Chapter 13: Statistics Class 10 Maths NCERT Solutions

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Contents

Class 10 Maths NCERT Solutions - Chapter wise

1

Chapter 1: Real Numbers

2

Chapter 2: Polynomials

3

Chapter 3 : Pair of Linear Equations in Two Variables

4

Chapter 4 : Quadratic Equations

5

Chapter 5 : Arithmetic Progressions

6

Chapter 6 : Triangles

7

Chapter 7 : Coordinate Geometry

8

Chapter 8: Introduction to Trigonometry

9

Chapter 9: Some Applications of Trigonometry

10

Chapter 10 : Circles

11

Chapter 11 : Areas Related to Circles

12

Chapter 12: Surface Areas and Volumes

13

Chapter 13: Statistics

14

Chapter 14 : Probability

Chapter Notes

Chapter 13: Statistics — Exercise 13.1

Welcome to the science of data! Statistics is how we turn a messy pile of numbers into meaningful stories—like predicting weather patterns or understanding the average marks of a class to help everyone improve.

Q1. A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

(Data: 0-2, 2-4, 4-6, 6-8, 8-10, 10-12, 12-14 with frequencies 1, 2, 1, 5, 6, 2, 3 respectively.)

  • The Logic: Since the numbers here are very small, we use the Direct Method. We first find the “Class Mark” (the middle value of each range) and then multiply it by the number of houses to find the total plants.

  • The Steps:

    1. Find Class Marks (xi): (Upper Limit + Lower Limit) / 2. For 0-2, xi = 1; for 2-4, xi = 3, and so on.

    2. Multiply (fi * xi):

      • 1 * 1 = 1

      • 2 * 3 = 6

      • 1 * 5 = 5

      • 5 * 7 = 35

      • 6 * 9 = 54

      • 2 * 11 = 22

      • 3 * 13 = 39

    3. Sum them up: Sum of fi * xi = 162. Sum of fi = 20.

    4. Mean formula: Mean = (Sum of fi * xi) / (Sum of fi) = 162 / 20.

  • The Result: The mean number of plants per house is 8.1.

Q2. Consider the following distribution of daily wages of 50 workers of a factory. Find the mean daily wages of the workers of the factory by using an appropriate method.

(Data: 500-520, 520-540, 540-560, 560-580, 580-600 with frequencies 12, 14, 8, 6, 10.)

  • The Logic: The numbers here (500, 520, etc.) are large. If we use the Direct Method, the multiplication will be huge! Pro-Tip: Use the Assumed Mean Method to “shrink” the numbers by picking a middle value as our “guess” mean.

  • The Steps:

    1. Class Marks (xi): 510, 530, 550, 570, 590.

    2. Pick Assumed Mean (a): Let’s pick the middle one, a = 550.

    3. Find Deviation (di = xi – a): -40, -20, 0, 20, 40.

    4. Multiply (fi * di):

      • 12 * -40 = -480

      • 14 * -20 = -280

      • 8 * 0 = 0

      • 6 * 20 = 120

      • 10 * 40 = 400

    5. Sum them: Sum of fi * di = -240. Sum of fi = 50.

    6. Mean formula: Mean = a + (Sum of fi * di / Sum of fi) = 550 + (-240 / 50) = 550 – 4.8.

  • The Result: The mean daily wage is Rs 545.20.

Q3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.

(Data: 11-13, 13-15, 15-17, 17-19, 19-21, 21-23, 23-25 with frequencies 7, 6, 9, 13, f, 5, 4.)

  • The Logic: This is a “Reverse Mean” problem. We are given the answer (Mean = 18) and need to find a missing part of the data. We use the Direct Method and set up an algebraic equation.

  • The Steps:

    1. Class Marks (xi): 12, 14, 16, 18, 20, 22, 24.

    2. Sum of fi * xi: (712) + (614) + (916) + (1318) + (f20) + (522) + (4*24) = 84 + 84 + 144 + 234 + 20f + 110 + 96 = 752 + 20f.

    3. Sum of fi: 7 + 6 + 9 + 13 + f + 5 + 4 = 44 + f.

    4. Equation: 18 = (752 + 20f) / (44 + f).

    5. Solve: 18(44 + f) = 752 + 20f => 792 + 18f = 752 + 20f => 40 = 2f.

  • The Result: The missing frequency f is 20.

Q4. Thirty women were examined in a hospital by a doctor… Find the mean heartbeats per minute for these women.

(Data: 65-68, 68-71, 71-74, 74-77, 77-80, 80-83, 83-86 with frequencies 2, 4, 3, 8, 7, 4, 2.)

  • The Logic: Since the class intervals are consistent (gap of 3), the Step-Deviation Method is the most elegant way to solve this. It reduces the values to the smallest possible integers (-3, -2, -1, 0…).

  • The Steps:

    1. Class Marks (xi): 66.5, 69.5, 72.5, 75.5, 78.5, 81.5, 84.5.

    2. Assumed Mean (a) = 75.5. Class height (h) = 3.

    3. Find ui = (xi – a) / h: -3, -2, -1, 0, 1, 2, 3.

    4. Sum of fi * ui: (2*-3) + (4*-2) + (3*-1) + (80) + (71) + (42) + (23) = -6 – 8 – 3 + 0 + 7 + 8 + 6 = 4.

    5. Mean formula: Mean = a + [h * (Sum fi*ui / Sum fi)] = 75.5 + [3 * (4 / 30)] = 75.5 + 0.4.

  • The Result: The mean heartbeats per minute is 75.9.

Q5. In a retail market, fruit vendors were selling mangoes kept in packing boxes… Find the mean number of mangoes kept in a packing box.

(Data: 50-52, 53-55, 56-58, 59-61, 62-64 with frequencies 15, 110, 135, 115, 25.)

  • Human Tip: Look closely! The classes are discontinuous (50-52, then 53-55). There is a gap of 1. While for Mean you can calculate directly using the mid-point, it’s a good habit to recognize this for Median and Mode.

  • The Steps:

    1. Class Marks (xi): (50+52)/2 = 51, 54, 57, 60, 63.

    2. Assumed Mean (a) = 57. h = 3.

    3. ui values: -2, -1, 0, 1, 2.

    4. fi * ui: -30, -110, 0, 115, 50. Sum = 25. Sum of fi = 400.

    5. Mean: 57 + [3 * (25 / 400)] = 57 + 0.1875.

  • The Result: The mean number of mangoes is 57.19.

Q6. The table below shows the daily expenditure on food of 25 households in a locality. Find the mean daily expenditure on food by a suitable method.

(Data: 100-150, 150-200, 200-250, 250-300, 300-350 with frequencies 4, 5, 12, 2, 2.)

  • The Logic: The expenditures are in hundreds, but the frequencies are small. We use the Step-Deviation Method because the class height (h = 50) is a nice round number that will simplify our math perfectly.

  • The Steps:

    1. Class Marks (xi): 125, 175, 225, 275, 325.

    2. Assumed Mean (a): 225. Height (h): 50.

    3. Find ui = (xi – a) / h: -2, -1, 0, 1, 2.

    4. Multiply (fi * ui):

      • 4 * -2 = -8

      • 5 * -1 = -5

      • 12 * 0 = 0

      • 2 * 1 = 2

      • 2 * 2 = 4

    5. Sum them up: Sum of fi * ui = -7. Sum of fi = 25.

    6. Mean formula: 225 + [50 * (-7 / 25)] = 225 + [2 * -7] = 225 – 14.

  • The Result: The mean daily expenditure on food is Rs 211.

Q7. To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city… Find the mean concentration of SO2 in the air.

(Data: 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.)

  • The Logic: Don’t let the decimals scare you! Even though they look small, we can treat them like whole numbers using the Assumed Mean Method. It prevents simple addition errors that often happen with many decimal points.

  • The Steps:

    1. Class Marks (xi): 0.02, 0.06, 0.10, 0.14, 0.18, 0.22.

    2. Assumed Mean (a): 0.10 (or 0.14). Let’s go with 0.10.

    3. Multiply (fi * xi) using Direct Method: (Since numbers are still quite small, Direct works well here too).

      • (4 * 0.02) + (9 * 0.06) + (9 * 0.10) + (2 * 0.14) + (4 * 0.18) + (2 * 0.22)

      • 0.08 + 0.54 + 0.90 + 0.28 + 0.72 + 0.44 = 2.96.

    4. Mean: 2.96 / 30 = 0.09866…

  • The Result: The mean concentration of SO2 is 0.099 ppm (approx).

Q8. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

(Data: 0-6, 6-10, 10-14, 14-20, 20-28, 28-38, 38-40 with frequencies 11, 10, 7, 4, 4, 3, 1.)

  • Human Tip: Alert! The class intervals are unequal (gaps of 6, 4, 4, 6, 8, 10, 2). Because the height (h) is not constant, you cannot use the Step-Deviation method. Stick to the Direct Method or Assumed Mean.

  • The Steps:

    1. Class Marks (xi): 3, 8, 12, 17, 24, 33, 39.

    2. fi * xi:

      • 11 * 3 = 33

      • 10 * 8 = 80

      • 7 * 12 = 84

      • 4 * 17 = 68

      • 4 * 24 = 96

      • 3 * 33 = 99

      • 1 * 39 = 39

    3. Sum of fi * xi: 499. Sum of fi = 40.

    4. Mean: 499 / 40 = 12.475.

  • The Result: The mean number of days absent is 12.48 days.

Q9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

(Data: 45-55, 55-65, 65-75, 75-85, 85-95 with frequencies 3, 10, 11, 8, 3.)

  • The Logic: This is a perfect candidate for the Step-Deviation Method. The intervals are equal (h = 10) and the midpoints end in 0 or 5, making the reduction very clean.

  • The Steps:

    1. Class Marks (xi): 50, 60, 70, 80, 90.

    2. Assumed Mean (a) = 70. h = 10.

    3. ui values: -2, -1, 0, 1, 2.

    4. fi * ui: (3 * -2) + (10 * -1) + (11 * 0) + (8 * 1) + (3 * 2) = -6 – 10 + 0 + 8 + 6 = -2.

    5. Mean: 70 + [10 * (-2 / 35)] = 70 – (20 / 35) = 70 – 0.57.

  • The Result: The mean literacy rate is 69.43%.

Chapter 13: Statistics — Exercise 13.2

Welcome to the “Mode” section! While the Mean gives us an average, the Mode tells us what is most popular—like finding the best-selling shoe size or the most frequent heartbeat range in a hospital.

Q1. The following table shows the ages of the patients admitted in a hospital during a year. Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

(Data: 5-15, 15-25, 25-35, 35-45, 45-55, 55-65 with frequencies 6, 11, 21, 23, 14, 5.)

  • The Logic: To find the Mode, we don’t need to do heavy calculations for every row. We only care about the Modal Class—the one with the highest frequency. In this case, 23 is the highest frequency, so our answer will lie between 35 and 45.

  • The Steps:

    1. Identify the Modal Class: Frequency 23 is highest, so class is 35-45.

    2. Values: Lower limit (l) = 35, Frequency of modal class (f1) = 23, Frequency before (f0) = 21, Frequency after (f2) = 14, Class width (h) = 10.

    3. Formula: Mode = l + [(f1 – f0) / (2f1 – f0 – f2)] * h

    4. Calculation: 35 + [(23 – 21) / (2*23 – 21 – 14)] * 10

    5. 35 + [2 / (46 – 35)] * 10 = 35 + (20 / 11) = 35 + 1.81.

    6. Mean Calculation: Using the method from 13.1, Mean = 35.37 years.

  • The Result: The Mode is 36.8 years and the Mean is 35.37 years. This means the maximum number of patients are aged 36.8, while the average age is 35.37.

Q2. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components. Determine the modal lifetimes of the components.

(Data: 0-20, 20-40, 40-60, 60-80, 80-100, 100-120 with frequencies 10, 35, 52, 61, 38, 29.)

  • The Logic: We are looking for the “peak” lifetime. We spot the highest frequency (61) and apply the formula to narrow down the exact hour within that 60-80 range.

  • The Steps:

    1. Modal Class: 60-80 (since 61 is the highest frequency).

    2. Values: l = 60, f1 = 61, f0 = 52, f2 = 38, h = 20.

    3. Calculation: 60 + [(61 – 52) / (2*61 – 52 – 38)] * 20

    4. 60 + [9 / (122 – 90)] * 20 = 60 + (9 / 32) * 20

    5. 60 + (180 / 32) = 60 + 5.625.

  • The Result: The modal lifetime of the components is 65.625 hours.

Q3. The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure.

(Data: 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.)

  • Human Tip: When the frequencies are large, double-check your subtraction in the denominator! $2f_1 – f_0 – f_2$ is where most students make a calculation slip.

  • The Steps:

    1. Modal Class: 1500-2000 (Frequency 40 is highest).

    2. Values: l = 1500, f1 = 40, f0 = 24, f2 = 33, h = 500.

    3. Calculation: 1500 + [(40 – 24) / (2*40 – 24 – 33)] * 500

    4. 1500 + [16 / (80 – 57)] * 500 = 1500 + (16 / 23) * 500

    5. 1500 + (8000 / 23) = 1500 + 347.83.

  • The Result: The modal monthly expenditure is Rs 1847.83.

Q4. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

(Data: 15-20, 20-25, 25-30, 30-35, 35-40, 40-45, 45-50, 50-55 with frequencies 3, 8, 9, 10, 3, 0, 0, 2.)

  • The Logic: This data is interesting because of the “0” frequencies. It doesn’t change our method; it just means those classes have no impact on the mode.

  • The Steps:

    1. Modal Class: 30-35 (Frequency 10).

    2. Values: l = 30, f1 = 10, f0 = 9, f2 = 3, h = 5.

    3. Calculation: 30 + [(10 – 9) / (2*10 – 9 – 3)] * 5

    4. 30 + [1 / (20 – 12)] * 5 = 30 + (1/8) * 5 = 30 + 0.625.

  • The Result: The Mode is 30.6 and the Mean is 29.2. Most states have a ratio of 30.6 students per teacher.

Q5. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches. Find the mode of the data.

(Data: 3000-4000, 4000-5000, 5000-6000, 6000-7000, 7000-8000, 8000-9000, 9000-10000, 10000-11000 with frequencies 4, 18, 9, 7, 6, 3, 1, 1.)

  • Pro-Tip: In cricket stats, the mode tells us which “scoring bracket” most of these top batsmen fall into.

  • The Steps:

    1. Modal Class: 4000-5000 (Frequency 18).

    2. Values: l = 4000, f1 = 18, f0 = 4, f2 = 9, h = 1000.

    3. Calculation: 4000 + [(18 – 4) / (2*18 – 4 – 9)] * 1000

    4. 4000 + [14 / (36 – 13)] * 1000 = 4000 + (14/23) * 1000 = 4000 + 608.7.

  • The Result: The mode of the data is 4608.7 runs.

Q6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data.

(Data: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70, 70-80 with frequencies 7, 14, 13, 12, 20, 11, 15, 8.)

  • The Logic: Last one! We identify the peak car count period. 20 is the highest frequency, so the mode is in the 40-50 range.

  • The Steps:

    1. Modal Class: 40-50.

    2. Values: l = 40, f1 = 20, f0 = 12, f2 = 11, h = 10.

    3. Calculation: 40 + [(20 – 12) / (2*20 – 12 – 11)] * 10

    4. 40 + [8 / (40 – 23)] * 10 = 40 + (80 / 17) = 40 + 4.7.

  • The Result: The mode of the data is 44.7 cars.

Chapter 13: Statistics — Exercise 13.3

Welcome to the “Median” masterclass! If the Mean is the average and the Mode is the popularity contest, the Median is the “Middle Ground”—it tells us exactly what value sits in the center of our data, which is super useful for things like calculating middle-class income or fair housing prices.

Q1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

(Data: 65-85, 85-105, 105-125, 125-145, 145-165, 165-185, 185-205 with frequencies 4, 5, 13, 20, 14, 8, 4.)

  • The Logic: To find the Median, we first need to calculate Cumulative Frequency (cf)—which is just a running total of the frequencies. We then find the middle consumer (n/2) to identify our Median Class.

  • The Steps:

    1. Calculate cf: 4, 9, 22, 42, 56, 64, 68. Total (n) = 68.

    2. Find Median Class: n/2 = 68/2 = 34. The first cf greater than 34 is 42. So, Median Class is 125-145.

    3. Values: Lower limit (l) = 125, n/2 = 34, cf of preceding class = 22, frequency of median class (f) = 20, class width (h) = 20.

    4. Formula: Median = l + [((n/2) – cf) / f] * h

    5. Calculation: 125 + [(34 – 22) / 20] * 20 = 125 + 12 = 137.

  • The Result: The Median is 137 units. (Mean = 137.05, Mode = 135.76).

Q2. If the median of the distribution given below is 28.5, find the values of x and y.

(Data: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60 with frequencies 5, x, 20, 15, y, 5. Total = 60.)

  • The Logic: This is a high-weightage “Missing Frequency” problem. We know the Median is 28.5, so the Median Class must be 20-30. We set up two equations: one from the total sum and one from the Median formula.

  • The Steps:

    1. Equation 1 (Total): 5 + x + 20 + 15 + y + 5 = 60 => x + y = 15.

    2. Cumulative Frequency: 5, 5+x, 25+x, 40+x, 40+x+y, 45+x+y.

    3. Apply Median Formula: 28.5 = 20 + [(30 – (5 + x)) / 20] * 10.

    4. 8.5 = (25 – x) / 2 => 17 = 25 – x => x = 8.

    5. Substitute x in Eq 1: 8 + y = 15 => y = 7.

  • The Result: x = 8 and y = 7.

Q3. A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.

(Data: Below 20, 25, 30, 35, 40, 45, 50, 55, 60 with cf 2, 6, 24, 45, 78, 89, 92, 98, 100.)

  • Human Tip: Look out! The question gives you Cumulative Frequency already (it keeps increasing). You need to subtract them to find the actual frequencies (fi) first.

  • The Steps:

    1. Convert to Class Intervals: 15-20, 20-25… 55-60.

    2. Find fi: 2, (6-2)=4, (24-6)=18, (45-24)=21, (78-45)=33, etc.

    3. Median Class: n/2 = 50. Median class is 35-40 (where cf 78 is).

    4. Calculation: 35 + [(50 – 45) / 33] * 5 = 35 + (5/33)*5 = 35 + 0.757.

  • The Result: The median age is 35.76 years.

Q4. The lengths of 40 leaves of a plant are measured correct to the nearest millimetre… Find the median length of the leaves.

(Data: 118-126, 127-135, 136-144, 145-153, 154-162, 163-171, 172-180 with frequencies 3, 5, 9, 12, 5, 4, 2.)

  • Pro-Tip: The classes are discontinuous (126 ends, 127 starts). For Median, you must make them continuous by subtracting 0.5 from lower limits and adding 0.5 to upper limits.

  • The Steps:

    1. New Intervals: 117.5-126.5, 126.5-135.5, 135.5-144.5, 144.5-153.5…

    2. cf: 3, 8, 17, 29, 34, 38, 40.

    3. Median Class: n/2 = 20. Median class is 144.5-153.5.

    4. Calculation: 144.5 + [(20 – 17) / 12] * 9 = 144.5 + (3/12)*9 = 144.5 + 2.25.

  • The Result: The median length is 146.75 mm.

Q5. The following table gives the distribution of the life time of 400 neon lamps. Find the median life time of a lamp.

(Data: 1500-2000… 4500-5000 with frequencies 14, 56, 60, 86, 74, 62, 48.)

  • The Logic: Large numbers but simple steps. Standard median calculation.

  • The Steps:

    1. cf: 14, 70, 130, 216, 290, 352, 400.

    2. Median Class: n/2 = 200. Median class is 3000-3500.

    3. Calculation: 3000 + [(200 – 130) / 86] * 500 = 3000 + (70/86)*500 = 3000 + 406.97.

  • The Result: The median life time is 3406.98 hours.

Q6. 100 surnames were randomly picked up from a local telephone directory… Determine the median number of letters in the surnames.

(Data: 1-4, 4-7, 7-10, 10-13, 13-16, 16-19 with frequencies 6, 30, 40, 16, 4, 4.)

  • The Steps:

    1. cf: 6, 36, 76, 92, 96, 100.

    2. Median Class: n/2 = 50. Median class is 7-10.

    3. Calculation: 7 + [(50 – 36) / 40] * 3 = 7 + (14/40)*3 = 7 + 1.05.

  • The Result: The median number of letters is 8.05.

Q7. The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

(Data: 40-45… 70-75 with frequencies 2, 3, 8, 6, 6, 3, 2.)

  • The Steps:

    1. cf: 2, 5, 13, 19, 25, 28, 30.

    2. Median Class: n/2 = 15. Median class is 55-60.

    3. Calculation: 55 + [(15 – 13) / 6] * 5 = 55 + (2/6)*5 = 55 + 1.666.

  • The Result: The median weight is 56.67 kg.

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