CBSE Class 7 Mathematics Connecting the Dots Assignment Set 01

Short Answer Questions

Question 1. What is a statistical question? Write one example of a statistical question and one example of a question that is not statistical, giving a reason for each.
Answer:
A statistical question is a question that can be answered only by collecting data, because its answer is expected to vary.
Statistical: “How many hours do the Grade 7 students of our school sleep at night?” Different students sleep for different lengths of time, so we must collect and analyse data.
Not statistical: “What is my roll number?” It has one fixed answer, so no data needs to be collected.
In simple words: A question is statistical only when its answer varies and data has to be collected.

Teacher's Note:
1. Test each question: does the answer change from case to case?
2. Questions about a group are often statistical.

Question 2. The numbers of story books read by five friends during a month are 4, 7, 5, 9 and 5. Find the arithmetic mean of this data and explain in one sentence what this mean tells us.
Answer:
Mean = (sum of all the values) ÷ (number of values).
Sum = 4 + 7 + 5 + 9 + 5 = 30; number of values = 5.
Mean = 30 ÷ 5 = 6 books.
This means that if all five friends had read the same number of books, each would have read 6 books.
In simple words: The five friends read 6 books each on average.

Teacher's Note:
1. The mean is the equal share of the total.
2. Check that 6 lies between 4 and 9.

Question 3. In Group A, 4 children together collected 6, 9, 5 and 8 oranges. In Group B, 5 children together collected 5, 4, 8, 6 and 7 oranges. Each group shares its oranges equally among its own members. In which group does a child get more oranges, and by how many?
Answer:
Group A: total = 6 + 9 + 5 + 8 = 28 oranges for 4 children, so each gets 28 ÷ 4 = 7.
Group B: total = 5 + 4 + 8 + 6 + 7 = 30 oranges for 5 children, so each gets 30 ÷ 5 = 6.
A child of Group A gets more, by 7 − 6 = 1 orange.
Notice that Group B collected more in total, but each member's share is smaller because the group is bigger.
In simple words: Each child in Group A gets 7 oranges and in Group B gets 6, so Group A gets 1 more.

Teacher's Note:
1. Compare equal shares, not totals, when groups differ in size.
2. Equal share is the mean.

Question 4. The numbers of pages Meera read on seven days are 12, 9, 15, 7, 11, 14 and 10. Find the median number of pages, showing the steps you follow.
Answer:
Step 1: Sort the data: 7, 9, 10, 11, 12, 14, 15.
Step 2: Count the values: there are 7 values, an odd number, so there is one middle value.
Step 3: The 4th value is the middle one, with 3 values below it and 3 above it.
Median = 11 pages.
In simple words: After sorting, the 4th of the 7 values, 11, is the median.

Teacher's Note:
1. For n values (n odd), the middle is the ((n + 1) ÷ 2)th value.
2. Always sort before finding the median.

Question 5. The runs scored by a batter in eight matches are 12, 18, 9, 25, 14, 20, 11 and 16. Find the median score. State clearly the rule used when the data has an even number of values.
Answer:
Sorted scores: 9, 11, 12, 14, 16, 18, 20, 25.
There are 8 values, an even number, so there is no single middle value.
Rule: when the number of values is even, the median is the average of the two middle values of the sorted data.
The two middle values are 14 and 16, so median = (14 + 16) ÷ 2 = 15 runs.
In simple words: With 8 values, average the 4th and 5th sorted values: (14 + 16) ÷ 2 = 15.

Teacher's Note:
1. The two middle values are the 4th and 5th here.
2. The median may not be a data value.

Question 6. What is an outlier? The numbers of mangoes plucked from a tree on six days are 12, 14, 11, 13, 45 and 12. Identify the outlier and say whether it lies at the higher end or the lower end of the data.
Answer:
An outlier is a value that differs a great deal from the rest of the values in the data.
Here five values lie between 11 and 14, while 45 is far above all of them.
So 45 is the outlier, and it lies at the higher end of the data.
In simple words: 45 is much larger than the other values, so it is a high-end outlier.

Teacher's Note:
1. Look for a value far away from the cluster.
2. A high outlier pulls the mean up.

Question 7. What is a dot plot? Write any two things about a collection of data that a dot plot helps us see quickly.
Answer:
A dot plot shows each data value as a dot above its position on an equally spaced number line; repeated values are shown as dots stacked one above another.
It quickly shows us:
(1) the smallest and the largest value, and hence the range;
(2) where the data is clustered and where it is spread out, including any outlier.
In simple words: A dot plot places a dot for each value on a number line so that extremes and clusters are easy to see.

Teacher's Note:
1. Taller stacks mean more frequent values.
2. A dot plot sorts the data automatically.

Question 8. What is a double bar graph (clustered column graph)? Describe one situation in which it is more useful than drawing two separate bar graphs.
Answer:
A double bar graph shows two sets of data for the same categories as two bars drawn side by side in each cluster, in two different colours or patterns.
It is more useful when we want to compare, for example, the monthly onion prices in two towns. The two bars of each month can be compared at a glance instead of looking from one graph to another.
In simple words: A double bar graph places the two groups' bars side by side so they can be compared at once.

Teacher's Note:
1. Both groups share one scale in a double bar graph.
2. A key must show which bar is which group.

Question 9. Player P scored 34, 0, 28, 50 and 18 runs in five matches, while Player Q scored 40, 22, 36 and 14 runs in four matches. Ravi says, “P played better because his total is larger.” Do you agree? Support your answer by finding the mean score of each player.
Answer:
No, I do not agree with Ravi.
P: total = 34 + 0 + 28 + 50 + 18 = 130 in 5 matches, so mean = 130 ÷ 5 = 26 runs.
Q: total = 40 + 22 + 36 + 14 = 112 in 4 matches, so mean = 112 ÷ 4 = 28 runs.
The players played different numbers of matches, so the total is not a fair measure. By the mean, Q played better.
In simple words: Q's mean of 28 runs is higher than P's 26, so Q did better per match.

Teacher's Note:
1. The 0 of P is a real score and is counted.
2. Use the mean when group sizes differ.

Question 10. A player’s scores in a six-match series are recorded as 45, 0, 62, —, 33, 50, where “—” shows that she did not play that match. Find her mean runs per match and explain why the total is not divided by 6.
Answer:
The dash means she did not play that match, so it has no value; it is not a score of 0.
Matches actually played = 5.
Total runs = 45 + 0 + 62 + 33 + 50 = 190.
Mean = 190 ÷ 5 = 38 runs per match.
Dividing by 6 would wrongly count a match she never played and would lower the mean to about 31.7.
In simple words: Only the 5 matches she played count, so the mean is 190 ÷ 5 = 38.

Teacher's Note:
1. 0 is counted; a dash is not.
2. Count the values before dividing.

Question 11. The heights (in cm) of the five members of a family are 158, 162, 165, 110 and 160.
(i) Find the mean height and the median height.
(ii) Which of the two represents this family’s heights better? Give a reason.
Answer:
(i) Sum = 158 + 162 + 165 + 110 + 160 = 755, so mean = 755 ÷ 5 = 151 cm.
Sorted heights: 110, 158, 160, 162, 165, so median = 160 cm.
(ii) The median represents the family better. The height 110 cm is an outlier that pulls the mean down, so the mean 151 cm is less than four of the five heights, while 160 cm lies truly in the middle.
In simple words: The low outlier 110 drags the mean to 151 cm, so the median 160 cm describes the family better.

Teacher's Note:
1. A low outlier makes mean < median.
2. A good representative should be close to most values.

Question 12. The numbers of birds seen at a pond on seven days are 8, 10, 12, 13, 14, 15 and 40.
(i) Find the mean and the median of this data.
(ii) Find the mean and the median again after leaving out the outlier, and write what you notice about the two changes.
Answer:
(i) Sum = 8 + 10 + 12 + 13 + 14 + 15 + 40 = 112, so mean = 112 ÷ 7 = 16 birds. The data is already sorted, so median = 4th value = 13 birds.
(ii) Leaving out 40: sum = 72, mean = 72 ÷ 6 = 12 birds; median = (12 + 13) ÷ 2 = 12.5 birds.
The mean falls by 4, but the median falls by only 0.5. This shows that the median is hardly affected by an outlier, while the mean is affected a lot.
In simple words: Removing the outlier 40 drops the mean from 16 to 12 but the median only from 13 to 12.5.

Teacher's Note:
1. The mean is sensitive to outliers.
2. The median depends only on the middle position.

Question 13. The rainfall (in mm) recorded on five days in two villages is — Village A: 12, 18, 25, 30, 15; Village B: 5, 40, 10, 35, 10. Show that both villages have the same mean rainfall, then compare the two data sets using their ranges. In which village was the rainfall more even?
Answer:
Village A: total = 12 + 18 + 25 + 30 + 15 = 100 mm, mean = 100 ÷ 5 = 20 mm.
Village B: total = 5 + 40 + 10 + 35 + 10 = 100 mm, mean = 100 ÷ 5 = 20 mm.
So both villages have the same mean, 20 mm.
Range of A = 30 − 12 = 18 mm; range of B = 40 − 5 = 35 mm.
Village A had more even rainfall, because its values are less spread out.
In simple words: Both means are 20 mm, but A's smaller range (18 mm against 35 mm) shows steadier rainfall.

Teacher's Note:
1. Equal means can hide very different spreads.
2. The range helps compare spread.

Question 14. Answer with a reason in each case.
(i) For one data set, the mean is 24 and the median is 30. At which end of the data would you expect an outlier?
(ii) For another data set, the mean is 48 and the median is 40. At which end would you expect an outlier?
Answer:
(i) Here mean < median (24 < 30). A very small value pulls the mean down towards itself, so the outlier is expected at the lower end.
(ii) Here mean > median (48 > 40). A very large value pulls the mean up, so the outlier is expected at the higher end.
The median stays near the centre in both cases, so the direction in which the mean shifts points to the outlier.
In simple words: The mean moves towards the outlier, so mean below median means a low outlier and mean above median means a high one.

Teacher's Note:
1. Compare the mean with the median to locate an outlier.
2. The median is the steady reference point.

Question 15. A bar graph has to be drawn for data whose largest value is 450, and only 10 equal divisions are available along the vertical line.
(i) What scale would you choose?
(ii) With that scale, how many divisions tall will the bars for 350 and for 275 be?
Answer:
(i) Each division must cover at least 450 ÷ 10 = 45. A convenient round choice is 1 division = 50, which lets the graph reach 500 and is easy to read.
(ii) Bar for 350: 350 ÷ 50 = 7 divisions.
Bar for 275: 275 ÷ 50 = 5.5 divisions, that is, five and a half divisions.
In simple words: Choose 1 division = 50; then 350 needs 7 divisions and 275 needs 5.5.

Teacher's Note:
1. Pick a round scale that makes all values fit.
2. Height in divisions = value ÷ scale.

Question 16. The double bar graph below shows the number of library books borrowed by two sections over four months.
(i) In which month was the difference between the two sections the greatest, and what was that difference?
(ii) Find the total number of books borrowed by each section during these four months.
Answer:
(i) Reading the bars, the monthly differences are: July 40 − 30 = 10; August 55 − 45 = 10; September 50 − 35 = 15 (Section B higher); October 60 − 25 = 35.
The greatest difference is in October, and it is 35 books.
(ii) Section A: 40 + 55 + 35 + 60 = 190 books.
Section B: 30 + 45 + 50 + 25 = 150 books.
In simple words: October shows the biggest gap (35 books); Section A borrowed 190 books and Section B 150.

Teacher's Note:
1. The scale here is 1 unit = 10 books.
2. Subtract the shorter bar from the taller one in each month.

Question 17. The mean of six numbers is 24. Five of them are 18, 30, 21, 27 and 19. Find the sixth number and verify your answer by recalculating the mean.
Answer:
Sum of six numbers = 24 × 6 = 144.
Sum of the five given numbers = 18 + 30 + 21 + 27 + 19 = 115.
Sixth number = 144 − 115 = 29.
Verification: 18 + 30 + 21 + 27 + 19 + 29 = 144 and 144 ÷ 6 = 24, which is the given mean.
So the sixth number is 29.
In simple words: The six numbers must total 144, so the missing one is 144 − 115 = 29.

Teacher's Note:
1. Total = mean × count.
2. Always verify by recomputing the mean.

Question 18. A school has two Grade 7 sections, each with 32 students. The mean height of the students of one section is 149.5 cm. What can be said with certainty about the mean height of the students of the other section? Justify your answer.
Answer:
Nothing can be said with certainty; the mean height of the other section cannot be found from this.
The mean of one group tells us only about the values of that group.
The other section has its own 32 students, whose heights are unknown; they could be shorter, taller or the same on average.
Equal group size does not force equal means. Only the heights of the second section can settle the question.
In simple words: One section's mean says nothing about the other section; we need the second section's own data.

Teacher's Note:
1. A mean belongs only to its own group.
2. Do not guess beyond the data.

Question 19. A survey table shows that the average height of 13-year-old girls is more than the average height of 11-year-old girls. Rahul concludes, “So every 13-year-old girl is taller than every 11-year-old girl.” Is his conclusion correct? Explain what an average can and cannot tell us here.
Answer:
No, Rahul's conclusion is not correct.
The table gives average heights, which compare the two age groups as a whole, not child with child.
An average balances out many values, so some 11-year-old girls can easily be taller than many 13-year-old girls.
The data supports a statement such as “13-year-old girls are taller on average”, but never a statement about every single girl.
In simple words: An average describes a group as a whole, not each member of it.

Teacher's Note:
1. Words like 'every' need much stronger evidence.
2. Averages hide individual differences.

Question 20. In a one-day match, the median of the runs scored by the eleven players of a team was 0, and yet the team scored 407 runs. Explain how both facts can be true at the same time, and state what this shows about the median.
Answer:
With 11 players, the median is the 6th score in the sorted list. If it is 0, then at least six players scored 0.
The remaining five batters, together with extras, can still build a large total; for example a few of them may score 80, 90 or 100 each.
So the 407 runs come from a small number of big scores.
This shows that the median tells us only about the middle of the data, and says nothing about the total or the extreme values.
In simple words: Six zeros make the median 0, while a few big scores make the total 407.

Teacher's Note:
1. The median ignores how big the large values are.
2. Use more than one measure to describe data.