CBSE • Class 10 • Mathematics • Chapter 13 (Statistics) • Exercise 13.1

Exercise 13.1: Statistics — NCERT Solutions

Mean of grouped data using the direct, assumed-mean and step-deviation methods.

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What this exercise covers

Direct methodAssumed meanStep deviation

Step-by-step solutions — Exercise 13.1

4 solved questions • Each solution includes a Socratic hint, full working and a common-mistake callout • Last reviewed 2026-09-03

Exercise 13.1 Q1 • 3 marks

The daily pocket money (in ₹) of 30 students is recorded in classes 10–20, 20–30, 30–40, 40–50, 50–60 with frequencies 4, 6, 10, 6, 4 respectively. Find the mean pocket money using the direct method.
Hint (Socratic — try this first)▾
What single value best represents each class interval when you have no individual data?
Step-by-step solution▾

Direct method: Mean xˉ=∑fixi∑fi\bar{x} = \dfrac{\sum f_i x_i}{\sum f_i}, where xix_i is the class mark.

| Class | fif_i | xix_i | fixif_i x_i | |---|---|---|---| | 10–20 | 4 | 15 | 60 | | 20–30 | 6 | 25 | 150 | | 30–40 | 10 | 35 | 350 | | 40–50 | 6 | 45 | 270 | | 50–60 | 4 | 55 | 220 | | Total | 30 | | 1050 |

xˉ=105030=35\bar{x} = \frac{1050}{30} = 35

The mean pocket money is ₹35.

Common mistake:
Using the lower or upper class limit instead of the class mark xi=lower+upper2x_i = \frac{\text{lower}+\text{upper}}{2}.
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Exercise 13.1 Q2 • 3 marks

The marks obtained by 50 students are grouped as 0–20 (f=8), 20–40 (f=12), 40–60 (f=15), 60–80 (f=10), 80–100 (f=5). Find the mean using the assumed-mean method.
Hint (Socratic — try this first)▾
If you subtract a convenient value aa from each class mark, how does the mean of those differences relate to the actual mean?
Step-by-step solution▾

Assumed-mean method: xˉ=a+∑fidi∑fi\bar{x} = a + \dfrac{\sum f_i d_i}{\sum f_i}, where di=xi−ad_i = x_i - a.

Take a=50a = 50.

| Class | fif_i | xix_i | di=xi−50d_i = x_i-50 | fidif_i d_i | |---|---|---|---|---| | 0–20 | 8 | 10 | -40 | -320 | | 20–40 | 12 | 30 | -20 | -240 | | 40–60 | 15 | 50 | 0 | 0 | | 60–80 | 10 | 70 | 20 | 200 | | 80–100 | 5 | 90 | 40 | 200 | | Total | 50 | | | -160 |

xˉ=50+−16050=50−3.2=46.8\bar{x} = 50 + \frac{-160}{50} = 50 - 3.2 = 46.8

The mean mark is 46.8.

Common mistake:
Forgetting to add aa back at the end, or making sign errors when computing negative deviations.
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Exercise 13.1 Q3 • 3 marks

The heights (in cm) of 40 plants are distributed as 120–130 (f=5), 130–140 (f=8), 140–150 (f=12), 150–160 (f=9), 160–170 (f=6). Find the mean height using the step-deviation method.
Hint (Socratic — try this first)▾
Since all class widths are equal, can you divide each deviation by the class size to make the numbers smaller?
Step-by-step solution▾

Step-deviation method: xˉ=a+h(∑fiui∑fi)\bar{x} = a + h\left(\dfrac{\sum f_i u_i}{\sum f_i}\right), where ui=xi−ahu_i = \dfrac{x_i - a}{h}.

Take a=145a = 145, h=10h = 10.

| Class | fif_i | xix_i | uiu_i | fiuif_i u_i | |---|---|---|---|---| | 120–130 | 5 | 125 | -2 | -10 | | 130–140 | 8 | 135 | -1 | -8 | | 140–150 | 12 | 145 | 0 | 0 | | 150–160 | 9 | 155 | 1 | 9 | | 160–170 | 6 | 165 | 2 | 12 | | Total | 40 | | | 3 |

xˉ=145+10(340)=145+0.75=145.75\bar{x} = 145 + 10\left(\frac{3}{40}\right) = 145 + 0.75 = 145.75

The mean height is 145.75 cm.

Common mistake:
Forgetting to multiply the fraction ∑fiui∑fi\frac{\sum f_iu_i}{\sum f_i} by the class size hh before adding aa.
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Exercise 13.1 Q4 • 4 marks

The mean of the following distribution is 18. Find the missing frequency ff: classes 11–13 (f=3), 13–15 (f=6), 15–17 (f=9), 17–19 (f=13), 19–21 (f=ff), 21–23 (f=5), 23–25 (f=4).
Hint (Socratic — try this first)▾
Can you set up the mean formula as an equation with ff as the unknown and solve?
Step-by-step solution▾

Using class marks xix_i: 12, 14, 16, 18, 20, 22, 24.

| xix_i | fif_i | fixif_i x_i | |---|---|---| | 12 | 3 | 36 | | 14 | 6 | 84 | | 16 | 9 | 144 | | 18 | 13 | 234 | | 20 | ff | 20f20f | | 22 | 5 | 110 | | 24 | 4 | 96 | | Total | 40+f40+f | 704+20f704 + 20f |

Mean =18= 18: 704+20f40+f=18\frac{704 + 20f}{40 + f} = 18 704+20f=720+18f704 + 20f = 720 + 18f 2f=16  ⟹  f=82f = 16 \implies f = 8

The missing frequency is 8.

Common mistake:
Cross-multiplying incorrectly or forgetting that ff also appears in the total frequency in the denominator.
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How to approach Exercise 13.1

  1. Re-read the chapter summary first. Open Statistics and refresh the key concepts: Grouped data, Mean, Median, Mode.
  2. Try each problem yourself before opening the solution. Ten focused minutes per problem usually beats reading two finished solutions.
  3. Use Guru AI for the questions you get stuck on. The Socratic AI tutor walks you through it with hints instead of dictating the answer.
  4. Mark the questions you got wrong and revisit them after 24 hours — the spacing is what locks the method into long-term memory.

All exercises in Statistics

  1. Exercise 13.1 — Mean of grouped data using the direct, assumed-mean and step-deviation methods.
  2. Exercise 13.2 — Mode of grouped data using the standard formula.
  3. Exercise 13.3 — Median of grouped data and cumulative-frequency interpretation.

Exercise 13.2 →

Mode of grouped data using the standard formula.

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