Chapter 14: Statistics (NCERT Solutions)

Exercise 14.1: Mean of Grouped Data

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.
0-2 (1), 2-4 (2), 4-6 (1), 6-8 (5), 8-10 (6), 10-12 (2), 12-14 (3)

We use Direct Method as values are small.
Class Mark (x): 1, 3, 5, 7, 9, 11, 13.
f: 1, 2, 1, 5, 6, 2, 3. Σf = 20.
fx: 1, 6, 5, 35, 54, 22, 39.
Σfx = 1+6+5+35+54+22+39 = 162.
Mean = Σfx / Σf = 162 / 20 = 8.1 plants.

Q2. Consider the following distribution of daily wages of 50 workers of a factory.
100-120 (12), 120-140 (14), 140-160 (8), 160-180 (6), 180-200 (10). Find the mean daily wages.

Using Assumed Mean Method (a=150):
x: 110, 130, 150, 170, 190.
d = x - 150: -40, -20, 0, 20, 40.
f: 12, 14, 8, 6, 10. Σf = 50.
fd: -480, -280, 0, 120, 400.
Σfd = -240.
Mean = a + Σfd/Σf = 150 + (-240/50) = 150 - 4.8 = Rs 145.20.

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

Mean = 18.
x: 12, 14, 16, 18, 20, 22, 24.
f: 7, 6, 9, 13, f, 5, 4. Σf = 44 + f.
fx: 84, 84, 144, 234, 20f, 110, 96.
Σfx = 752 + 20f.
18 = (752 + 20f) / (44 + f).
18(44 + f) = 752 + 20f.
792 + 18f = 752 + 20f.
2f = 40 ⇒ f = 20.

Q4. Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.
65-68 (2), 68-71 (4), 71-74 (3), 74-77 (8), 77-80 (7), 80-83 (4), 83-86 (2)

Using Step Deviation (a=75.5, h=3):
x: 66.5, 69.5, 72.5, 75.5, 78.5, 81.5, 84.5.
u = (x-a)/h: -3, -2, -1, 0, 1, 2, 3.
f: 2, 4, 3, 8, 7, 4, 2. Σf = 30.
fu: -6, -8, -3, 0, 7, 8, 6. Σfu = 4.
Mean = 75.5 + (4/30)×3 = 75.5 + 0.4 = 75.9 beats/min.

Q5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.
50-52 (15), 53-55 (110), 56-58 (135), 59-61 (115), 62-64 (25)

Convert to continuous: 49.5-52.5, etc.
x: 51, 54, 57, 60, 63.
Using Step Deviation (a=57, h=3):
u: -2, -1, 0, 1, 2.
f: 15, 110, 135, 115, 25. Σf = 400.
fu: -30, -110, 0, 115, 50. Σfu = 25.
Mean = 57 + (25/400)×3 = 57 + 0.1875 = 57.19 mangoes.

Q6. The table below shows the daily expenditure on food of 25 households in a locality.
100-150 (4), 150-200 (5), 200-250 (12), 250-300 (2), 300-350 (2)

Using Step Deviation (a=225, h=50):
x: 125, 175, 225, 275, 325.
u: -2, -1, 0, 1, 2.
f: 4, 5, 12, 2, 2. Σf = 25.
fu: -8, -5, 0, 2, 4. Σfu = -7.
Mean = 225 + (-7/25)×50 = 225 - 14 = 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.
0.00-0.04 (4), 0.04-0.08 (9), 0.08-0.12 (9), 0.12-0.16 (2), 0.16-0.20 (4), 0.20-0.24 (2)

x: 0.02, 0.06, 0.10, 0.14, 0.18, 0.22.
f: 4, 9, 9, 2, 4, 2. Σf = 30.
Direct method: Σfx = 0.08 + 0.54 + 0.90 + 0.28 + 0.72 + 0.44 = 2.96.
Mean = 2.96 / 30 = 0.0986... = 0.099 ppm.

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

Class sizes vary, use Direct Method.
x: 3, 8, 12, 17, 24, 33, 39.
f: 11, 10, 7, 4, 4, 3, 1.
fx: 33, 80, 84, 68, 96, 99, 39.
Σfx = 499. Σf = 40.
Mean = 499 / 40 = 12.475 days.

Q9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
45-55 (3), 55-65 (10), 65-75 (11), 75-85 (8), 85-95 (3)

Step Deviation (a=70, h=10):
u: -2, -1, 0, 1, 2.
f: 3, 10, 11, 8, 3. Σf = 35.
fu: -6, -10, 0, 8, 6. Σfu = -2.
Mean = 70 + (-2/35)×10 = 70 - 0.57 = 69.43 %.

Exercise 14.2: Mode of Grouped Data

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

Mode: Max freq = 23 (Class 35-45).
l=35, f1=23, f0=21, f2=14, h=10.
Mode = l + [(f1-f0)/(2f1-f0-f2)]×h
= 35 + [(2)/ (46-21-14)]×10
= 35 + (20/11) = 35 + 1.81 = 36.8 years.
Mean (Step dev): 35.37 years.

Q2. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components. Determine the modal lifetimes.
0-20 (10), 20-40 (35), 40-60 (52), 60-80 (61), 80-100 (38), 100-120 (29)

Max freq = 61 (Class 60-80).
l=60, f1=61, f0=52, f2=38, h=20.
Mode = 60 + [(9)/(122-52-38)]×20
= 60 + (9/32)×20 = 60 + 5.625 = 65.625 hours.

Q3. 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.
3000-4000 (4), 4000-5000 (18), 5000-6000 (9), 6000-7000 (7), 7000-8000 (6), 8000-9000 (3), 9000-10000 (1), 10000-11000 (1)

Max freq = 18 (Class 4000-5000).
l=4000, f1=18, f0=4, f2=9, h=1000.
Mode = 4000 + [(14)/(36-4-9)]×1000
= 4000 + (14000/23) = 4000 + 608.7 = 4608.7 runs.

Q4. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean.
15-20 (3), 20-25 (8), 25-30 (9), 30-35 (10), 35-40 (3), 40-45 (0), 45-50 (0), 50-55 (2)

Mode: Max freq = 10 (Class 30-35).
l=30, f1=10, f0=9, f2=3, h=5.
Mode = 30 + [(1)/(20-9-3)]×5 = 30 + 5/8 = 30.6.
Mean: 29.2.

Q5. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches. (Data same type as Q3). Find the mode.

[Repeated concept] Find max freq, identity modal class, apply formula.

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

Max freq = 20 (Class 40-50).
l=40, f1=20, f0=12, f2=11, h=10.
Mode = 40 + [(8)/(40-12-11)]×10
= 40 + (80/17) = 40 + 4.7 = 44.7 cars.

Exercise 14.3: Median of Grouped Data

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

N=68, N/2=34.
cf: 4, 9, 22, 42 (>34), 56, 64, 68.
Median Class: 125-145. l=125, cf=22, f=20, h=20.
Median = 125 + [(34-22)/20]×20 = 125 + 12 = 137 units.
Mean = 137.05 units.
Mode = 135.76 units.

Q2. If the median of the distribution given below is 28.5, find the values of x and y. Total frequency = 60.
0-10 (5), 10-20 (x), 20-30 (20), 30-40 (15), 40-50 (y), 50-60 (5)

N=60. 45+x+y = 60 ⇒ x+y = 15.
Median = 28.5 (Class 20-30). l=20, f=20, cf=5+x, h=10.
28.5 = 20 + [(30 - (5+x))/20]×10.
8.5 = (25-x)/2.
17 = 25 - x ⇒ x = 8.
y = 15 - 8 = 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.
Below 20 (2), Below 25 (6), Below 30 (24)... (Cumulative Frequency given)

Convert to classes: 15-20 (2), 20-25 (4), 25-30 (18), etc.
N=100, N/2=50.
Median Class: 35-40.
l=35, cf=45, f=33, h=5.
Median = 35 + [(50-45)/33]×5 = 35 + 0.76 = 35.76 years.

Q4. The lengths of 40 leaves of a plant are measured correct to the nearest millimetre... Find the median length. (118-126, 127-135...)

Make continuous: 117.5-126.5, etc.
N=40, N/2=20.
Median Class: 144.5-153.5.
Median = 144.5 + [(20-17)/12]×9 = 144.5 + 2.25 = 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.

Median = 3406.98 hours.

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

Median = 8.05 letters.

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

Median = 56.67 kg.

Statistics - RD Sharma Important Questions

Q1. If the mean of the following distribution is 6, find the value of p.
x: 2, 4, 6, 10, p+5
f: 3, 2, 3, 1, 2

Σf = 11.
Σfx = 6 + 8 + 18 + 10 + 2(p+5) = 42 + 2p + 10 = 52 + 2p.
Mean = Σfx / Σf ⇒ 6 = (52+2p)/11.
66 = 52 + 2p.
2p = 14 ⇒ p = 7.

Q2. The mean of the following frequency distribution is 53. But the frequencies f1 and f2 in the classes 20-40 and 60-80 are missing. Find the missing frequencies.
0-20 (15), 20-40 (f1), 40-60 (21), 60-80 (f2), 80-100 (17). Total = 100.

15 + f1 + 21 + f2 + 17 = 100 ⇒ f1 + f2 = 47.
x: 10, 30, 50, 70, 90.
Σfx = 150 + 30f1 + 1050 + 70f2 + 1530 = 2730 + 30f1 + 70f2.
Mean = 53 ⇒ 5300 = 2730 + 30f1 + 70f2.
30f1 + 70f2 = 2570 ⇒ 3f1 + 7f2 = 257.
Solving simultaneously with f1 + f2 = 47:
f1 = 18, f2 = 29.

Q3. Compute the mode for the following frequency distribution.
Less than 20 (4), Less than 40 (12), Less than 60 (35), Less than 80 (56), Less than 100 (84), Less than 120 (100).

Convert to classes:
0-20 (4), 20-40 (8), 40-60 (23), 60-80 (21), 80-100 (28), 100-120 (16).
Max freq = 28 (Class 80-100).
Mode = 80 + [(28-21)/(56-21-16)]×20
= 80 + (7/19)×20 = 80 + 7.37 = 87.37.

Q4. Calculate the median from the following data.
Rent: 15-25 (8), 25-35 (10), 35-45 (15), 45-55 (25), 55-65 (40), 65-75 (20), 75-85 (15), 85-95 (7).

N = 140. N/2 = 70.
cf check: 8, 18, 33, 58, 98 (Class 55-65).
l=55, f=40, cf=58, h=10.
Median = 55 + [(70-58)/40]×10
= 55 + 3 = 58.

Q5. Find the mean, mode and median of the following data:
Class: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70
Freq: 5, 8, 15, 20, 14, 8, 5

Mean = 35.6
Mode = 33.33
Median = 34.5
Verify Empirical Relationship: 3 Median = Mode + 2 Mean.
3(34.5) = 103.5.
33.33 + 2(35.6) = 33.33 + 71.2 = 104.53 (Approx holds).

Q6-Q15. [Additional problems on missing frequencies in median and mode, step-deviation method, and empirical relationship]

Key concepts:
• Missing frequency problems usually require two equations.
• Empirical Formula: 3 Median = Mode + 2 Mean.
• Cumulative frequency curves (Ogives).

Statistics - Formulas & PYQs

Key Formulas

1. Mean (&xmacr;)

Direct Method: &xmacr; = Σf_ix_i / Σf_i

Assumed Mean Method: &xmacr; = a + Σf_id_i / Σf_i (where d = x - a)

Step Deviation Method: &xmacr; = a + (Σf_iu_i / Σf_i) × h (where u = (x-a)/h)

2. Mode

Mode = l + [(f₁ - f₀) / (2f₁ - f₀ - f₂)] × h

l = lower limit of modal class
f₁ = frequency of modal class
f₀ = frequency of preceding class
f₂ = frequency of succeeding class
h = class size

3. Median

Median = l + [(N/2 - cf) / f] × h

N = sum of frequencies
cf = cumulative frequency of preceding class
f = frequency of median class

4. Empirical Relationship

3 Median = Mode + 2 Mean

Previous Year Questions (CBSE/JKBOSE)

Q1. Find the mean of the following distribution: (CBSE 2020)
Class: 3-5, 5-7, 7-9, 9-11, 11-13
Freq: 5, 10, 10, 7, 8

x: 4, 6, 8, 10, 12.
fx: 20, 60, 80, 70, 96. Sum = 326.
N = 40.
Mean = 326/40 = 8.15.

Q2. If the mode of a distribution is 8 and its mean is 8, then find its median. (CBSE 2019)

3 Median = Mode + 2 Mean.
3 Median = 8 + 16 = 24.
Median = 8.

Q3. The median of the following data is 525. Find the values of x and y, if the total frequency is 100. (CBSE 2018)

Standard missing frequency problem suitable for 5 marks. x=9, y=15 (values vary based on exact data).

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