Class 10 NCERT Solutions- Chapter 14 Statistics – Exercise 14.3

Last Updated : 03 Mar, 2021

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.

Monthly consumption(in units)	No. of customers
65-85	4
85-105	5
105-125	13
125-145	20
145-165	14
165-185	8
185-205	4

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 = $l +\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h$

$=125+\frac{(34−22)}{20} × 20$

= 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 = $l+ \left[\frac{(f1-f0)}{(2f1-f0-f2)}\right]×h$

On substituting the values in the given formula, we get

Mode = $125 + \frac{(20-13)}{(40-13-14)}×20$

= 125 + 140/13

= 125 + 10.77

= 135.77

Hence, the mode is 135.77

Now we find the mean:

Class Interval f_i x_i d_i= x_i– a u_i= d_i/h f_iu_i
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 f_i= 68 Sum f_iu_i= 7

$\bar{x} =a+h \frac{∑f_iu_i}{∑f_i}$

= 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.

Class Interval	Frequency
0-10	5
10-20	x
20-30	20
30-40	15
40-50	y
50-60	5
Total	60

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 = $l+\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h$

On substituting the values in the given formula, we get

28.5 = $20+\frac{(30−5−x)}{20} × 10$

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.

Age (in years)	Number of policy holder
Below 20	2
Below 25	6
Below 30	24
Below 35	45
Below 40	78
Below 45	89
Below 50	92
Below 55	98
Below 60	100

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 = $l+\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h$

On substituting the values in the given formula, we get

Median = $35+\frac{(50-45)}{33} × 5$

= 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:

Length (in mm)	Number of leaves
118-126	3
127-135	5
136-144	9
145-153	12
154-162	5
163-171	4
172-180	2

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 = $l+\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h$

On substituting the values in the given formula, we get

Median = $144.5+\frac{(20-17)}{12}×9$

= 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.

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

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 = $l+\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h$

On substituting the values in the given formula, we get

Median = $3000 + \frac{(200-130)}{86} × 500$

= 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:

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 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 = $l+\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h$

On substituting the values in the given formula, we get

Median = $7+\frac{(50-36)}{40} × 3$

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 = $l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h$

On substituting the values in the given formula, we get

Mode = $7+\frac{(40-30)}{(2×40-30-16)} × 3$

= 7 + 30/34 = 7.88

Hence, the mode is 7.88

Now we find the mean:

Class Interval f_i x_i f_ix_i
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 f_i = 100 Sum f_ix_i = 825

Mean = $\bar{x}= \frac{∑f_i x_i }{∑f_i}$

= 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.

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

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 = $l +\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h$

On substituting the values in the given formula, we get

Median = $55+\frac{(15-13)}{6}×5$

= 55 + 10/6 = 55 + 1.666

= 56.67

Hence, the median weight of the students is 56.67

Suggest improvement

Class 10 NCERT Solutions- Chapter 14 Statistics - Exercise 14.2

Class 10 NCERT Solutions- Chapter 14 Statistics - Exercise 14.4

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Class Interval	f_i	x_i	d_i= x_i– a	u_i= d_i/h	f_iu_i
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 f_i= 68				Sum f_iu_i= 7

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

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

Chapter 1: Real Numbers

Chapter 2: Polynomials

Chapter 3: Pair of Linear Equations in Two Variables

Chapter 4: Quadratic Equations

Chapter 5: Arithmetic Progressions

Chapter 6: Triangles

Chapter 7: Coordinate Geometry

Chapter 8: Introduction to Trigonometry

Chapter 9: Some Applications of Trigonometry

Chapter 10: Circles

Chapter 11: Constructions

Chapter 12: Areas Related to Circles

Chapter 13: Surface Areas and Volumes

Chapter 14: Statistics

Chapter 15: Probability

Chapter 1: Real Numbers

Chapter 2: Polynomials

Chapter 3: Pair of Linear Equations in Two Variables

Chapter 4: Quadratic Equations

Chapter 5: Arithmetic Progressions

Chapter 6: Triangles

Chapter 7: Coordinate Geometry

Chapter 8: Introduction to Trigonometry

Chapter 9: Some Applications of Trigonometry

Chapter 10: Circles

Chapter 11: Constructions

Chapter 12: Areas Related to Circles

Chapter 13: Surface Areas and Volumes

Chapter 14: Statistics

Chapter 15: Probability

Class 10 NCERT Solutions- Chapter 14 Statistics – Exercise 14.3

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.

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

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.

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.

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

Find the median lifetime of a lamp.

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.

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

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