NCERT Solutions for Class 7th Maths Chapter 5 Figure it Out — Data Detective

First read the dots off the two plots.

Only statement (b) is true.

(a) A's average.

(b) Divide C's total by 3, and B's total by 4.

(c) Who is the best performer?

A is the best performer. A has the highest mean per game (12.5), the highest total (50) and the highest single score (16), and A never dropped below 10 points.

Group 2 did slightly better on both measures, and Group 1's marks are far more spread out.

Describing the two groups.

• Group 2 is ahead on both mean and median, but only just — by 2.67 marks on the mean and 0.5 on the median.
• In both groups mean < median (73 < 78.5 and 75.67 < 79). That tells us both groups have a few low marks pulling the mean down. In Group 1 the low marks are 39, 48 and 56; in Group 2 just 47 and 59.
• Group 1 is more spread out — its marks stretch over 56 marks against Group 2's 46. Group 1 also holds the highest single mark, 95.
• In Group 1, 6 of 10 students scored 75 or more. In Group 2, 5 of 9 scored 79 or more. The top halves of the two groups are very similar; the difference lies in the weak tail.

The largest value is 1240, so take 1 unit length = 200 people and run the vertical line to 1400.

Observations.

• More people watch than play, in every single sport. Not one orange bar reaches its blue partner.
• Cricket dominates both. 1240 watchers — more than the other four sports put together (470 + 510 + 430 + 250 = 1660, so almost as many). And its 620 participants are more than double any other sport's.
• Cricket's participants are exactly half its watchers (620 out of 1240). Athletics is the opposite extreme: only 105 of its 250 watchers take part, well under half.
• Swimming beats basketball for watching (510 against 470) but the two are exactly level for participating (320 each).
• Athletics is last on both counts — 250 watching, 105 participating.

Exactly equal groups are impossible here — and the data shows why twice over.

• Problem 1: 17 is odd. Two equal groups would need 8½ students each. The closest possible split is 8 and 9.
• Problem 2: five students are exactly 115 cm tall. No cut-off height can separate two students of the same height. Cutting at 115 gives: 7 students below (101, 102, 106, 109, 110, 110, 112) and 5 above (117, 120, 120, 123, 125) — with 5 students stranded on the line itself.

What to suggest. Use the median, 115 cm, as the dividing height, and then place the five students who are exactly 115 cm to balance the groups:

Or, if the teacher truly needs two groups of the same size, ask one more student to join, making 18.

Guessing the age.

Now look up the 2019 column of the heights table:

A mean of 113.5 cm sits almost exactly on the age-6 figures, and the median of 115 cm falls between age 6 and age 7. So these students are probably about 6 to 7 years old — a Grade 1 or Grade 2 class.

This is a data-collection activity for your own class. Here is the method, then a worked sample.

• Measure every student against a wall, without shoes, and record to the nearest centimetre. Use the same measuring tape for everybody.
• Note down all the heights, including anyone who is unusually short or tall — those are the values that make the exercise interesting.
• Add them up and divide by the number of students to get the mean.
• Sort the heights and pick the middle one to get the median. If your class has an even number of students, average the two middle values.
• Draw a dot plot with a number line covering your class's range, in steps of 1 or 2 cm, and stack one dot per student. Mark the mean and the median on it.

Sample answer for a Grade 7 class of 20 students:

Describing it: the class runs from 138 cm to 164 cm, a range of 26 cm. Most students cluster between 144 and 154 cm. The mean (149) and the median (148.5) are only half a centimetre apart, so the data is fairly balanced — although the single student at 164 cm pulls the mean a little above the median.

(d) The mean height of students in the other section cannot be determined.

Here are three possible second sections, all perfectly consistent with what we are told:

(a) Estimating the three unlabelled bars. Use the labelled bars on either side as a ruler.

(b) Are the statements valid?

This is a hands-on activity. The method matters more than the numbers.

• Write down every estimate first, for all five objects, before you touch a ruler. If you measure one object first, your next estimate will not be honest.
• Measure to the nearest centimetre with the same ruler each time.
• The positive difference means "always subtract the smaller from the larger", so it is never negative. That way an over-estimate and an under-estimate do not cancel each other out.

Sample answer:

The double bar graph. Put the five objects along the bottom. For each object draw two bars side by side — one for your estimate, one for your measurement. The longest object is about 29 cm, so a scale of 1 unit = 5 cm works well.

How accurate were the estimates? In this sample, four of the five were within 4 cm, but the palm was 7 cm out — the estimate was much too small. Removing that one outlier drops the average difference from 3 cm to (1 + 1 + 2 + 4) ÷ 4 = 2 cm.

(a) The two weeks. The first nine values are Week 1; the remaining eight are Week 2.

(b) Mean, median and observations.

• Aditi got much faster. Her typical time dropped by about 85 seconds — nearly a minute and a half — in a single week.
• Week 1's mean and median are both exactly 360 s, so that week's times were nicely balanced with no outlier.
• Week 2 has an outlier at 380 s — the third puzzle of that week. Every other Week 2 time is 320 s or less. That is why Week 2's mean (278.75) sits above its median (275).
• Her last four puzzles (270, 230, 220, 240) are all faster than anything she managed in Week 1. The improvement is real, not an accident of averaging.

Pick one part. Here is how to run each properly, with a worked sample.

(a) How Long is a Sentence?

• Choose one full page of running text from each book — say a page of this maths book and a page of your science book. Skip headings, captions and lists.
• Decide your rule before you start: a sentence ends at a full stop, a question mark or an exclamation mark; count every word including "a" and "the"; count a number like 144.5 as one word. Use the same rule for both pages.
• Make one dot plot for each page, with sentence length along the number line.

Sample answer. Maths page: 9, 12, 7, 15, 11, 6, 18, 10, 13, 9 words (10 sentences). Science page: 14, 22, 17, 25, 19, 16, 28, 21 words (8 sentences).

In this sample the science sentences are roughly twice as long, and their dot plot is also more spread out. Both books have mean ≈ median, so neither page has a runaway sentence.

(b) What is in a Name?

(i) and (ii) Name lengths. Sample for 12 classmates: Aarav 5, Diya 4, Ishaan 6, Meera 5, Rohan 5, Kavya 5, Sanjay 6, Anaya 5, Vikram 6, Priya 5, Tanvi 5, Nikhil 6.

The dot plot would show a tall stack at 5, a shorter stack at 6, and a single dot at 4 — very little variability, with a range of only 2 letters. Mean and median almost agree, so there is no outlier.

(iii) Starting letters. Count how many names begin with each letter and find the tallest stacks. In the sample, A appears twice (Aarav, Anaya); most other letters appear once.

(iv) The median starting letter. Write the initials in alphabetical order and take the middle one: A, A, D, I, K, M, N, P, R, S, T, V — the 6th and 7th are M and N, so the median starting letter sits right at the M/N boundary. That means about half the names start with A–M and about half with N–Z. If instead the median starting letter had been, say, D, it would tell you that names in this class crowd into the early part of the alphabet.

(v) The double-bar graph. Sort every name into one of four boxes and count boys and girls separately.

A month-long project. What makes it work is being consistent.

• Fix the rule on day one. Does going to the gate and coming back count? Does going out and returning twice count as 2? Write your rule down and keep to it for all 30 days.
• Record every day, including 0s. A day when you never left is a real data point, exactly like a batsman's 0.
• Also note the day of the week beside each count. That extra column is what makes the later analysis interesting.

(i) Sample answer for one month (30 days):

The dot plot would run from 0 to 7 with the tallest stack at 2, a good stack at 3, and single dots at 0 and 7. Mean ≈ median, so the data is balanced.

(ii) Interesting observations from the sample:

• The counts of 6 and 7 all fall on weekends; the 0 and the 1s fall on days when it rained or when there was a test.
• School days settle into a steady rhythm of 2 or 3 — school, and one errand.
• The mode (the commonest value) is 2, a little below both the mean and the median.
• There is no true outlier: 7 is only one more than 6, so the counts step up smoothly.

(iii) If a family member records the same thing, draw both dot plots one above the other on the same number line — as the book does for Group A and Group B on page 110 — and compare the two central tendencies and the two spreads.

Work in groups of 8 to 10. Each of you collects your own data first; then the group pools it.

(a) Our heights vs. our family's heights

(i) Sample family of 5: 168, 155, 172, 149, 161 cm.

Mean and median agree exactly, so this family has no outlier. If a young child were included, the mean would drop well below the median, just as it did in Poovizhi's family on page 105.

(ii) The double-bar graph. Along the bottom put each student's name. For each name draw two bars — the student's own height and that student's family mean. All the values lie between about 130 and 175 cm, so start the vertical line at 120 cm and use 1 unit = 10 cm, exactly the trick the book uses on page 128 where the axis starts at 145 cm. Label clearly that the axis does not start at 0.

(iii) What to look for across the whole group:

• Nearly every student's bar will be shorter than their family's mean bar, because the family mean includes grown-ups.
• Families with a small child will have a surprisingly low mean — that is the outlier effect again.
• Ask whether the taller students tend to come from the families with the taller means. Even if they do, remember that 8–10 families is a very small sample.

(b) Estimating time

(i) and (ii) Sample family — 1-minute estimates: 52, 58, 61, 47, 65 seconds. 3-minute estimates: 150, 172, 195, 138, 205 seconds.

(iii) The double-bar graph. One cluster per family, with the mean 1-minute estimate and the mean 3-minute estimate side by side. Because 3-minute values are about three times the size, the 3-minute bars will tower over the others — that is fine and is itself worth pointing out.

(iv) Observations to look for:

• The 3-minute estimates spread much wider — a range of 67 s against 18 s here. The longer the interval, the more room there is to drift.
• A fairer comparison is the error as a fraction of the target: 3.4 s out of 60 is about 6%, while 8 s out of 180 is about 4%. By that measure the family was actually better at 3 minutes.
• Most people underestimate — they open their eyes early. Check whether that holds across all 8–10 families before claiming it as a general finding.