Class 10 NCERT Solutions- Chapter 14 Statistics - Exercise 14.3

Chapter 14 of Class 10 NCERT Mathematics is dedicated to Statistics which is a crucial part of data handling. Exercise 14.3 focuses on the concepts of Mean, Median and Mode for the grouped data essential for summarizing and understanding the large datasets effectively. In this exercise, students are required to calculate these central tendencies using various methods like direct assumed mean and step-deviation methods which provide deeper insight into the data representation.

Statistics

Statistics is the branch of mathematics that deals with the collection, classification, analysis and interpretation of numerical data. It helps in making sense of complex datasets by summarizing them into meaningful insights. In this chapter, students will learn how to compute key measures such as the Mean, Median and Mode for the grouped and ungrouped data using the different methods to arrive at the most accurate result.

Question 1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers in a locality. Find the median, mean, and mode of the data and compare them.

Solution:

 Total number of consumer n = 68 

n/2 =34

So, the median class is 125-145 with cumulative frequency = 42

Here, l = 125, n = 68, C_f = 22, f = 20, h = 20

Now we find the median:

Median = 

= 125 + 12 = 137

Hence, the median is 137

Now we find the mode:

Modal class = 125 - 145,

Frequencies are

f₁ = 20, f₀ = 13, f₂ = 14 & h = 20

Mode = 

On substituting the values in the given formula, we get

Mode = 

= 125 + 140/13 

= 125 + 10.77

= 135.77

Hence, the mode is 135.77

Now we find the mean:

Class Interval	fi	xi	di = xi - a	ui = di/h	fiui
65-85	4	75	-60	-3	-12
85-105	5	95	-40	-2	-10
105-125	13	115	-20	-1	-13
125-145	20	135	0	0	0
145-165	14	155	20	1	14
165-185	8	175	40	2	16
185-205	4	195	60	3	12
Sum fi = 68	Sum fiui = 7

= 135 + 20(7/68)

= 137.05

Hence, the mean is 137.05

Now, on comparing the median, mean, and mode, we found that mean, median and mode are more/less equal in this distribution.

Question 2. If the median of a distribution given below is 28.5 then, find the value of x & y.

Solution:

According to the question

The total number of observations are n = 60

Median of the given data = 28.5

n/2 = 30  

Median class is 20 – 30 with a cumulative frequency = 25 + x

Lower limit of median class, l = 20,

C_f = 5 + x,

f = 20 & h = 10

Now we find the median:

Median = 

On substituting the values in the given formula, we get

28.5 = 

8.5 = (25 - x)/2

17 = 25 - x

Therefore, x = 8

From the cumulative frequency, we can identify the value of x + y as follows:

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

On substituting the values of x, we will find the value of y

60 = 5 + 20 + 15 + 5 + 8 + y

y = 60 - 53

y = 7

So the value of a is 8 and y is 7

Question 3. The Life insurance agent found the following data for the distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to the persons whose age is 18 years onwards but less than the 60 years.

Solution: 

According to the given question the table is 

Class interval	Frequency	Cumulative frequency
15-20	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

Given data: n = 100 and n/2 = 50

Median class = 35 - 45

Then, l = 35, c_f = 45, f = 33 & h = 5

Now we find the median:

Median = 

On substituting the values in the given formula, we get

Median = 

= 35 + 5(5/33) 

= 35.75

Hence, the median age is 35.75 years.

Question 4. The lengths of 40 leaves in a plant are measured correctly to the nearest millimeter, and the data obtained is represented as in the following table:

Find the median length of leaves.             

Solution:

The data in the given table are not continuous to reduce 0.5 in the lower limit and add 0.5 in the upper limit.

We get a new table:

Class Interval	Frequency	Cumulative frequency
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

From the given table

n = 40 and n/2 = 20

Median class = 144.5 - 153.5

l = 144.5,

cf = 17, f = 12 & h = 9

Now we find the median:

Median = 

On substituting the values in the given formula, we get

Median = 

= 144.5 + 9/4 

= 146.75 mm

Hence, the median length of the leaves is 146.75 mm.

Question 5. The following table gives the distribution of a life time of 400 neon lamps.

Find the median lifetime of a lamp.

Solution:

According to the question

Class Interval	Frequency	Cumulative
1500-2000	14	14
2000-2500	56	70
2500-3000	60	130
3000-3500	86	216
3500-4000	74	290
4000-4500	62	352
4500-5000	48	400

n = 400 and n/2 = 200

Median class = 3000 – 3500

l = 3000, C_f = 130,

f = 86 & h = 500

Now we find the median:

Median = 

On substituting the values in the given formula, we get

Median = 

= 3000 + 35000/86 = 3000 + 406.97

= 3406.97

Hence, the median lifetime of the lamps is 3406.97 hours

Question 6. In this 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in English alphabets in the surnames was obtained as follows:

Determine the number of median letters in the surnames. Find the number of mean letters in the surnames and also, find the size of modal in the surnames.

Solution:

According to the question

Class Interval	Frequency	Cumulative Frequency
1-4	6	6
4-7	30	36
7-10	40	76
10-13	16	92
13-16	4	96
16-19	4	100

n = 100 and n/2 = 50

Median class = 7 - 10

Therefore, l = 7, C_f = 36, f = 40 & h = 3

Now we find the median:

Median = 

On substituting the values in the given formula, we get

Median = 

Median = 7 + 42/40 = 8.05

Hence, the median is 8.05

Now we find the mode:

Modal class = 7 - 10,

Where, l = 7, f₁ = 40, f₀ = 30, f₂ = 16 & h = 3

Mode = 

On substituting the values in the given formula, we get

Mode = 

= 7 + 30/34 = 7.88

Hence, the mode is 7.88

Now we find the mean:

Class Interval	fi	xi	fixi
1-4	6	2.5	15
4-7	30	5.5	165
7-10	40	8.5	340
10-13	16	11.5	184
13-16	4	14.5	51
16-19	4	17.5	70
Sum fi = 100	Sum fixi = 825

Mean = 

= 825/100 = 8.25

Hence, the mean is 8.25

Question 7. The distributions of below give a weight of 30 students of a class. Find the median weight of a student.

Solution:

According to the question

Class Interval	Frequency	Cumulative 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

n = 30 and n/2 = 15

Median class = 55 - 60

l = 55, C_f = 13, f = 6 & h = 5

Now we find the median:

Median = 

On substituting the values in the given formula, we get

Median = 

= 55 + 10/6 = 55 + 1.666

= 56.67

Hence, the median weight of the students is 56.67

Conclusion

Exercise 14.3 is designed to help students master the techniques of computing the central tendencies such as the mean, median and mode of the grouped data. By practicing these questions students not only improve their problem-solving skills but also develop a better understanding of how statistical measures can be applied to the real-world data analysis. Solving this exercise prepares students for the more advanced data handling concepts in the higher education.