STATISTICS

EXERCISE 13.1

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.

Which method did you use for finding the mean, and why?

Solution:

We use the direct method to calculate the mean.

1. First, find the mid-point (x) of each class:

• For 0–2, x = (0 + 2)/2 = 1

• For 2–4, x = (2 + 4)/2 = 3

• For 4–6, x = (4 + 6)/2 = 5

• For 6–8, x = (6 + 8)/2 = 7

• For 8–10, x = (8 + 10)/2 = 9

• For 10–12, x = (10 + 12)/2 = 11

• For 12–14, x = (12 + 14)/2 = 13

2. Multiply frequency (f) with mid-point (x) to get f × x.

Class Interval Mid-point (x) Frequency (f) f × x

0–2 1 1 1

2–4 3 2 6

4–6 5 1 5

6–8 7 5 35

8–10 9 6 54

10–12 11 2 22

12–14 13 3 39

3. Now, add frequencies and f × x values:

• Σf = 1 + 2 + 1 + 5 + 6 + 2 + 3 = 20

• Σf × x = 1 + 6 + 5 + 35 + 54 + 22 + 39 = 162

4. Mean = Σf × x / Σf = 162 ÷ 20 = 8.1

Final Answer:

The mean number of plants per house = 8.1

Method used: Direct Method, because the data values are simple and the calculation can be done directly without assuming deviations.

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Q2. Consider the following distribution (refer book) 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.

Solution:

We will use the direct method to calculate the mean.

1. Find the mid-point (x) of each class:

• For 500–520, x = (500 + 520)/2 = 510

• For 520–540, x = (520 + 540)/2 = 530

• For 540–560, x = (540 + 560)/2 = 550

• For 560–580, x = (560 + 580)/2 = 570

• For 580–600, x = (580 + 600)/2 = 590

2. Multiply frequency (f) with mid-point (x) to get f × x.

Daily wages (₹) Mid-point (x) No. of workers (f) f × x

500–520 510 12 6120

520–540 530 14 7420

540–560 550 8 4400

560–580 570 6 3420

580–600 590 10 5900

3. Add frequencies and f × x values:

• Σf = 12 + 14 + 8 + 6 + 10 = 50

• Σf × x = 6120 + 7420 + 4400 + 3420 + 5900 = 27,260

4. Mean = Σf × x ÷ Σf = 27,260 ÷ 50 = 545.2

Final Answer:

The mean daily wages of the workers = ₹545.20

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Q3: The following distribution (refer book) shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹18. Find the missing frequency f.

Solution:

We use the direct method to calculate the mean.

1. Find the mid-point (x) of each class:

• 11–13 → 12

• 13–15 → 14

• 15–17 → 16

• 17–19 → 18

• 19–21 → 20

• 21–23 → 22

• 23–25 → 24

2. Multiply frequency (f) with mid-point (x) to get f × x.

Daily allowance (₹) Mid-point (x) Frequency (f) f × x

11–13 12 7 84

13–15 14 6 84

15–17 16 9 144

17–19 18 13 234

19–21 20 f 20f

21–23 22 5 110

23–25 24 4 96

3. Totals:

• Σf = 7 + 6 + 9 + 13 + f + 5 + 4 = 44 + f

• Σf × x = 84 + 84 + 144 + 234 + 20f + 110 + 96 = 752 + 20f

4. Mean formula:

Mean = (Σf × x) ÷ (Σf)

18 = (752 + 20f) ÷ (44 + f)

5. Solve:

18(44 + f) = 752 + 20f

792 + 18f = 752 + 20f

792 – 752 = 20f – 18f

40 = 2f

f = 20

Final Answer:

The missing frequency = 20

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Q4: Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarized as follows (refer book). Find the mean heartbeats per minute for these women, choosing a suitable method.

Solution:

We use the assumed mean method to simplify the calculation.

1. Find the mid-point (x) of each class:

• 65–68 → 66.5

• 68–71 → 69.5

• 71–74 → 72.5

• 74–77 → 75.5

• 77–80 → 78.5

• 80–83 → 81.5

• 83–86 → 84.5

2. Choose an assumed mean (A).

Let A = 75.5 (the mid-point of the central class 74–77).

Class size (h) = 3

3. Calculate deviation d = (x – A)/h and then f × d.

Heartbeats (per min) Mid-point (x) f d = (x – 75.5)/3 f × d

65–68 66.5 2 -3 -6

68–71 69.5 4 -2 -8

71–74 72.5 3 -1 -3

74–77 75.5 8 0 0

77–80 78.5 7 +1 +7

80–83 81.5 4 +2 +8

83–86 84.5 2 +3 +6

4. Totals:

• Σf = 30

• Σf × d = (-6) + (-8) + (-3) + 0 + 7 + 8 + 6 = 4

5. Formula (Assumed mean method):

Mean = A + (Σf × d / Σf) × h

= 75.5 + (4/30) × 3

= 75.5 + 0.4

= 75.9

Final Answer:

The mean number of heartbeats per minute = 75.9

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Q5: In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following (refer book) was the distribution of mangoes according to the number of boxes. Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Solution:

We will use the assumed mean method because the numbers are large and deviations make the calculation easier.

1. Find the mid-point (x) of each class:

• 50–52 → 51

• 53–55 → 54

• 56–58 → 57

• 59–61 → 60

• 62–64 → 63

2. Choose an assumed mean (A).

Let A = 57 (the central value).

Class size (h) = 3

3. Calculate deviation d = (x – A)/h and then f × d.

No.of mangoes Mid-point (x) f d = (x – 57)/3 f × d

50–52 51 15 -2 -30

53–55 54 110 -1 -110

56–58 57 135 0 0

59–61 60 115 +1 115

62–64 63 25 +2 50

4. Totals:

• Σf = 15 + 110 + 135 + 115 + 25 = 400

• Σf × d = -30 - 110 + 0 + 115 + 50 = 25

5. Formula (Assumed mean method):

Mean = A + (Σf × d / Σf) × h

= 57 + (25/400) × 3

= 57 + 0.1875

= 57.19

Final Answer:

The mean number of mangoes in a packing box = 57.2 (approx.)

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

Solution:

We will use the assumed mean method for easy calculation.

1. Find the mid-point (x) of each class:

• 100–150 → 125

• 150–200 → 175

• 200–250 → 225

• 250–300 → 275

• 300–350 → 325

2. Choose an assumed mean (A).

Let A = 225 (the central value).

Class size (h) = 50

3. Calculate deviation d = (x – A)/h and then f × d.

Expenditure (₹) Mid-point (x) f d = (x – 225)/50 f × d

100–150 125 4 -2 -8

150–200 175 5 -1 -5

200–250 225 12 0 0

250–300 275 2 +1 +2

300–350 325 2 +2 +4

4. Totals:

• Σf = 25

• Σf × d = -8 - 5 + 0 + 2 + 4 = -7

5. Formula (Assumed mean method):

Mean = A + (Σf × d / Σf) × h

= 225 + (-7/25) × 50

= 225 - 14

= 211

Final Answer:

The mean daily expenditure on food = ₹211

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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 and is presented below (refer book). Find the mean concentration of SO₂ in the air.

Solution:

We will use the assumed mean method for easier calculation.

1. Find the mid-point (x) of each class:

• 0.00–0.04 → 0.02

• 0.04–0.08 → 0.06

• 0.08–0.12 → 0.10

• 0.12–0.16 → 0.14

• 0.16–0.20 → 0.18

• 0.20–0.24 → 0.22

2. Choose an assumed mean (A).

Let A = 0.10 (the central value).

Class size (h) = 0.04

3. Calculate deviation d = (x – A)/h and then f × d.

Concentration (ppm) Mid-point (x) f d = (x – 0.10)/0.04 f × d

0.00–0.04 0.02 4 -2 -8

0.04–0.08 0.06 9 -1 -9

0.08–0.12 0.10 9 0 0

0.12–0.16 0.14 2 +1 +2

0.16–0.20 0.18 4 +2 +8

0.20–0.24 0.22 2 +3 +6

4. Totals:

• Σf = 30

• Σf × d = -8 - 9 + 0 + 2 + 8 + 6 = -1

5. Formula (Assumed mean method):

Mean = A + (Σf × d / Σf) × h

= 0.10 + (-1/30) × 0.04

= 0.10 - 0.0013

= 0.0987

Final Answer:

The mean concentration of SO₂ in the air = 0.099 ppm (approx.)

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

Solution:

Step 1: Find the midpoints

For each class interval, midpoint = (lower limit + upper limit) ÷ 2

Class Interval Mid-point (x) f f × x

0–6 3 11 33

6–10 8 10 80

10–14 12 7 84

14–20 17 4 68

20–28 24 4 96

28–38 33 3 99

38–40 39 1 39

Step 2: Totals

• Total frequency (Σf) = 40

• Total f × x = 499

Step 3: Mean formula

Mean = (Σ f × x) ÷ (Σ f)

= 499 ÷ 40

= 12.475

Final Answer

The mean number of days a student was absent = 12.48 days (approx).

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

Solution:

We will use the Assumed Mean Method.

1. Find mid-points (x):

• 45–55 → 50

• 55–65 → 60

• 65–75 → 70

• 75–85 → 80

• 85–95 → 90

2. Choose an assumed mean (A).

Let A = 70 (middle value).

Class size (h) = 10.

3. Prepare the table:

Literacy rate (x) f d = (x – 70)/10 f × d

50 3 -2 -6

60 10 -1 -10

70 11 0 0

80 8 1 8

90 3 2 6

4. Totals:

Σf = 35

Σ(f × d) = -2

5. Mean formula (Assumed Mean Method):

Mean = A + (Σf × d / Σf) × h

Mean = 70 + (-2 / 35) × 10

Mean = 70 – 0.57

Mean = 69.43

Final Answer:

The mean literacy rate of the cities = 69.43% (approx.)

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EXERCISE 13.2

Q1: The following table (refer book) 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.

Solution:

Step 1: Find the Mean (using Assumed Mean method)

1. Class size (h) = 10.

2. Midpoints (x):

• 5–15 → 10

• 15–25 → 20

• 25–35 → 30

• 35–45 → 40

• 45–55 → 50

• 55–65 → 60

3. Choose assumed mean A = 30.

4. Prepare table:

Class Interval f x d = (x–30)/10 f × d

5–15 6 10 -2 -12

15–25 11 20 -1 -11

25–35 21 30 0 0

35–45 23 40 1 23

45–55 14 50 2 28

55–65 5 60 3 15

Totals:

Σf = 80

Σ(f × d) = 43

5. Formula:

Mean = A + (Σf × d / Σf) × h

= 30 + (43 / 80) × 10

= 30 + 5.375

= 35.38 years

Step 2: Find the Mode

Formula:

Mode = L + [(f₁ – f₀) / (2f₁ – f₀ – f₂)] × h

Here,

• Modal class = 35–45 (highest frequency = 23)

• L = 35

• f₁ = 23 (frequency of modal class)

• f₀ = 21 (frequency before modal class)

• f₂ = 14 (frequency after modal class)

• h = 10

Substitute:

Mode = 35 + [(23 – 21) / (2×23 – 21 – 14)] × 10

= 35 + [2 / (46 – 35)] × 10

= 35 + (2/11) × 10

= 35 + 1.82

= 36.82 years

Final Answer:

• Mean age = 35.38 years

• Mode age = 36.82 years

Comparison and Interpretation:

• The mean gives the average age of patients.

• The mode shows the most frequently occurring age group of patients (35–45 years).

• Since both values are close (35.38 ≈ 36.82), the data is fairly symmetrical around the central values.

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

Solution:

Step 1: Identify the modal class

The highest frequency is 61 (class 60–80).

So, modal class = 60–80.

Step 2: Formula for Mode

Mode = L + [(f₁ – f₀) / (2f₁ – f₀ – f₂)] × h

Where,

• L = 60 (lower boundary of modal class)

• h = 20 (class size)

• f₁ = 61 (frequency of modal class)

• f₀ = 52 (frequency before modal class)

• f₂ = 38 (frequency after modal class)

Step 3: Substitute values

Mode = 60 + [(61 – 52) / (2 × 61 – 52 – 38)] × 20

= 60 + [9 / (122 – 90)] × 20

= 60 + [9 / 32] × 20

= 60 + 180 / 32

= 60 + 5.625

= 65.625 hours

Final Answer:

The modal lifetime of the components = 65.625 hours

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Q3: The following data (refer book) gives the distribution of total monthly household expenditure of 200 families of a village. Find:

1. The modal monthly expenditure

2. The mean monthly expenditure

Solution:

1. Modal Monthly Expenditure

Step 1: Identify modal class

The highest frequency is 40 in class 1500–2000.

So, modal class = 1500–2000.

Step 2: Formula for Mode

Mode = L + [(f₁ – f₀) / (2f₁ – f₀ – f₂)] × h

Where:

• L = 1500 (lower limit of modal class)

• h = 500 (class size)

• f₁ = 40 (frequency of modal class)

• f₀ = 24 (frequency before modal class)

• f₂ = 33 (frequency after modal class)

Step 3: Substitute values

Mode = 1500 + [(40 – 24) / (2×40 – 24 – 33)] × 500

= 1500 + [16 / (80 – 57)] × 500

= 1500 + [16 / 23] × 500

= 1500 + 347.83

≈ 1847.83

Modal monthly expenditure = ₹1847.83 (approx.)

2. Mean Monthly Expenditure

We use the assumed mean method.

Expenditure (₹) Frequency (f) Midpoint (x) d = (x–A)/h f·d

1000–1500 24 1250 –2 –48

1500–2000 40 1750 –1 –40

2000–2500 33 2250 0 0

2500–3000 28 2750 1 28

3000–3500 30 3250 2 60

3500–4000 22 3750 3 66

4000–4500 16 4250 4 64

4500–5000 7 4750 5 35

Total 200 165

Where,

• Assumed mean (A) = 2250

• Class size (h) = 500

• Σf = 200

• Σf·d = 165

Formula:

Mean = A + (Σf·d / Σf) × h

= 2250 + (165 / 200) × 500

= 2250 + 412.5

= 2662.5

Mean monthly expenditure = ₹2662.5

Final Answer:

• Modal monthly expenditure = ₹1847.83

• Mean monthly expenditure = ₹2662.5

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Q4: The following distribution (refer book) 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.

Total states / U.T. = 35

Solution - Mode

1. Modal class = class with highest frequency = 30–35 (f₁ = 10).

f₀ = frequency of previous class = 9 (for 25–30).

f₂ = frequency of next class = 3 (for 35–40).

Lower limit L of modal class = 30.

Class size h = 5.

2. Mode formula (grouped data):

Mode = L + [(f₁ − f₀) / (2f₁ − f₀ − f₂)] × h

3. Substitute values:

Mode = 30 + [(10 − 9) / (2×10 − 9 − 3)] × 5

= 30 + [1 / 8] × 5

= 30 + 0.625

= 30.625

Mode ≈ 30.63 students per teacher

Solution — Mean (Assumed Mean Method)

1. Mid-points x of classes and f given:

Class Mid-point x f

15–20 17.5 3

20–25 22.5 8

25–30 27.5 9

30–35 32.5 10

35–40 37.5 3

40–45 42.5 0

45–50 47.5 0

50–55 52.5 2

Choose assumed mean A = 32.5 (mid-point of 30–35). Class size h = 5.

2. Compute d = (x − A)/h and f × d:

x f d = (x−32.5)/5 f × d

17.5 3 -3 -9

22.5 8 -2 -16

27.5 9 -1 -9

32.5 10 0 0

37.5 3 1 3

42.5 0 2 0

47.5 0 3 0

52.5 2 4 8

Σf = 35

Σ(f × d) = -23

3. Mean = A + (Σf × d / Σf) × h

= 32.5 + (-23 / 35) × 5

= 32.5 − 3.285714...

= 29.2142857

Mean ≈ 29.21 students per teacher

Interpretation and comparison

• Mean ≈ 29.21 is the average number of students per teacher across the states/UTs.

• Mode ≈ 30.63 indicates the most typical (most frequently occurring) students-per-teacher value (around the 30–35 class).

• The mean is slightly less than the mode (29.21 < 30.63), which suggests the distribution is slightly negatively skewed (a few lower class-values pull the average down a bit).

• Both measures are close, so central tendency lies around 29–31 students per teacher; mode highlights the most common class, while mean gives the overall average.

Final values: Mode = 30.63 (approx.), Mean = 29.21 (approx.)

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Q5: The given distribution (refer book) 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.

Solution:

1. Identify the modal class (class with highest frequency).

Highest frequency = 18, so modal class = 4000 – 5000.

2. Note values needed for the mode formula:

L = lower limit of modal class = 4000

h = class width = 1000

f₁ = frequency of modal class = 18

f₀ = frequency of class before modal class = 4 (3000–4000)

f₂ = frequency of class after modal class = 9 (5000–6000)

3. Use the formula for mode for grouped data:

Mode = L + [(f₁ − f₀) / (2f₁ − f₀ − f₂)] × h

4. Substitute values and calculate:

f₁ − f₀ = 18 − 4 = 14

2f₁ − f₀ − f₂ = 2×18 − 4 − 9 = 36 − 13 = 23

(f₁ − f₀) / (2f₁ − f₀ − f₂) = 14 / 23 ≈ 0.60869565

Multiply by h: 0.60869565 × 1000 ≈ 608.69565

Mode = 4000 + 608.69565 = 4608.69565

5. Round appropriately: Mode ≈ 4608.7 runs

Final Answer:

The modal runs scored ≈ 4608.7 runs.

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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 below(refer book). Find the mode of the data.

Solution:

1. Identify the modal class (class with highest frequency).

Highest frequency = 20, so modal class = 40 – 50.

2. Note values for formula:

L = lower limit of modal class = 40

h = class width = 10

f₁ = frequency of modal class = 20

f₀ = frequency of class before modal class = 12 (30–40)

f₂ = frequency of class after modal class = 11 (50–60)

3. Mode formula:

Mode = L + [(f₁ − f₀) / (2f₁ − f₀ − f₂)] × h

4. Substitute values:

f₁ − f₀ = 20 − 12 = 8

2f₁ − f₀ − f₂ = 40 − 12 − 11 = 17

(f₁ − f₀) / (2f₁ − f₀ − f₂) = 8 / 17 ≈ 0.4706

Multiply by h: 0.4706 × 10 = 4.706

Mode = 40 + 4.706 ≈ 44.7

Final Answer:

The modal number of cars passing through the spot = 44.7 cars

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EXERCISE 13.3

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

Solution

All class widths h = 20.

1. FIND THE MEAN (direct table method)

Mid-points x and f × x:

Class Mid-point x f f × x

65–85 75 4 300

85–105 95 5 475

105–125 115 13 1495

125–145 135 20 2700

145–165 155 14 2170

165–185 175 8 1400

185–205 195 4 780

Total 68 9320

Σf = 68, Σ(f x) = 9320

Mean = Σ(f x) ÷ Σf = 9320 ÷ 68 = 137.0588235…

Mean ≈ 137.06 units

2. FIND THE MEDIAN

n = 68, n/2 = 34. Cumulative frequencies:

• up to 65–85: 4

• up to 85–105: 9

• up to 105–125: 22

• up to 125–145: 42 ← first cumulative ≥ 34, so median class = 125–145

For median class 125–145:

• L = 125 (lower limit)

• cfb = cumulative frequency before median class = 22

• fm = frequency of median class = 20

• h = 20

Median = L + [(n/2 − cfb) / fm] × h

= 125 + [(34 − 22) / 20] × 20

= 125 + (12/20) × 20

= 125 + 12

= 137 units

3. FIND THE MODE

Modal class = class with highest frequency = 125–145 (fm = 20).

Let f0 = frequency of previous class = 13 (105–125)

Let f2 = frequency of next class = 14 (145–165)

L = 125, h = 20

Mode = L + [(fm − f0) / (2fm − f0 − f2)] × h

= 125 + [(20 − 13) / (40 − 13 − 14)] × 20

= 125 + [7 / 13] × 20

= 125 + 10.7692307…

Mode ≈ 135.77 units

Final Answers:

• Median = 137.00 units

• Mean = 137.06 units

• Mode = 135.77 units

Comparison and interpretation:

• All three measures are very close (mean ≈ 137.06, median = 137, mode ≈ 135.77), so the distribution is roughly symmetric around the centre.

• Mode < Median < Mean (mode slightly smallest, mean slightly largest) — this indicates a very slight positive (right) skew, but the skew is negligible given how close the values are.

• Median (137) is a good measure of the “typical” monthly consumption; mean (137.06) gives the exact average; mode (135.77) shows the most frequent central value.

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Q2: If the median of the distribution given below (refer book) is 28.5, find the values of x and y.

Solution:

1. Total frequency = 60, so

5 + x + 20 + 15 + y + 5 = 60

⇒ x + y + 45 = 60

⇒ x + y = 15. ......(1)

2. Median position = n/2 = 60/2 = 30.

The cumulative frequencies from the start:

o up to 0–10 = 5

o up to 10–20 = 5 + x

o up to 20–30 = 5 + x + 20 = 25 + x

For the median (30th value) to lie in class 20–30 we need 25 + x ≥ 30 and 5 + x < 30. This will hold for reasonable x; proceed assuming median class = 20–30.

3. Use the median formula for grouped data:

Median = L + [(n/2 − cfb) / fm] × h

Where for class 20–30:

o L = 20 (lower limit of median class)

o cfb = cumulative frequency before median class = 5 + x

o fm = frequency of median class = 20

o h = class width = 10

o n/2 = 30

Given Median = 28.5

Substitute:

28.5 = 20 + [(30 − (5 + x)) / 20] × 10

Simplify:

8.5 = (25 − x)/2

17 = 25 − x

x = 8

4. From (1), x + y = 15 ⇒ 8 + y = 15 ⇒ y = 7.

Answer:

x = 8 and y = 7.

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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 years.

Solution:

Step 1: Convert cumulative frequency into class intervals

We subtract successive cumulative frequencies:

Class Interval Frequency (f)

18 – 20 2

20 – 25 4

25 – 30 18

30 – 35 21

35 – 40 33

40 – 45 11

45 – 50 3

50 – 55 6

55 – 60 2

Step 2: Prepare cumulative frequency (CF)

Class Interval f CF

18 – 20 2 2

20 – 25 4 6

25 – 30 18 24

30 – 35 21 45

35 – 40 33 78

40 – 45 11 89

45 – 50 3 92

50 – 55 6 98

55 – 60 2 100

Step 3: Find the median class

• Total frequency, n = 100

• Median position = n/2 = 50

• From the CF column:

o 45 < 50 ≤ 78, so the median class is 35 – 40.

Step 4: Apply median formula

Median = L + [(n/2 – cfb) / f] × h

Here:

• L = 35 (lower limit of median class)

• n/2 = 50

• cfb = 45 (cumulative frequency before median class)

• f = 33 (frequency of median class)

• h = 5 (class width)

Median = 35 + [(50 – 45) / 33] × 5

= 35 + (5/33) × 5

= 35 + 25/33

= 35 + 0.76

≈ 35.8

Final Answer:

The median age is 35.8 years.

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Q4: The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table (refer book). Find the median length of the leaves.

(Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 – 126.5, 126.5 – 135.5, …, 171.5 – 180.5.)

Solution:

First, convert the classes into continuous intervals:

Class interval Frequency (f) Cumulative frequency (CF)

117.5 – 126.5 3 3

126.5 – 135.5 5 8

135.5 – 144.5 9 17

144.5 – 153.5 12 29

153.5 – 162.5 5 34

162.5 – 171.5 4 38

171.5 – 180.5 2 40

Total number of leaves (n) = 40.

So, n/2 = 20.

The 20th observation lies in the class interval 144.5 – 153.5.

Therefore, the median class = 144.5 – 153.5.

Median formula:

Median = l + [(n/2 – C) / f] × h

Where:

• l = lower boundary of the median class = 144.5

• C = cumulative frequency before the median class = 17

• f = frequency of the median class = 12

• h = class size = 9

Calculation:

Median = 144.5 + [(20 – 17) / 12] × 9

= 144.5 + (3/12 × 9)

= 144.5 + 2.25

= 146.75 mm

Final Answer:

The median length of the leaves is 146.75 mm.

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Q5: The following table (refer book) gives the distribution of the life time of 400 neon lamps. Find the median life time of a lamp.

Solution:

First convert the classes into continuous intervals (assuming class boundaries half a unit beyond each end):

Class interval (continuous) Frequency (f) Cumulative frequency (CF)

1499.5 – 2000.5 14 14

2000.5 – 2500.5 56 70

2500.5 – 3000.5 60 130

3000.5 – 3500.5 86 216

3500.5 – 4000.5 74 290

4000.5 – 4500.5 62 352

4500.5 – 5000.5 48 400

Total number of lamps, n = 400.

So n/2 = 200.

The 200th observation falls in the class 3000.5 – 3500.5 (because cumulative frequency just before this class is 130 and up to this class is 216).

Therefore median class = 3000.5 – 3500.5.

Use the grouped median formula:

Median = l + [(n/2 – C) / f] × h

Where

l = lower boundary of median class = 3000.5

C = cumulative frequency before median class = 130

f = frequency of median class = 86

h = class width = 500

Calculation:

n/2 – C = 200 – 130 = 70

(70 / 86) × 500 ≈ 0.81395349 × 500 ≈ 406.97674

Median = 3000.5 + 406.97674 ≈ 3407.47674

Final answer (rounded to two decimal places): Median life time = 3407.48 hours

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Q6: 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the surnames was obtained as follows (refer book):

Determine the median, the mean, and the modal size of the surnames.

Solution:

Step 1: Median

Total number of surnames = 100 → n/2 = 50

The cumulative frequencies are:

Class interval Frequency (f) Cumulative frequency (CF)

1 – 4 6 6

4 – 7 30 36

7 – 10 40 76

10 – 13 16 92

13 – 16 4 96

16 – 19 4 100

The 50th value lies in the class 7 – 10.

Here,

• Median class = 7 – 10

• Lower boundary (l) = 7

• Cumulative frequency before this class (C) = 36

• Frequency of median class (f) = 40

• Class size (h) = 3

Median = l + [(n/2 – C) / f] × h

= 7 + [(50 – 36) / 40] × 3

= 7 + (14 / 40) × 3

= 7 + 1.05

= 8.05

Step 2: Mean

We find the midpoints of each class:

Class interval Midpoint (x) Frequency (f) f × x

1 – 4 2.5 6 15

4 – 7 5.5 30 165

7 – 10 8.5 40 340

10 – 13 11.5 16 184

13 – 16 14.5 4 58

16 – 19 17.5 4 70

Σ f × x = 832

Σ f = 100

Mean = Σ f × x / Σ f = 832 / 100 = 8.32

Step 3: Mode

The class with the highest frequency is 7 – 10. This is the modal class.

Here,

• l = 7

• h = 3

• fₘ (frequency of modal class) = 40

• f₁ (frequency of class before modal class) = 30

• f₂ (frequency of class after modal class) = 16

Mode = l + [(fₘ – f₁) / (2fₘ – f₁ – f₂)] × h

= 7 + [(40 – 30) / (80 – 30 – 16)] × 3

= 7 + (10 / 34) × 3

= 7 + 0.88

= 7.88

Final Answer

• Median number of letters = 8.05

• Mean number of letters = 8.32

• Modal size of surnames = 7.88

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Q7: The distribution below (refer book) gives the weights of 30 students of a class. Find the median weight of the students.

Solution:

First, we find the cumulative frequencies.

Weight (in kg) No. of students Cum. frequency

40–45 2 2

45–50 3 5

50–55 8 13

55–60 6 19

60–65 6 25

65–70 3 28

70–75 2 30

Total number of students (N) = 30

Median class = the class where the cumulative frequency ≥ N/2 = 30/2 = 15.

So, the median class = 55–60.

Here,

• Lower boundary (L) = 55

• Cumulative frequency before median class (CF) = 13

• Frequency of median class (f) = 6

• Class width (h) = 5

Median = L + [(N/2 – CF) / f] × h

= 55 + [(15 – 13) / 6] × 5

= 55 + (2/6) × 5

= 55 + 1.67

= 56.67

Answer:

Median weight = 56.67 kg

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