NCERT Solutions Class 10 Maths Chapter 14 Statistics Exercise 14.4

NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.4 help students understand and solve questions based on finding the mode of grouped data. This exercise introduces an important statistical measure used to determine the value that occurs most frequently in a data set.

Prepared as per the latest CBSE Class 10 Maths syllabus, Exercise 14.4 focuses on calculating the mode using the modal class and mode formula. This concept is important for board exams as it tests understanding of frequency distribution and formula application.

NCERT Solutions Class 10 Maths Chapter 14 Statistics Exercise 14.4

Statistics

Medium

Q.

The following table gives the distribution of the life time of 400 neon lamps:

Life time (in hours)	Number of lamps
1500– 2000 2000– 2500 2500–3000 3000–3500 3500–4000 4000–4500 4500–5000	14 56 60 86 74 62 48

Find the median life time of a lamp.

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Statistics

Medium

Q.

100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

Number of letters	1-4	4-7	7-10	10-13	13-16	16-19
Number of surnames	6	30	40	16	4	4

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.

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Statistics

Medium

Q.

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

Weight (in kg)	40-45	45-50	50-55	55-60	60-65	65-70	70-75
Number of students	2	3	8	6	6	3	2

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Statistics

Medium

Q.

The following distribution gives the daily income of 50 workers of a factory.

Daily income  (in ₹)	100-120	120-140	140-160	160-180	180-200
Number of workers	12	14	8	6	10

Convert the distribution above to a less than type cumulative frequency distribution, and draw its graph.

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Statistics

Medium

Q.

During the medical check-up of 35 students of a class, their weights were recorded as follows:

Weight (in kg)	Number of students
Less than 38 Less than 40 Less than 42 Less than 44 Less than 46 Less than 48 Less than 50 Less than 52	0 3 5 9 14 28 32 35

Draw a less than type graph for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.

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Statistics

Medium

Q.

The following table gives production yield per hectare of wheat of 100 farms of a village.

Production yield  (in kg/ha)	50-55	55-60	60-65	65-70	70-75	75-80
Number of farms	2	8	12	24	38	16

Change the distribution to a more than type distribution, and draw its graph.

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The solutions are presented in a clear, step-by-step format so students can easily identify the modal class, apply the formula correctly, and avoid calculation errors in exams.

Q1. The following distribution gives the daily income of 50 workers of a factory. Convert it to a less than type cumulative frequency distribution and draw its ogive.

Difficulty Level: Medium
Known: Daily income distribution of 50 workers
Unknown: Less than type cumulative frequency table + ogive points

Given Frequency Distribution
Daily income (₹): 100–120, 120–140, 140–160, 160–180, 180–200
No. of workers: 12, 14, 8, 6, 10

Reasoning

Cumulative frequency curve (ogive) banane ke liye less than type cumulative frequencies nikalte hain.
Ogive mein upper class limits x-axis par aur cumulative frequencies y-axis par li jaati hain.

Solution

(A) Less than type cumulative frequency distribution

✅ Less than type cumulative frequency table ready.

(B) Ogive (plotting points)

Ogive ke points (Upper class limit, Cumulative frequency):

• (120, 12)

• (140, 26)

• (160, 34)

• (180, 40)

• (200, 50)

In points ko graph par plot karke smooth curve join kar do — यही less than ogive hai.

Q2. Given less than type ogive data, draw ogive and find median weight from graph, then verify by formula.

Difficulty Level: Medium
Known: Less than type cumulative frequency data of 35 students
Unknown: Ogive + median (graph se) + formula se verify

Given (Less than type):
Less than 38 → 0
Less than 40 → 3
Less than 42 → 5
Less than 44 → 9
Less than 46 → 14
Less than 48 → 28
Less than 50 → 32
Less than 52 → 35

Reasoning

• Ogive mein upper class limits x-axis par, cumulative frequency y-axis par.

• Median nikalne ke liye:

   

  $$\frac{n}{2} = \frac{35}{2} = 17.5$$2n​=235​=17.5

Ogive par y = 17.5 se horizontal line kheench kar curve ko cut karo, phir x-axis par drop karo. Jo x-value mile, wahi median.

Solution

(A) Ogive plotting points

Points (Upper limit, Cumulative frequency):

• (38, 0)

• (40, 3)

• (42, 5)

• (44, 9)

• (46, 14)

• (48, 28)

• (50, 32)

• (52, 35)

In points ko plot karke smooth curve join karo.

(B) Median from graph

$$n = 35 \Rightarrow \frac{n}{2} = 17.5$$

n=35⇒2n​=17.5

Graph par y = 17.5 mark karke curve se intersect karvao, x-axis par projection lo.
Median (graph se) = 46.5 kg

✅ Median weight = 46.5 kg

(C) Verification by formula

Step 1: Continuous class intervals aur frequency nikaalna (difference 2 ka hai):

Now:

$$\frac{n}{2} = 17.5$$

2n​=17.5

cf just greater than 17.5 is 28, so median class = 46–48

• $l = 46$

  l=46

• $h = 2$

  h=2

• $f = 14$

  f=14

• $cf = 14$

  cf=14 (preceding cumulative frequency)

$$\text{Median} = l + \left(\frac{\frac{n}{2} - cf}{f}\right)h = 46 + \left(\frac{17.5 - 14}{14}\right)2$$

Median=l+(f2n​−cf​)h=46+(1417.5−14​)2

$$= 46 + \left(\frac{3.5}{14}\right)2 = 46 + 0.5 = 46.5$$

=46+(143.5​)2=46+0.5=46.5

✅ Median = 46.5 kg (verified)

Q3. Convert the distribution to a more than type cumulative frequency distribution and draw ogive.

Difficulty Level: Medium
Known: Wheat yield per hectare data for 100 farms
Unknown: More than type cumulative frequency table + ogive points

Given Frequency Distribution
Yield (kg/ha): 50–55, 55–60, 60–65, 65–70, 70–75, 75–80
No. of farms: 2, 8, 12, 24, 38, 16
Total = 100

Reasoning

More than type cumulative frequency banane ke liye lower class limits se start karte hain.
Start: “More than or equal to 50” = total farms = 100
Phir har step par previous CF se frequency subtract karte jao.

Solution

(A) More than type cumulative frequency distribution

✅ More than type table ready.

(B) Ogive (plotting points)

More than ogive ke points (Lower class limit, Cumulative frequency):

• (50, 100)

• (55, 98)

• (60, 90)

• (65, 78)

• (70, 54)

• (75, 16)

• (80, 0)

FAQs – Class 10 Maths Chapter 14 Exercise 14.4

Q1. What is the focus of Exercise 14.4?
Exercise 14.4 focuses on finding the mode of grouped data using the mode formula.

Q2. What is a modal class?
The modal class is the class interval with the highest frequency in the grouped data.

Q3. Why is Exercise 14.4 important for board exams?
Questions based on mode calculation are commonly asked in CBSE board exams and are scoring when solved with proper steps.

Q4. How can students prepare effectively for Exercise 14.4?
Students should practice identifying the modal class correctly, apply the mode formula carefully, and revise previous years’ board questions for better preparation.