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NCERT Class 10 Mathematics Exercise 13.1 Solutions: Statistics

AI-written, checked on 23 Oct 2026

Solutions written by Shishya, checked — not the board's official answers

Step-by-step working for every question of Exercise 13.1 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.

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  1. Ex 13.1 Q1

    Plants in 20 houses — plants 0-2, 2-4, 4-6, 6-8, 8-10, 10-12, 12-14 with 1, 2, 1, 5, 6, 2, 3 houses. Find the mean number of plants per house and say which method you used and why.

    1. Identify the class intervals and their corresponding frequencies from the given data.
    2. Calculate the midpoint (x) for each class interval: 1, 3, 5, 7, 9, 11, and 13.
    3. Multiply each midpoint by its frequency to find the product fx for each class.
    4. Sum the frequencies to get total frequency N = 1 + 2 + 1 + 5 + 6 + 2 + 3 = 20.
    5. Sum the products fx to get total sum = 1 + 6 + 5 + 35 + 54 + 22 + 39 = 162.
    6. Use the direct method for mean since the class intervals are small and data is not large.
    7. Calculate the mean using the formula Mean = Sum(fx) / N = 162 / 20.
    8. The calculated mean is 8.1.

    Answer: mean = 8.1

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  2. Ex 13.1 Q2

    Daily wages of 50 workers — ₹500-520, 520-540, 540-560, 560-580, 580-600 with 12, 14, 8, 6, 10 workers. Find the mean daily wage.

    1. Identify the class intervals and their corresponding frequencies: 500-520 (12), 520-540 (14), 540-560 (8), 560-580 (6), 580-600 (10).
    2. Calculate the midpoint (x) for each class interval: 510, 530, 550, 570, 590.
    3. Multiply each midpoint by its frequency to get fx: 6120, 7420, 4400, 3420, 5900.
    4. Sum the frequencies to get N = 50 and sum the fx values to get 27260.
    5. Calculate the mean using the formula Mean = Sum of fx / N = 27260 / 50 = 545.2.

    Answer: ₹545.2

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  3. Ex 13.1 Q3

    Daily pocket allowance — ₹11-13, 13-15, 15-17, 17-19, 19-21, 21-23, 23-25 with 7, 6, 9, 13, f, 5, 4 children; the mean is ₹18. Find f.

    1. Class marks (mid-points) xᵢ: 12, 14, 16, 18, 20, 22, 24 with frequencies 7, 6, 9, 13, f, 5, 4.
      why?

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

    2. Σf = 7 + 6 + 9 + 13 + f + 5 + 4 = 44 + f.
    3. Σfx = 7×12 + 6×14 + 9×16 + 13×18 + 20f + 5×22 + 4×24 = 84 + 84 + 144 + 234 + 20f + 110 + 96 = 752 + 20f.
    4. Mean = Σfx/Σf, so 18 = (752 + 20f)/(44 + f).
      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.

    5. 18(44 + f) = 752 + 20f gives 792 + 18f = 752 + 20f, so 2f = 40 and f = 20.

    Answer: f = 20

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  4. Ex 13.1 Q4

    Heartbeats per minute of 30 women — 65-68, 68-71, 71-74, 74-77, 77-80, 80-83, 83-86 with 2, 4, 3, 8, 7, 4, 2 women. Find the mean.

    1. Class intervals are 65-68, 68-71, 71-74, 74-77, 77-80, 80-83, 83-86 with frequencies 2, 4, 3, 8, 7, 4, 2.
    2. Calculate the midpoint (x) for each interval: 66.5, 69.5, 72.5, 75.5, 78.5, 81.5, 84.5.
    3. Multiply each midpoint by its frequency (fx): 133, 278, 217.5, 604, 549.5, 326, 169.
    4. Sum the frequencies (f) to get 30 and sum the products (fx) to get 2277.
    5. Apply the mean formula: Mean = Sum(fx) / Sum(f) = 2277 / 30.
    6. Divide 2277 by 30 to get 75.9.

    Answer: 75.9

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  5. Ex 13.1 Q5

    Mangoes per box — 50-52, 53-55, 56-58, 59-61, 62-64 with 15, 110, 135, 115, 25 boxes. Find the mean number of mangoes per box and name the method.

    1. Use the step deviation method with assumed mean l = 57 and class size h = 3.
      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.

    2. Calculate the midpoint x_i for each class: 51, 54, 57, 60, 63.
      why?

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

    3. Find the deviations d_i = (x_i - l) / h: -2, -1, 0, 1, 2.
    4. Compute the products f_i d_i: -30, -110, 0, 115, 50.
    5. Sum the frequencies: 15 + 110 + 135 + 115 + 25 = 400.
    6. Sum the products: -30 - 110 + 0 + 115 + 50 = 25.
    7. Apply the formula: mean = l + (sum of f_i d_i / sum of f_i) * h.
      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.

    8. Substitute values: 57 + (25 / 400) * 3.
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

    9. Calculate the result: 57 + 0.1875 = 57.1875.
    10. Round to two decimal places to get 57.19.

    Answer: The mean number of mangoes per box is 57.19, found using the step deviation method.

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  6. Ex 13.1 Q6

    Daily food expenditure of 25 households — ₹100-150, 150-200, 200-250, 250-300, 300-350 with 4, 5, 12, 2, 2 households. Find the mean.

    1. Identify the class intervals and their corresponding frequencies from the given data.
    2. Calculate the midpoint (x) for each class interval: 125, 175, 225, 275, and 325.
    3. Multiply each midpoint by its frequency (f) to get fx: 500, 875, 2700, 550, and 650.
    4. Sum the frequencies to find N = 4 + 5 + 12 + 2 + 2 = 25.
    5. Sum the fx values to find Σfx = 500 + 875 + 2700 + 550 + 650 = 5275.
    6. Apply the mean formula: Mean = Σfx / N.
      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.

    7. Substitute the values: Mean = 5275 / 25.
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

    8. Calculate the final result: 5275 ÷ 25 = 211.

    Answer: ₹211

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  7. Ex 13.1 Q7

    SO₂ concentration (ppm) in 30 localities — 0.00-0.04, 0.04-0.08, 0.08-0.12, 0.12-0.16, 0.16-0.20, 0.20-0.24 with frequencies 4, 9, 9, 2, 4, 2. Find the mean concentration.

    1. Identify the class intervals and their corresponding frequencies from the given data.
    2. Calculate the midpoint (x) for each class interval by adding the lower and upper limits and dividing by 2.
    3. The midpoints are 0.02, 0.06, 0.10, 0.14, 0.18, and 0.22 respectively.
    4. Multiply each midpoint (x) by its frequency (f) to get fx for each class.
    5. The values of fx are 0.08, 0.54, 0.90, 0.28, 0.72, and 0.44.
    6. Sum the frequencies to find the total number of localities, which is 4 + 9 + 9 + 2 + 4 + 2 = 30.
    7. Sum the products fx to get the total sum, which is 0.08 + 0.54 + 0.90 + 0.28 + 0.72 + 0.44 = 2.96.
    8. Calculate the mean by dividing the sum of fx by the sum of frequencies, which is 2.96 ÷ 30.
      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.

    9. The mean concentration is 0.09866..., which rounds to 0.099 ppm.

    Answer: 0.099 ppm

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  8. Ex 13.1 Q8

    Days absent for 40 students — 0-6, 6-10, 10-14, 14-20, 20-28, 28-38, 38-40 with 11, 10, 7, 4, 4, 3, 1 students. Find the mean number of days absent.

    1. Identify the class intervals and their corresponding frequencies from the given data.
    2. Calculate the class mark (midpoint) for each interval by adding the lower and upper limits and dividing by 2.
      why?

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

    3. The class marks are 3, 8, 12, 17, 24, 33, and 39 respectively.
      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. Multiply each class mark by its frequency to get the product (f × x).
      why?

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

    5. The products are 33, 80, 84, 68, 96, 99, and 39.
    6. Sum the frequencies to find the total number of students, which is 40.
    7. Sum the products to find the total number of days absent, which is 499.
    8. Divide the total days absent by the total number of students to calculate the mean.
    9. The mean is 499 ÷ 40 = 12.475.

    Answer: 12.475 days

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  9. Ex 13.1 Q9

    Literacy rate (%) of 35 cities — 45-55, 55-65, 65-75, 75-85, 85-95 with 3, 10, 11, 8, 3 cities. Find the mean literacy rate.

    1. Identify the class intervals and frequencies: 45-55 (3), 55-65 (10), 65-75 (11), 75-85 (8), 85-95 (3).
    2. Calculate the class marks (midpoints) for each interval: 50, 60, 70, 80, 90.
      why?

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

    3. Multiply each class mark by its frequency to get fx: 150, 600, 770, 640, 270.
      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. Sum the frequencies to find N: 3 + 10 + 11 + 8 + 3 = 35.
    5. Sum the products fx to find the total: 150 + 600 + 770 + 640 + 270 = 2430.
    6. Apply the mean formula: Mean = (Sum of fx) / N = 2430 / 35.
    7. Compute the division: 2430 ÷ 35 ≈ 69.43.

    Answer: 69.43%

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