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NCERT Class 10 Mathematics Exercise 13.3 Solutions: Statistics

Step-by-step working for every question of Exercise 13.3 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 13.3 Q1

   Monthly electricity use of 68 consumers — 65-85, 85-105, 105-125, 125-145, 145-165, 165-185, 185-205 units with 4, 5, 13, 20, 14, 8, 4. Find the median, mean and mode and compare them.

   1. Identify the class intervals and frequencies: 65-85 (4), 85-105 (5), 105-125 (13), 125-145 (20), 145-165 (14), 165-185 (8), 185-205 (4).
   2. Calculate the cumulative frequencies: 4, 9, 22, 42, 56, 64, 68.
      why?

      For grouped data, Median = l + ((n/2 − cf) / f) × h, where the median class is the one holding the (n/2)th value, l, f and h are its lower limit, frequency and width, and cf is the cumulative frequency of the class before it.

   3. Find the median class where cumulative frequency is just greater than N/2 = 68/2 = 34, which is 125-145.
      why?

      For grouped data, Median = l + ((n/2 − cf) / f) × h, where the median class is the one holding the (n/2)th value, l, f and h are its lower limit, frequency and width, and cf is the cumulative frequency of the class before it.

   4. Calculate the median using the formula L + ((N/2 - cf) / f) * h, where L=125, cf=22, f=20, h=20, giving 125 + (12/20)*20 = 137.
      why?

      For grouped data, Median = l + ((n/2 − cf) / f) × h, where the median class is the one holding the (n/2)th value, l, f and h are its lower limit, frequency and width, and cf is the cumulative frequency of the class before it.

   5. Determine the modal class with the highest frequency 20, which is 125-145.
      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. Calculate the mode using the formula L + ((f1 - f0) / (2f1 - f0 - f2)) * h, where L=125, f1=20, f0=13, f2=14, h=20, giving 125 + (7/16)*20 = 135.77.
   7. Mean by the step-deviation method with assumed mean a = 135 and h = 20: uᵢ = (xᵢ − 135)/20 = −3, −2, −1, 0, 1, 2, 3, so Σfᵢuᵢ = −12 − 10 − 13 + 0 + 14 + 16 + 12 = 7 and N = 68.
      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. Mean = 135 + (7/68) × 20 = 135 + 2.06 = 137.06.
   9. Compare the results: Mean (137.06) is slightly higher than Median (137), which is higher than Mode (135.77), indicating a slight positive skew.

   Answer: median = 137 units; mean = 137.06 units; mode = 135.77 units

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2. Ex 13.3 Q2

   Classes 0-10, 10-20, 20-30, 30-40, 40-50, 50-60 have frequencies 5, x, 20, 15, y, 5 with total 60 and median 28.5. Find x and y.

   1. The total frequency is the sum of all class frequencies, so 5 + x + 20 + 15 + y + 5 = 60.
   2. Simplifying the equation gives x + y + 45 = 60, which means x + y = 15.
      why?

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

   3. The median is 28.5, which lies in the class interval 20-30, so this is the median class.
      why?

      For grouped data, Median = l + ((n/2 − cf) / f) × h, where the median class is the one holding the (n/2)th value, l, f and h are its lower limit, frequency and width, and cf is the cumulative frequency of the class before it.

   4. For the median class 20-30, the lower limit l is 20, the class size h is 10, and the frequency f is 20.
      why?

      For grouped data, Median = l + ((n/2 − cf) / f) × h, where the median class is the one holding the (n/2)th value, l, f and h are its lower limit, frequency and width, and cf is the cumulative frequency of the class before it.

   5. The cumulative frequency of the class preceding the median class is 5 + x.
      why?

      For grouped data, Median = l + ((n/2 − cf) / f) × h, where the median class is the one holding the (n/2)th value, l, f and h are its lower limit, frequency and width, and cf is the cumulative frequency of the class before it.

   6. The total frequency N is 60, so N/2 is 30.
   7. Using the median formula Median = l + (h × (N/2 - cf)) / f, substitute the values: 28.5 = 20 + (10 × (30 - (5 + x))) / 20.
      why?

      For grouped data, Median = l + ((n/2 − cf) / f) × h, where the median class is the one holding the (n/2)th value, l, f and h are its lower limit, frequency and width, and cf is the cumulative frequency of the class before it.

   8. Simplify the equation to 8.5 = (10 × (25 - x)) / 20, which becomes 8.5 = (25 - x) / 2.
      why?

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

   9. Multiply both sides by 2 to get 17 = 25 - x, so x = 25 - 17 = 8.
      why?

      Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

   10. Substitute x = 8 into the equation x + y = 15 to find y: 8 + y = 15, so y = 7.
       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.

   Answer: x = 8, y = 7

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3. Ex 13.3 Q3

   Ages of 100 policy holders (18 to below 60), cumulative: below 20: 2, below 25: 6, below 30: 24, below 35: 45, below 40: 78, below 45: 89, below 50: 92, below 55: 98, below 60: 100. Find the median age.

   1. The total number of policy holders is N = 100.
   2. The median position is N/2 = 100/2 = 50.
   3. The cumulative frequency just greater than 50 is 78, which corresponds to the class 35-40.
      why?

      For grouped data, Median = l + ((n/2 − cf) / f) × h, where the median class is the one holding the (n/2)th value, l, f and h are its lower limit, frequency and width, and cf is the cumulative frequency of the class before it.

   4. So, the median class is 35-40, with lower limit l = 35 and class size h = 5.
      why?

      For grouped data, Median = l + ((n/2 − cf) / f) × h, where the median class is the one holding the (n/2)th value, l, f and h are its lower limit, frequency and width, and cf is the cumulative frequency of the class before it.

   5. The cumulative frequency of the class preceding the median class is cf = 45.
      why?

      For grouped data, Median = l + ((n/2 − cf) / f) × h, where the median class is the one holding the (n/2)th value, l, f and h are its lower limit, frequency and width, and cf is the cumulative frequency of the class before it.

   6. Using the formula, Median = l + ((N/2 - cf)/f) × h.
      why?

      For grouped data, Median = l + ((n/2 − cf) / f) × h, where the median class is the one holding the (n/2)th value, l, f and h are its lower limit, frequency and width, and cf is the cumulative frequency of the class before it.

   7. Substituting the values, Median = 35 + ((50 - 45)/33) × 5.
      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. Calculating the term, (5/33) × 5 = 25/33 ≈ 0.7576.
   9. Adding this to the lower limit, 35 + 0.7576 = 35.7576.
   10. Rounding to two decimal places, the median age is 35.76 years.

   Answer: 35.76 years

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4. Ex 13.3 Q4

   Lengths (mm) of 40 leaves — 118-126, 127-135, 136-144, 145-153, 154-162, 163-171, 172-180 with 3, 5, 9, 12, 5, 4, 2. Find the median length (use continuous classes 117.5-126.5 and so on).

   1. Write the continuous class intervals: 117.5-126.5, 126.5-135.5, 135.5-144.5, 144.5-153.5, 153.5-162.5, 162.5-171.5, 171.5-180.5.
   2. List the frequencies: 3, 5, 9, 12, 5, 4, 2 and calculate the cumulative frequencies: 3, 8, 17, 29, 34, 38, 40.
      why?

      For grouped data, Median = l + ((n/2 − cf) / f) × h, where the median class is the one holding the (n/2)th value, l, f and h are its lower limit, frequency and width, and cf is the cumulative frequency of the class before it.

   3. Total number of observations n = 40, so n/2 = 20.
   4. The cumulative frequency just greater than 20 is 29, which corresponds to the class 144.5-153.5.
      why?

      For grouped data, Median = l + ((n/2 − cf) / f) × h, where the median class is the one holding the (n/2)th value, l, f and h are its lower limit, frequency and width, and cf is the cumulative frequency of the class before it.

   5. Identify the median class as 144.5-153.5, so l = 144.5, h = 9, cf = 17, and f = 12.
      why?

      For grouped data, Median = l + ((n/2 − cf) / f) × h, where the median class is the one holding the (n/2)th value, l, f and h are its lower limit, frequency and width, and cf is the cumulative frequency of the class before it.

   6. Use the median formula: Median = l + ((n/2 - cf)/f) × h.
      why?

      For grouped data, Median = l + ((n/2 − cf) / f) × h, where the median class is the one holding the (n/2)th value, l, f and h are its lower limit, frequency and width, and cf is the cumulative frequency of the class before it.

   7. Substitute the values: Median = 144.5 + ((20 - 17)/12) × 9.
      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 term: (3/12) × 9 = 0.25 × 9 = 2.25.
   9. Add to the lower limit: 144.5 + 2.25 = 146.75.

   Answer: 146.75 mm

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5. Ex 13.3 Q5

   Lifetimes (hours) of 400 neon lamps — 1500-2000, 2000-2500, 2500-3000, 3000-3500, 3500-4000, 4000-4500, 4500-5000 with 14, 56, 60, 86, 74, 62, 48. Find the median lifetime.

   1. Calculate cumulative frequencies: 14, 70, 130, 216, 290, 352, 400.
      why?

      For grouped data, Median = l + ((n/2 − cf) / f) × h, where the median class is the one holding the (n/2)th value, l, f and h are its lower limit, frequency and width, and cf is the cumulative frequency of the class before it.

   2. Identify the median class as 3000-3500 since the cumulative frequency 216 is the first one greater than N/2 = 200.
      why?

      For grouped data, Median = l + ((n/2 − cf) / f) × h, where the median class is the one holding the (n/2)th value, l, f and h are its lower limit, frequency and width, and cf is the cumulative frequency of the class before it.

   3. Determine the values: l = 3000, h = 500, cf = 130, and f = 86.
      why?

      For grouped data, Median = l + ((n/2 − cf) / f) × h, where the median class is the one holding the (n/2)th value, l, f and h are its lower limit, frequency and width, and cf is the cumulative frequency of the class before it.

   4. Apply the median formula: Median = l + ((N/2 - cf)/f) × h.
      why?

      For grouped data, Median = l + ((n/2 − cf) / f) × h, where the median class is the one holding the (n/2)th value, l, f and h are its lower limit, frequency and width, and cf is the cumulative frequency of the class before it.

   5. Substitute the values: 3000 + ((200 - 130)/86) × 500.
      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.

   6. Simplify the term inside the bracket: (70/86) × 500 ≈ 406.98.
      why?

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

   7. Add this to the lower limit: 3000 + 406.98 = 3406.98.

   Answer: 3406.98 hours

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6. Ex 13.3 Q6

   Letters in 100 surnames — 1-4, 4-7, 7-10, 10-13, 13-16, 16-19 with 6, 30, 40, 16, 4, 4. Find the median, the mean and the modal size.

   1. Identify the median class where the cumulative frequency exceeds N/2 = 50, which is the 7-10 class.
      why?

      For grouped data, Median = l + ((n/2 − cf) / f) × h, where the median class is the one holding the (n/2)th value, l, f and h are its lower limit, frequency and width, and cf is the cumulative frequency of the class before it.

   2. Calculate the median using the formula L + ((N/2 - cf) / f) * h, giving 7 + ((50 - 36) / 40) * 3 = 8.05.
      why?

      For grouped data, Median = l + ((n/2 − cf) / f) × h, where the median class is the one holding the (n/2)th value, l, f and h are its lower limit, frequency and width, and cf is the cumulative frequency of the class before it.

   3. Determine the mean by summing the products of class midpoints and frequencies, which is 832, and dividing by the total frequency 100.
      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.

   4. The mean is 832 / 100 = 8.32.
   5. Identify the modal class with the highest frequency, which is 7-10 with a frequency of 40.
      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. Calculate the mode using the formula L + ((f1 - f0) / (2f1 - f0 - f2)) * h, giving 7 + ((40 - 30) / (80 - 30 - 16)) * 3.
   7. Simplify the mode calculation to 7 + (10 / 34) * 3, which equals 7 + 30/34 or 7.88.
      why?

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

   Answer: median = 8.05, mean = 8.32, mode = 7.88

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7. Ex 13.3 Q7

   Weights (kg) of 30 students — 40-45, 45-50, 50-55, 55-60, 60-65, 65-70, 70-75 with 2, 3, 8, 6, 6, 3, 2. Find the median weight.

   1. Class intervals: 40-45, 45-50, 50-55, 55-60, 60-65, 65-70, 70-75
   2. Frequencies: 2, 3, 8, 6, 6, 3, 2
   3. Cumulative frequencies: 2, 5, 13, 19, 25, 28, 30
      why?

      For grouped data, Median = l + ((n/2 − cf) / f) × h, where the median class is the one holding the (n/2)th value, l, f and h are its lower limit, frequency and width, and cf is the cumulative frequency of the class before it.

   4. Total number of observations n = 30
   5. Median position = n/2 = 30/2 = 15
   6. The 15th observation lies in the class 55-60 (since cf 13 < 15 ≤ 19)
      why?

      For grouped data, Median = l + ((n/2 − cf) / f) × h, where the median class is the one holding the (n/2)th value, l, f and h are its lower limit, frequency and width, and cf is the cumulative frequency of the class before it.

   7. Median class is 55-60
      why?

      For grouped data, Median = l + ((n/2 − cf) / f) × h, where the median class is the one holding the (n/2)th value, l, f and h are its lower limit, frequency and width, and cf is the cumulative frequency of the class before it.

   8. Lower limit L = 55
   9. Cumulative frequency before median class cf = 13
      why?

      For grouped data, Median = l + ((n/2 − cf) / f) × h, where the median class is the one holding the (n/2)th value, l, f and h are its lower limit, frequency and width, and cf is the cumulative frequency of the class before it.

   10. Frequency of median class f = 6
       why?

       For grouped data, Median = l + ((n/2 − cf) / f) × h, where the median class is the one holding the (n/2)th value, l, f and h are its lower limit, frequency and width, and cf is the cumulative frequency of the class before it.

   11. Class width h = 5
   12. Median formula: L + (n/2 − cf)/f × h
       why?

       For grouped data, Median = l + ((n/2 − cf) / f) × h, where the median class is the one holding the (n/2)th value, l, f and h are its lower limit, frequency and width, and cf is the cumulative frequency of the class before it.

   13. Substitute values: 55 + (15 − 13)/6 × 5
       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.

   14. Calculate: 55 + (2/6) × 5
   15. Simplify: 55 + (1/3) × 5
       why?

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

   16. Result: 55 + 5/3 = 55 + 1.67 = 56.67

   Answer: 56.67 kg

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