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NCERT Class 10 Mathematics Exercise 13.2 Solutions: Statistics

Step-by-step working for every question of Exercise 13.2 in NCERT's Class 10 Mathematics book, chapter “Statistics”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

1. Ex 13.2 Q1

   Ages of patients — 5-15, 15-25, 25-35, 35-45, 45-55, 55-65 years with 6, 11, 21, 23, 14, 5 patients. Find the mode and the mean, and compare them.

   1. Identify the modal class as 35-45 since it has the highest frequency of 23.
      why?

      Mode = l + ((f₁ − f₀) / (2f₁ − f₀ − f₂)) × h, where the modal class has the highest frequency f₁, and f₀, f₂ are the frequencies of the classes just before and after it.

   2. Note the lower limit l = 35, modal frequency f1 = 23, preceding frequency f0 = 21, and succeeding frequency f2 = 14.
   3. Calculate the difference f1 - f0 = 23 - 21 = 2 and f1 - f2 = 23 - 14 = 9.
   4. Use the formula Mode = l + ((f1 - f0) / (2f1 - f0 - f2)) × h with class width h = 10.
      why?

      Mode = l + ((f₁ − f₀) / (2f₁ − f₀ − f₂)) × h, where the modal class has the highest frequency f₁, and f₀, f₂ are the frequencies of the classes just before and after it.

   5. Substitute values to get Mode = 35 + (2 / (46 - 21 - 14)) × 10.
      why?

      Mode = l + ((f₁ − f₀) / (2f₁ − f₀ − f₂)) × h, where the modal class has the highest frequency f₁, and f₀, f₂ are the frequencies of the classes just before and after it.

   6. Simplify the denominator to 46 - 35 = 11, giving Mode = 35 + (2/11) × 10.
      why?

      Mode = l + ((f₁ − f₀) / (2f₁ − f₀ − f₂)) × h, where the modal class has the highest frequency f₁, and f₀, f₂ are the frequencies of the classes just before and after it.

   7. Calculate the fraction part as 20/11 ≈ 1.82, so Mode = 35 + 1.82 = 36.82 years.
      why?

      Mode = l + ((f₁ − f₀) / (2f₁ − f₀ − f₂)) × h, where the modal class has the highest frequency f₁, and f₀, f₂ are the frequencies of the classes just before and after it.

   8. Create a table with class midpoints: 10, 20, 30, 40, 50, 60 corresponding to frequencies 6, 11, 21, 23, 14, 5.
   9. Calculate the product of midpoint and frequency for each class: 60, 220, 630, 920, 700, 300.
   10. Sum the frequencies to get total N = 6 + 11 + 21 + 23 + 14 + 5 = 80.
   11. Sum the products to get total Σfx = 60 + 220 + 630 + 920 + 700 + 300 = 2830.
   12. Calculate the mean using the formula Mean = Σfx / N = 2830 / 80.
       why?

       Mean of grouped data = Σfᵢxᵢ / Σfᵢ: each class mark counts as many times as its frequency. The assumed-mean and step-deviation methods only shift and scale the numbers and give the same mean.

   13. Divide to find Mean = 35.375, which rounds to 35.38 years.
       why?

       Mean of grouped data = Σfᵢxᵢ / Σfᵢ: each class mark counts as many times as its frequency. The assumed-mean and step-deviation methods only shift and scale the numbers and give the same mean.

   14. Compare the results to see that the mode (36.82) is slightly higher than the mean (35.38).

   Answer: mode = 36.82 years; mean = 35.38 years

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2. Ex 13.2 Q2

   Lifetimes (hours) of 225 components — 0-20, 20-40, 40-60, 60-80, 80-100, 100-120 with 10, 35, 52, 61, 38, 29. Find the modal lifetime.

   1. The modal class is 60-80 since it has the highest frequency of 61.
      why?

      Mode = l + ((f₁ − f₀) / (2f₁ − f₀ − f₂)) × h, where the modal class has the highest frequency f₁, and f₀, f₂ are the frequencies of the classes just before and after it.

   2. Here, l = 60, h = 20, f1 = 61, f0 = 52, and f2 = 38.
   3. The formula for mode is Mode = l + (f1 - f0) / (2f1 - f0 - f2) × h.
      why?

      Mode = l + ((f₁ − f₀) / (2f₁ − f₀ − f₂)) × h, where the modal class has the highest frequency f₁, and f₀, f₂ are the frequencies of the classes just before and after it.

   4. Substituting the values: Mode = 60 + (61 - 52) / (2 × 61 - 52 - 38) × 20.
      why?

      Mode = l + ((f₁ − f₀) / (2f₁ − f₀ − f₂)) × h, where the modal class has the highest frequency f₁, and f₀, f₂ are the frequencies of the classes just before and after it.

   5. Simplify the numerator: 61 - 52 = 9.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   6. Simplify the denominator: 122 - 52 - 38 = 32.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   7. Calculate the fraction: 9 / 32 = 0.28125.
   8. Multiply by h: 0.28125 × 20 = 5.625.
   9. Add to l: 60 + 5.625 = 65.625.
   10. Rounding to two decimal places gives 65.63 hours.

   Answer: 65.63 hours

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3. Ex 13.2 Q3

   Monthly expenditure of 200 families — ₹1000-1500, 1500-2000, 2000-2500, 2500-3000, 3000-3500, 3500-4000, 4000-4500, 4500-5000 with 24, 40, 33, 28, 30, 22, 16, 7 families. Find the modal and the mean monthly expenditure.

   1. Class intervals are 1000-1500, 1500-2000, 2000-2500, 2500-3000, 3000-3500, 3500-4000, 4000-4500, 4500-5000 with frequencies 24, 40, 33, 28, 30, 22, 16, 7.
   2. The highest frequency is 40, so the modal class is 1500-2000.
      why?

      Mode = l + ((f₁ − f₀) / (2f₁ − f₀ − f₂)) × h, where the modal class has the highest frequency f₁, and f₀, f₂ are the frequencies of the classes just before and after it.

   3. For the modal class, l = 1500, h = 500, f1 = 40, f0 = 24, f2 = 33.
      why?

      Mode = l + ((f₁ − f₀) / (2f₁ − f₀ − f₂)) × h, where the modal class has the highest frequency f₁, and f₀, f₂ are the frequencies of the classes just before and after it.

   4. Mode = l + ((f1 - f0) / (2f1 - f0 - f2)) × h = 1500 + ((40 - 24) / (80 - 24 - 33)) × 500.
      why?

      Mode = l + ((f₁ − f₀) / (2f₁ − f₀ − f₂)) × h, where the modal class has the highest frequency f₁, and f₀, f₂ are the frequencies of the classes just before and after it.

   5. Mode = 1500 + (16 / 23) × 500 = 1500 + 347.83 = 1847.83.
      why?

      Mode = l + ((f₁ − f₀) / (2f₁ − f₀ − f₂)) × h, where the modal class has the highest frequency f₁, and f₀, f₂ are the frequencies of the classes just before and after it.

   6. To find the mean, take class marks (midpoints) as 1250, 1750, 2250, 2750, 3250, 3750, 4250, 4750.
      why?

      The class mark (midpoint of a class) = (lower limit + upper limit) / 2; in grouped data it stands for every value in that class.

   7. Calculate f × x for each class: 24×1250=30000, 40×1750=70000, 33×2250=74250, 28×2750=77000, 30×3250=97500, 22×3750=82500, 16×4250=68000, 7×4750=33250.
   8. Sum of frequencies N = 24 + 40 + 33 + 28 + 30 + 22 + 16 + 7 = 200.
   9. Sum of f × x = 30000 + 70000 + 74250 + 77000 + 97500 + 82500 + 68000 + 33250 = 532500.
   10. Mean = (Sum of f × x) / N = 532500 / 200 = 2662.5.

   Answer: mode = ₹1847.83; mean = ₹2662.5

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4. Ex 13.2 Q4

   Students per teacher in states/UTs — 15-20, 20-25, 25-30, 30-35, 35-40, 40-45, 45-50, 50-55 with 3, 8, 9, 10, 3, 0, 0, 2. Find the mode and the mean and interpret them.

   1. Mode: the highest frequency is 10, so the modal class is 30-35, with l = 30, h = 5, f₁ = 10, f₀ = 9 (class before), f₂ = 3 (class after).
      why?

      Mode = l + ((f₁ − f₀) / (2f₁ − f₀ − f₂)) × h, where the modal class has the highest frequency f₁, and f₀, f₂ are the frequencies of the classes just before and after it.

   2. Mode = l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h = 30 + [(10 − 9)/(20 − 9 − 3)] × 5 = 30 + 5/8 = 30.625 ≈ 30.63.
      why?

      Mode = l + ((f₁ − f₀) / (2f₁ − f₀ − f₂)) × h, where the modal class has the highest frequency f₁, and f₀, f₂ are the frequencies of the classes just before and after it.

   3. Mean: class marks 17.5, 22.5, 27.5, 32.5, 37.5, 42.5, 47.5, 52.5 with frequencies 3, 8, 9, 10, 3, 0, 0, 2 (Σf = 35).
      why?

      The class mark (midpoint of a class) = (lower limit + upper limit) / 2; in grouped data it stands for every value in that class.

   4. Σfx = 52.5 + 180 + 247.5 + 325 + 112.5 + 0 + 0 + 105 = 1022.5.
   5. Mean = 1022.5 ÷ 35 ≈ 29.21.
   6. Interpretation: the largest number of states/UTs have about 30.6 students per teacher (the mode), while on average a state/UT has about 29.2 students per teacher (the mean).

   Answer: mode = 30.63, mean = 29.21 students per teacher

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5. Ex 13.2 Q5

   ODI runs of top batsmen — 3000-4000, 4000-5000, 5000-6000, 6000-7000, 7000-8000, 8000-9000, 9000-10000, 10000-11000 with 4, 18, 9, 7, 6, 3, 1, 1 batsmen. Find the mode.

   1. Identify the modal class as 4000-5000 since it has the highest frequency of 18.
      why?

      Mode = l + ((f₁ − f₀) / (2f₁ − f₀ − f₂)) × h, where the modal class has the highest frequency f₁, and f₀, f₂ are the frequencies of the classes just before and after it.

   2. Note the lower limit l = 4000, frequency f1 = 18, previous frequency f0 = 4, and next frequency f2 = 9.
   3. Calculate the class width h as 5000 − 4000 = 1000.
   4. Substitute values into the mode formula: Mode = l + [(f1 − f0) / (2f1 − f0 − f2)] × h.
      why?

      Mode = l + ((f₁ − f₀) / (2f₁ − f₀ − f₂)) × h, where the modal class has the highest frequency f₁, and f₀, f₂ are the frequencies of the classes just before and after it.

   5. Compute the numerator: 18 − 4 = 14.
   6. Compute the denominator: 2 × 18 − 4 − 9 = 36 − 13 = 23.
   7. Calculate the fraction: 14 / 23 ≈ 0.6087.
   8. Multiply by the class width: 0.6087 × 1000 = 608.7.
   9. Add to the lower limit: 4000 + 608.7 = 4608.7.

   Answer: 4608.7 runs

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6. Ex 13.2 Q6

   Cars passing a spot in 100 periods of 3 minutes — 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70, 70-80 with 7, 14, 13, 12, 20, 11, 15, 8. Find the mode.

   1. Identify the modal class as 40-50 since it has the highest frequency of 20.
      why?

      Mode = l + ((f₁ − f₀) / (2f₁ − f₀ − f₂)) × h, where the modal class has the highest frequency f₁, and f₀, f₂ are the frequencies of the classes just before and after it.

   2. Note the lower limit l = 40, modal frequency f1 = 20, previous frequency f0 = 12, and next frequency f2 = 11.
   3. Calculate the class width h as 50 - 40 = 10.
   4. Apply the mode formula: Mode = l + ((f1 - f0) / (2f1 - f0 - f2)) × h.
      why?

      Mode = l + ((f₁ − f₀) / (2f₁ − f₀ − f₂)) × h, where the modal class has the highest frequency f₁, and f₀, f₂ are the frequencies of the classes just before and after it.

   5. Substitute the values: Mode = 40 + ((20 - 12) / (2 × 20 - 12 - 11)) × 10.
      why?

      Mode = l + ((f₁ − f₀) / (2f₁ − f₀ − f₂)) × h, where the modal class has the highest frequency f₁, and f₀, f₂ are the frequencies of the classes just before and after it.

   6. Simplify the numerator: 20 - 12 = 8.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   7. Simplify the denominator: 40 - 12 - 11 = 17.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   8. Calculate the fraction: 8 / 17 ≈ 0.4706.
   9. Multiply by class width: 0.4706 × 10 = 4.706.
   10. Add to lower limit: 40 + 4.706 = 44.706.
   11. Round to two decimal places to get 44.71.

   Answer: 44.71 cars

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