Spine-tailed swift ($170 \text{ kph}$): Since $88 \times 2 = 176$, the Spine-tailed swift is approximately twice as fast as the Pronghorn antelope.

(d) Aquatic Animal Comparison:

Speed of Sailfish $= 109 \text{ kph}$

Speed of Humpback whale $= 26 \text{ kph}$

Calculation: $26 \times 4 = 104$

Since $109$ is very close to $104$, yes, we can say that a sailfish is about $4$ times faster than a humpback whale.

To Find:

A double-bar graph comparing the superpower preferences of Grade 5 and Grade 9 students.

Step 1: Organizing Data (Frequency Count)

We count the occurrences for each category: Aerial ($a$), Aquatic ($w$), Spaceborne ($s$), and None ($n$).

Step 2: Representation on Graph

To construct the double-bar graph:

1. Scale: We can take $1 \text{ unit} = 2 \text{ students}$ on the vertical (Y) axis.

2. X-axis: Represent the four categories: Aerial, Aquatic, Spaceborne, and None.

3. Y-axis: Number of students, ranging from $0$ to $14$.

4. Bars: For each category, draw two side-by-side bars. Use different shading or colors (e.g., solid for Grade 5 and striped for Grade 9).

Given:

Temperature readings for Jodhpur at $8$ different time intervals for two separate days.

Scale required: $1 \text{ unit} = 4^\circ\text{C}$

To Find:

1. Double-bar graph representation.

2. Identification of the likely months for Day 1 and Day 2.

Solution:

To draw the double-bar graph, we represent the time intervals on the x-axis and the temperature on the y-axis.

For Day 1, the temperatures range from $16^\circ\text{C}$ to $34^\circ\text{C}$. For Day 2, the temperatures range from $30^\circ\text{C}$ to $43^\circ\text{C}$.

Graph Construction: On the y-axis, we mark points at intervals of $4$ ($0, 4, 8, 12, 16, \dots, 44$). For every time slot, two bars are drawn side-by-side: one for Day 1 and one for Day 2.

Month Identification:

Jodhpur is located in Rajasthan, which experiences extreme climatic variations.

1. Day 1: The temperatures are moderate, with a minimum of $16^\circ\text{C}$ and a maximum of $34^\circ\text{C}$. This indicates a transition period. It could likely be the month of February (end of winter) or November (beginning of winter).

2. Day 2: The temperatures are very high, with a minimum of $30^\circ\text{C}$ (even at night) and a peak of $43^\circ\text{C}$. This is characteristic of the peak summer season in Rajasthan. It most likely belongs to the month of May or June.

(a) Marking the bars for Gujarat and Delhi:

To accurately mark the bars for part (a), we first determine the value of the grid lines on the y-axis.

There are $5$ small horizontal grid lines between $0$ and $25000$. Thus, the value of each small grid line is:

Using this scale, we can calculate the heights for the bars of Gujarat and Delhi:

For Gujarat:

1. 2022 ($69000$): Since $70000$ is the $14^{th}$ small line ($14 \times 5000$), the bar for $2022$ should be drawn just slightly below the $14^{th}$ line.

2. 2023 ($89000$): Since $90000$ is the $18^{th}$ small line ($18 \times 5000$), the bar for $2023$ should be drawn just below the $18^{th}$ line.

3. 2024 ($78000$): This bar should be placed slightly above the $15^{th}$ line ($75000$) but below the $16^{th}$ line ($80000$).

For Delhi:

1. 2022 ($62000$): This bar should be drawn between the $12^{th}$ line ($60000$) and the $13^{th}$ line ($65000$).

2. 2023 ($74000$): This bar should end almost at the $15^{th}$ line ($75000$).

3. 2024 ($81000$): This bar should be drawn just above the $16^{th}$ line ($80000$).

(b) Organisation, Scale, and Patterns:

Organisation: The graph is a clustered-bar graph where data for three years is grouped together for each state along the x-axis.

Scale: The vertical scale (y-axis) is $1 \text{ unit} = 25000$ registrations, with sub-grid lines at every $5000$ registrations.

Patterns: Most states show a consistent increase in EV registrations over the three-year period, indicating a growing adoption of electric vehicles in India.

(c) Describing the change between $2022$ and $2024$:

Between $2022$ and $2024$, all states shown (Uttarakhand, West Bengal, Andhra Pradesh, Odisha, and Assam) experienced an upward trend. West Bengal and Andhra Pradesh show particularly significant growth, with registrations nearly doubling or more over the two-year span.

(d) Increase in Assam registrations ($2023$ vs $2022$):

From the graph:

Registrations in Assam ($2022$) $\approx 40000$

Registrations in Assam ($2023$) $\approx 60000$

Assam received approximately $20000$ more registrations in $2023$ compared to $2022$.

(e) Growth factor for West Bengal ($2022$ to $2024$):

From the graph:

Registrations in West Bengal ($2022$) $\approx 11000$

Registrations in West Bengal ($2024$) $\approx 44000$

$\text{Growth Factor} = 4$

The registrations in West Bengal increased by approximately $4$ times.

(f) Evaluating the statement about Uttarakhand:

The statement is correct. In the graph, the heights of the bars for Uttarakhand in $2022$, $2023$, and $2024$ are very close to each other. This indicates that the incremental increase in the number of new electric vehicles registered each year was minimal compared to the significant growth seen in other states.

Analysis:

Boys' Data: $3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6$ (Total $12$ students)

Girls' Data: $0, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 6$ (Total $13$ students)

Evaluating Statements:

(a) False: The girls' data ranges from $0$ to $6$ (variation $= 6$), while boys' data ranges from $3$ to $6$ (variation $= 3$). Thus, girls' data varies more.

(b) True: The median for boys (average of $6^{th}$ and $7^{th}$ term) is $\frac{4+5}{2} = 4.5$. The median for girls ($7^{th}$ term) is $4$. Since $4.5 > 4$, the median for boys is more.

(c) False: By observation, the boys' dots are concentrated towards higher numbers ($4$ and $5$) compared to girls ($3$ and $4$). The boys' mean will be higher.

(d) False: Both groups have a maximum of $6$ pockets.

Conclusion: The correct statement is (b).

(a) Average of Player A:

Points scored by A = $14, 16, 10, 10$

$\text{Total points} = 14 + 16 + 10 + 10 = 50$

$\text{Mean for A} = \frac{50}{4} = 12.5$

(b) Mean calculation for C and B:

For Player C, we would divide the total points by $3$ because he played only $3$ games. The game he "did not play" is not counted as an observation.

For Player B, we would divide the total points by $4$ because he played all $4$ games. Scoring $0$ points in Game 1 is still a valid score for a played game.

(c) Identifying the Best Performer:

Mean of A = $12.5$

Mean of B = $\frac{0 + 8 + 6 + 4}{4} = \frac{18}{4} = 4.5$

Mean of C = $\frac{8 + 11 + 13}{3} = \frac{32}{3} \approx 10.67$

Since $12.5 > 10.67 > 4.5$, Player A is the best performer.

Group 1 Analysis:

Scores: $85, 76, 90, 85, 39, 48, 56, 95, 81, 75$ ($10$ students)

$\text{Mean}_1 = \frac{85+76+90+85+39+48+56+95+81+75}{10} = \frac{730}{10} = 73$

Sorted: $39, 48, 56, 75, 76, 81, 85, 85, 90, 95$. Median is between $5^{th}$ and $6^{th}$ term: $\frac{76+81}{2} = 78.5$

Group 2 Analysis:

Scores: $68, 59, 73, 86, 47, 79, 90, 93, 86$ ($9$ students)

$\text{Mean}_2 = \frac{68+59+73+86+47+79+90+93+86}{9} = \frac{681}{9} \approx 75.67$

Sorted: $47, 59, 68, 73, 79, 86, 86, 90, 93$. Median ($5^{th}$ term) = $79$

Comparison:

Group 2 performed better on average (Mean $= 75.67$ vs $73$) and also had a higher median score ($79$ vs $78.5$). Group 1's performance was pulled down by the very low score of $39$.

Scale: Let $1 \text{ unit length} = 100 \text{ people}$.

Observations:

1. Cricket is the most popular sport both for watching ($1240$) and participating ($620$).

2. Athletics is the least popular sport in this colony.

3. In every category, the number of people who enjoy watching the sport is significantly higher than the number of people who participate in it.

4. Swimming and Basketball have an equal number of participants ($320$).

Given:

Heights of $17$ students in cm: $106, 110, 123, 125, 117, 120, 112, 115, 110, $$ 120, 115, 102, 115, 115, 109, 115, 101$

Number of students ($n$) = $17$

To Find:

1. A particular height (Median) to divide the class into two equal groups.

2. The heights of students in each group.

3. Guess the age of the students based on Indian growth data.

Solution:

To divide the students into two equal groups, we must first arrange the heights in ascending order to identify the central value or the Median.

Step 1: Ascending Order

$101, 102, 106, 109, 110, 110, 112, 115, 115, 115, 115, 115, 117, 120, 120, $$ 123, 125$

Step 2: Identifying the Median

Since the number of observations ($n = 17$) is odd, the median is the $\left(\frac{n+1}{2}\right)^{th}$ term.

Counting the $9^{th}$ term in our ordered list:

Group Formation:

With $17$ students, the teacher can form two groups of $8$ students each, using the $9^{th}$ student (the median) as the reference point.

Group 1: Students with height $\leq$ Median ($115\text{ cm}$)

Group 2: Students with height $\geq$ Median ($115\text{ cm}$)

Age Estimation:

A height range of $101\text{ cm}$ to $125\text{ cm}$ with a median of $115\text{ cm}$ corresponds to children in the age group of $6$ to $7$ years.

This suggests that these students are likely in Class 1 or Class 2.

This is an activity-based question. To describe the mean and median, you must first collect the heights of all students in your class (e.g., $30$ students).

Procedure:

1. Mean: Add the heights of all students and divide the sum by the total number of students. The mean represents the average height of the class.

2. Median: Arrange the heights in ascending order. The height of the middle student is the median. It represents the height such that half the class is shorter and half the class is taller than this value.

Sample Observation:

If a class has heights like $130, 132, 135, 135, 138, 140, 142 \text{ cm}$, the mean would be the average of these, and the median would be the middle value ($135 \text{ cm}$).

Dot Plot: You can create a dot plot by drawing a number line and placing a dot for each student's height above the corresponding number. This helps visualise if the heights are clustered together or spread out.

Given:

Section A: Total students = $30$ ($15$ boys and $15$ girls), Mean height = $154.2 \text{ cm}$.

Section B: Total students = $30$ ($15$ boys and $15$ girls).

Solution:

The mean height of a group depends entirely on the specific heights of the individuals within that group. Knowing the average height of one section does not provide any mathematical data about the heights of students in a different section.

Students in Section B could be taller, shorter, or of the same average height as Section A, but there is no rule stating they must be the same.

Therefore, the correct option is (d) The mean height of students in the other section cannot be determined.

Given:

An infographic bar graph showing the number of skyscrapers (buildings taller than $150\text{ m}$) in various global cities. The values for some cities are provided numerically (e.g., Hong Kong: $553$, Shenzhen: $367$, Dubai: $251$, Mumbai: $86$).

To Find:

1. Estimated number of skyscrapers for New York, Tokyo, and London.

2. Validity of statements regarding Mumbai's rank and the location of the world's tallest building.

Solution:

(a) Estimating values based on the bar lengths:

We can estimate the values by comparing the lengths of the bars for the requested cities with the lengths of the bars for which values are already provided.

New York: The bar for New York is between Shenzhen ($367$) and Dubai ($251$). Visually, it is roughly in the middle but slightly closer to the $300$ mark.
Estimated Value $\approx 300$

Tokyo: The bar for Tokyo is between Shanghai ($183$) and Kuala Lumpur ($154$).
Estimated Value $\approx 165$

London: London has the shortest bar in the list, positioned below Moscow ($46$).
Estimated Value $\approx 30$

(b) Evaluating the statements:

(i) Only $12$ cities have more skyscrapers than Mumbai.

By counting the cities listed above Mumbai in the chart: (1) Hong Kong, (2) Shenzhen, (3) New York, (4) Dubai, (5) Guangzhou, (6) Shanghai, (7) Tokyo, (8) Kuala Lumpur, (9) Chongqing, (10) Jakarta, (11) Bangkok, and (12) Singapore.

There are exactly $12$ cities with more skyscrapers than Mumbai according to this list.

Statement (i) is Valid.

(ii) Only $7$ cities have fewer skyscrapers than Mumbai.

By counting the cities listed below Mumbai in the chart: (1) Seoul, (2) Toronto, (3) Melbourne, (4) Miami, (5) Istanbul, (6) Moscow, and (7) London.

There are exactly $7$ cities with fewer skyscrapers than Mumbai according to this list.

Statement (ii) is Valid.

(iii) The tallest building in the world is in Hong Kong.

The graph shows the number of skyscrapers (count) in each city, not the height of individual buildings. While Hong Kong has the highest count of skyscrapers ($553$), this does not mean it has the single tallest building in the world (the tallest building, Burj Khalifa, is in Dubai).

Statement (iii) is Invalid.

Given:

A list of common stationery and personal objects to be estimated and then measured using a ruler or measuring tape.

To Find/To Do:

1. Record estimated and actual measured values for each object.

2. Calculate the Positive Difference for each observation.

3. Represent the data using a Double Bar Graph.

4. Calculate the Average Difference to check accuracy.

Solution:

Step 1: Data Collection

The student should first look at the object and write down an Estimate. After that, use a ruler to find the actual Measure. The Positive Difference is the absolute value of the subtraction between the two.

Step 2: Calculating Positive Difference

For each object, use the following formula:

Step 3: Calculating Average Difference

To find how accurate the estimates were on average, sum all the values in the "Positive Difference" column and divide by the number of objects ($5$).

A smaller average difference indicates higher accuracy in estimation.

Step 4: Visualisation (Double Bar Graph)

To draw the double bar graph, follow these steps:

1. Draw the Horizontal Axis ($x$-axis) and label it "Objects".

2. Draw the Vertical Axis ($y$-axis) and label it "Length (in cm)". Choose an appropriate scale (e.g., $1\text{ cm} = 5\text{ cm}$ on the graph).

3. For each object, draw two bars side-by-side: one representing the Estimate and the other representing the Measure.

4. Use different colours or patterns to distinguish between the Estimated and Measured bars and provide a Legend/Key.

Final Observations:

The student should conclude whether they tend to overestimate (Estimate > Measure) or underestimate (Measure > Estimate) based on the visual height of the bars in the graph.

Given:

Week 1 data ($9$ values): $410, 400, 370, 340, 360, 400, 320, 330, 310$

Week 2 data ($8$ values): $320, 290, 380, 280, 270, 230, 220, 240$

(a) Dot Plot Construction:

To construct the dot plot, we place a blue triangle for Week 1 values and a red circle for Week 2 values on the provided time scale (from $200$ to $420$ seconds).

(b) Mean and Median Calculation:

For Week 1:

$\text{Sum} = 410 + 400 + 370 + 340 + 360 + 400 + 320 + 330 + 310 $$ = 3240$

$\text{Mean}_{\text{W1}} = \frac{3240}{9} = 360 \text{ seconds}$

Sorted data: $310, 320, 330, 340, \mathbf{360}, 370, 400, 400, 410$

$\text{Median}_{\text{W1}} = 360 \text{ seconds}$

For Week 2:

$\text{Sum} = 320 + 290 + 380 + 280 + 270 + 230 + 220 + 240 = 2230$

$\text{Mean}_{\text{W2}} = \frac{2230}{8} = 278.75 \text{ seconds}$

Sorted data: $220, 230, 240, 270, 280, 290, 320, 380$

$\text{Median}_{\text{W2}} = \frac{270 + 280}{2} = 275 \text{ seconds}$

Observations:

1. The average time in Week 2 ($278.75 \text{ s}$) is significantly lower than in Week 1 ($360 \text{ s}$), indicating that Aditi's speed has improved as she practiced.

2. The median also dropped from $360 \text{ s}$ to $275 \text{ s}$.

3. The Week 2 data is more spread out compared to the cluster in Week 1, but most values are shifted towards the faster (left) side of the scale.

Note: These projects are designed for students to collect their own data. Below is the methodology and step-by-step guide on how to perform the analysis and present the findings.

Project (a): How Long is a Sentence?

Step 1: Data Collection

Pick two different textbooks (e.g., English Literature and Science). Select one page from each and count the number of words in at least $10$ to $15$ consecutive sentences. Record your data in a table:

Step 2: Visualisation (Dot Plot)

For each book, draw a horizontal line with numbers representing the word counts. Place a dot ($\bullet$) for each sentence above its corresponding word count. If two sentences have the same number of words, stack the dots vertically.

Step 3: Statistical Analysis

Calculate the Mean and Median for both sets of data.

For the Median, arrange the word counts in ascending order and find the middle value. If the number of sentences is even, find the average of the two middle values.

Project (b): What is in a Name?

Step 1: Listing and Counting

List the names of all your classmates. Count the number of letters in each name (excluding surnames if needed for simplicity).

Step 2: Central Tendency (Mean and Median)

1. Mean Length: Add the number of letters of all names and divide by the total number of students.

2. Median Length: Arrange the letter counts in ascending order and identify the middle value.

Step 3: Alphabetical Analysis

1. Popularity: Use tally marks to find the frequency of each starting letter. The letter with the highest frequency is the Mode (most popular).

2. Median Starting Letter: Arrange the starting letters of all names in alphabetical order (A to Z). Find the middle letter in this list. If it falls between $A$ and $M$, most names start with letters from the first half of the alphabet.

Step 4: Categorisation and Double-Bar Graph

Classify every student into one of the four categories mentioned in the question (Vowel-Vowel, Vowel-Consonant, etc.). Count the number of Boys and Girls in each category.

Draw a double-bar graph with these four categories on the x-axis and the count of students on the y-axis.

Final Observations:

1. Variability: Mention the Range (Difference between the longest and shortest name/sentence).

2. Interpretation: Describe if the data is "spread out" or "clustered" around the mean/median.

To complete this project, follow these steps:

1. Data Collection: Keep a small diary near your door and record every time you exit. Do this for $30$ days.

2. Calculation: At the end of the month, add up all the counts and divide by $30$ to find your daily mean. Arrange the $30$ values in order to find the median.

3. Variability: Observe the difference between your highest count (maybe on a weekend) and your lowest count (maybe on a holiday or rainy day). This is the range.

Visualisation: Draw a number line from $0$ to your maximum count and place a dot for every day you stepped out that many times.

Note: This project is to be performed by the student based on actual data collection. Below is the step-by-step framework and methodology to complete this project successfully.

Project (a): Our heights vs. our family’s heights

Step 1: Data Collection

Collect the heights of yourself and at least four other family members (parents, siblings, etc.) in centimetres (cm). Use the table below to record your data:

Step 2: Analysis of Family Data

(i) Dot Plot: Draw a horizontal number line marked with height values. For every family member's height, place a dot ($\bullet$) above the corresponding value. If heights repeat, stack the dots vertically.

(ii) Central Tendency: Calculate the Mean height of your family.

(iii) Variability: Calculate the Range of the heights.

Step 3: Group Collaboration and Visualisation

Combine your Mean Family Height with your group mates' data and represent it using a double-bar graph. On the x-axis, write student names. For each student, draw two bars: one for the Student's height and one for their Family's Mean height.

Project (b): Estimating Time

Step 1: Experiment and Recording

Perform the time estimation activity with your family members. Use a stopwatch to record the actual seconds elapsed when they think the time is up. Do not let them count or look at a clock.

Step 2: Analysis and Visualisation

(i) Dot Plots: Create two separate dot plots. One for $1$-minute estimates (Goal: $60\text{s}$) and one for $3$-minute estimates (Goal: $180\text{s}$).

(ii) Comparison: Calculate the Mean for both sets of estimates. Observe if the mean value is close to the actual goal ($60\text{s}$ or $180\text{s}$).

(iii) Double-Bar Graph: Plot a graph showing the Mean $1$-minute estimate versus the Mean $3$-minute estimate for every family in your group.

Final Step: Group Observations

Discuss the following within your group:

1. For Heights: Is there a strong correlation between the student's height and their family's average height?

2. For Time: Does the variability increase when the time duration is longer ($3$ minutes) compared to $1$ minute? Do people generally under-estimate or over-estimate the time?