NCERT Solutions for Class 10 Maths Chapter 14 Statistics

NCERT Solutions Class 10 Maths Chapter 14 Statistics – This article is a useful resource containing free NCERT Solutions for Class 10 Maths Chapter 14 Statistics. These NCERT solutions have been developed by the subject matter experts at GFG to assist students in easily solving questions related to Statistics from the NCERT textbook.

These NCERT Solutions for Class 10 Maths Chapter 14 Statistics cover all four exercises of the NCERT Class 10 Maths Chapter 14 according to the latest CBSE syllabus 2023-24 and guidelines, which are as follows:

NCERT Class 10 Maths Chapter 14 Statistics will help the students learn important statistical concepts like mean, mode, standard deviation, and the graphical depiction of cumulative frequency distribution.

The solutions to all the problems in this Chapter 14 Statistics exercises from the NCERT textbook have been properly covered in the NCERT Solutions for Class 10 Maths.

Question 1. A survey was conducted by a group of students as a part of their environment awareness program, 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: 

Step 1: Let us find out the Class Mark (x_i) for the following class intervals by using the formula

Class Mark = (Upper class Limit + Lower Class Limit)/2

Step 2: Now, we will multiply the classmark with the number of times they have occurred, i.e, with the frequency.

Step 3: Now we will apply the general formula to calculate the mean

 

Now, Let’s see the detailed solution: 

Now, after creating this table we will be able to find the mean very easily – 

= 16

= 8.1

Hence, we come to the conclusion that the number of plants per house is 8.1. Since the numeral value of frequency(fi) and the class mark(xi) is small so we use DIRECT METHOD to find the mean number of plants per house.

Question 2. Consider the following distribution 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:

Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula

Class Mark = (Upper class Limit + Lower Class Limit)/2

Step 2: In this case, the value of mid-point (xi) is very large, so let us assume the mean value, A = 150, and the class interval is h = 20.

u_i = (x_i – A)/h 

=> u_i = (xi – 150)/20

Step 3: Now we will apply the Assumed Mean Formula to calculate the mean

Now, Let’s see the detailed solution: 

So, the formula to find out the mean is:

Mean = 

           = 150 + (20 × -12/50) 

           = 150 – 4.8

           = 145.20

Thus, mean daily wage of the workers = Rs. 145.20.

Question 3. 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.

Solution:

Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula

Class Mark = (Upper class Limit + Lower Class Limit)/2

Step 2: Now, we will multiply the classmark with the number of times they have occurred, i.e, with the frequency. As a certain frequency is missing and we have an odd number of class intervals hence, we will assume the middle-Class Mark as our Assumed Mean(A).

Step 3: Now we will apply the general formula to calculate the mean

Now, Let’s see the detailed solution: 

The mean formula is

Mean = 

          = (752 + 20f)/(44 + f)

Now substitute the values and equate to find the missing frequency (f)

⇒ 18 = (752 + 20f)/(44 + f)

⇒ 18(44 + f) = (752 + 20f)

⇒ 792 + 18f = 752 + 20f

⇒ 792 + 18f = 752 + 20f

⇒ 792 – 752 = 20f – 18f

⇒ 40 = 2f

⇒ f = 20

So, the missing frequency, f = 20.

Question 4. Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute was recorded and summarized as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

Solution:

Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula

Class Mark = (Upper class Limit + Lower Class Limit)/2

Step 2: In this case, the value of mid-point (xi) is very large, so let us assume the mean value, A = 75.5 and class size is h = 3.

d_i = (x_i – A) 

=> d_i = (x_i – 75.5)

Step 3: Now we will apply the Assumed Mean Formula to calculate the mean

Now, Let’s see the detailed solution: 

Mean = 

= 75.5 + (12/30)

= 75.5 + 2/5

= 75.5 + 0.4

= 75.9

Therefore, the mean heartbeats per minute for these women is 75.9

Question 5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying numbers of mangoes. The following 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:

Step 1: In the above table we find that the class intervals are not continuous and hence to make them a continuous set of data we add  0.5 to the upper limit and subtract 0.45 from the lower limit as the gap between two intervals is 1. Then find the Mid Point by using the formula 

Class Mark = (Upper class Limit + Lower Class Limit)/2

Step 2: In this case, let us assume the mean value, A = 57 and class size is h = 3.

Step 3: Since the frequency values are big, hence we are using the STEP-DEVIATION METHOD.

Now, Lets see the detailed solution: 

Mean = 

= 57 + 3 * (25/400)

= 57 + 0.1875

= 57.19

Therefore, the mean number of mangoes kept in a packing box is 57.19

Question 6. The table below shows the daily expenditure on the food of 25 households in a locality. Find the mean daily expenditure on food by a suitable method.

Solution:

Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula

Class Mark = (Upper class Limit + Lower Class Limit)/2

Step 2: In this case, the value of mid-point (xi) is very large, so let us assume the mean value, A = 225 and class size is h = 50.

d_i = (x_i – A) 

=> d_i = (x_i – 225)

u_i = (x_i – A)/h

=> u_i = (x_i – 225)/50

Step 3: Now we will apply the Step Deviation Formula to calculate the mean

Now, Let’s see the detailed solution: 

Mean = 

= 225 + 50 (-7/25)

= 225 – 14

= 211

Therefore, the mean daily expenditure on food is ₹211

Question 7. 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:

Find the mean concentration of SO₂ in the air.

Solution:

Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula

Class Mark = (Upper class Limit + Lower Class Limit)/2

Step 2: Now, we will multiply the classmark with the number of times they have occurred, i.e, with the frequency.

Step 3: Now we will apply the general formula to calculate the mean

Now, Let’s see the detailed solution: 

The formula to find out the mean is

Mean = 

= 2.96/30

= 0.099 ppm

Therefore, the mean concentration of SO₂ in the air is 0.099 ppm.

Question 8. 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.

Solution:

Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula

Class Mark = (Upper class Limit + Lower Class Limit)/2

Step 2: Now, we will multiply the classmark with the number of times they have occurred, i.e, with the frequency.

Step 3: Now we will apply the general formula to calculate the mean

Now, Let’s see the detailed solution: 

The mean formula is,

Mean = 

= 499/40

= 12.48 days

Therefore, the mean number of days a student was absent = 12.48.

Question 9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Solution:

Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula

Class Mark = (Upper class Limit + Lower Class Limit)/2

Step 2: In this case, the value of mid-point (xi) is very large, so let us assume the mean value, A = 70 and class size is h = 10.

d_i = (x_i – A) 

=> d_i = (x_i – 70)

u_i = (x_i – A)/h

=> u_i = (x_i – 70)/10

Step 3: Now we will apply the Step Deviation Formula to calculate the mean

Now, Let’s see the detailed solution: 

So, 

Mean = 

= 70 + (-2/35) × 10

= 69.42

Therefore, the mean literacy rate = 69.42%.

Question 1. 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. Compare and interpret the two measures of central tendency.

Solution:

The greatest frequency in the given table is 23, so the modal class = 35 – 45,

l = 35,

Class width = 10, and the frequencies are

f_m = 23, f₁ = 21 and f₂ = 14

Now, we find the mode using the given formula

Mode = 

On substituting the values in the formula, we get

Mode = 

= 35 + (20/11) = 35 + 1.8

= 36.8

Hence, the mode of the given data is 36.8 year

Now, we find the mean. So for that first we need to find the midpoint.

x_i = (upper limit + lower limit)/2

Class Interval	Frequency (fi)	Mid-point (xi)	fixi
5-15	6	10	60
15-25	11	20	220
25-35	21	30	630
35-45	23	40	920
45-55	14	50	700
55-65	5	60	300
	Sum fi = 80		Sum fixi = 2830

Mean =  = ∑f_ix_i /∑f_i

= 2830/80

= 35.37 years

Question 2. The following data gives information on the observed lifetimes (in hours) of 225 electrical components:

Determine the modal lifetimes of the components.

Solution:

According to the given question

The modal class is 60 – 80

l = 60, and the frequencies are

f_m = 61, f₁ = 52, f₂ = 38 and h = 20

Now, we find the mode using the given formula

Mode = 

On substituting the values in the formula, we get

Mode = 

= 

= 60 + 45/8 = 60 + 5.625

Hence, the modal lifetime of the components is 65.625 hours.

Question 3. The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Solution:

According to the question

Modal class = 1500-2000,

l = 1500,and the frequencies are

f_m = 40 f₁ = 24, f₂ = 33 and

h = 500

Now, we find the mode using the given formula

Mode = 

On substituting the values in the formula, we get

Mode = 

= 

= 1500 + 8000/23 = 1500 + 347.83

So, the modal monthly expenditure of the families is 1847.83 Rupees 

Now, we find the mean. So for that first we need to find the midpoint.

x_i = (upper limit + lower limit)/2

Let us considered a mean, A be 2750

Class Interval	fi	xi	di = xi – a	ui = di/h	fiui
1000-1500	24	1250	-1500	-3	-72
1500-2000	40	1750	-1000	-2	-80
2000-2500	33	2250	-500	-1	-33
2500-3000	28	2750	0	0	0
3000-3500	30	3250	500	1	30
3500-4000	22	3750	1000	2	44
4000-4500	16	4250	1500	3	48
4500-5000	7	4750	2000	4	28
	fi = 200				fiui = -35

Mean = 

On substituting the values in the given formula

= 

= 2750 – 87.50

= 2662.50

Hence, the mean monthly expenditure of the families is  2662.50 Rupees

Question 4. The following distribution 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

Solution:

According to the question

Modal class = 30 – 35,

l = 30,

Class width (h) = 5, and the frequencies are

f_m = 10, f₁ = 9 and f₂ = 3

Now, we find the mode using the given formula

Mode = 

On substituting the values in the formula, we get

Mode = 

= 30 + 5/8 = 30 + 0.625

= 30.625

Hence, the mode of the given data is 30.625

Now, we find the mean. So for that first we need to find the midpoint.

x_i = (upper limit + lower limit)/2

Class Interval	Frequency (fi)	Mid-point (xi)	fixi
15-20	3	17.5	52.5
20-25	8	22.5	180.0
25-30	9	27.5	247.5
30-35	10	32.5	325.0
35-40	3	37.5	112.5
40-45	0	42.5	0
45-50	0	47.5	0
50-55	2	52.5	105.5
	Sum fi = 35		Sum fixi = 1022.5

Mean = 

= 1022.5/35 

= 29.2

Hence, the mean is 29.2

Question 5. 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.

Solution:

According to the question

Modal class = 4000 – 5000,

l = 4000,

class width (h) = 1000, and the frequencies are

f_m = 18, f₁ = 4 and f₂ = 9

Now, we find the mode using the given formula

Mode = 

On substituting the values in the formula, we get

Mode = 

Mode = 4000 + 14000/23 = 4000 + 608.695

= 4608.695

Hence, the mode of the given data is 4608.7 runs

Question 6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarized it in the table given below. Find the mode of the data:

Solution:

According to the question

Modal class = 40 – 50, l = 40,

Class width (h) = 10, and the frequencies are

f_m = 20, f₁ = 12 and f₂ = 11

Now, we find the mode using the given formula

Mode = 

On substituting the values in the formula, we get

Mode = 

Mode = 40 + 80/17 = 40 + 4.7 = 44.7

Hence, the mode of the given data is 44.7 cars

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 policyholders. Calculate the median age, if policies are given only to persons whose age is 18 years onwards but less than 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 lifetime 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

Question 1. The following distribution gives the daily income of 50 workers if a factory. Convert the distribution above to a less-than-type cumulative frequency distribution and draw its ogive.

Solution:

According to the question, we convert the given distribution to a less than type cumulative frequency distribution, 

Daily income	Frequency	Cumulative Frequency
Less than 120	12	12
Less than 140	14	26
Less than 160	8	34
Less than 180	6	40
Less than 200	10	50

Now according to the table we plot the points that are corresponding to the ordered pairs (120, 12), (140, 26), (160, 34), (180, 40), and (200, 50) on a graph paper. Here x-axis represents the upper limit and y-axis represent the frequency. The curve obtained from the graph is known as less than type ogive curve. 

Question 2. During the medical check-up of 35 students of a class, their weights were recorded as follows:

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

Solution:

According to the given table, we get the points (38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32) and (52, 35). Now using these points we draw an ogive, where the x-axis represents the upper limit and y-axis represents the frequency. The curve obtained is known as less than type ogive.

Now, locate the point 17.5 on the y-axis and draw a line parallel to the x-axis cutting the curve at a point. From this point, now we draw a perpendicular line to the x-axis and the intersection point which is perpendicular to x-axis is the median of the given data. After, locating point now we create a table to find the mode:

Class interval	Number of students(Frequency)	Cumulative Frequency
Less than 38	0	0
Less than 40	3 – 0 = 3	3
Less than 42	5 – 3 = 2	8
Less than 44	9 – 5 = 4	9
Less than 46	14 – 9 = 5	14
Less than 48	28 – 14 = 14	28
Less than 50	32 – 28 = 4	32
Less than 52	35 – 22 = 3	35

The class 46 – 48 has the maximum frequency, hence, this is the modal class

l = 46, h = 2, f₁ = 14, f₀ = 5 and f₂ = 4

Now we find the mode:

Mode = 

On substituting the values in the given formula, we get

= 46 + 0.95 = 46.95

Hence, the mode is verified.

Question 3. The following tables give the production yield per hectare of wheat of 100 farms of a village.

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

Solution:

According to the question, we change the distribution to a more than type distribution.

Production Yield (kg/ha)	Number of farms
More than or equal to 50	100
More than or equal to 55	100 – 2 = 98
More than or equal to 60	98 – 8 = 90
More than or equal to 65	90 – 12 = 78
More than or equal to 70	78 – 24 = 54
More than or equal to 75	54 – 38 = 16

Now, according to the table we draw the ogive by plotting the points. Here, the a-axis represents the upper limit and y-axis represents the cumulative frequency. And the points are(50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) on this graph paper. The graph obtained is known as more than type ogive curve.

• These NCERT solutions are developed by the GFG team, with a focus on students’ benefit.
• These solutions are entirely accurate and can be used by students to prepare for their board exams. 
• Each solution is presented in a step-by-step format with comprehensive explanations of the intermediate steps.

Also Check:

• NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers
• NCERT Solutions for Class 10 Maths Chapter 2 Polynomials
• NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables
• NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equation
• NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions
• NCERT Solutions for Class 10 Maths Chapter 6 Triangles
• NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Q1: Why is it important to learn Statistics in NCERT Class 10 Maths Chapter 14?

Statistics help people and groups in resolving issues regardless of the industry, market, economic sector, etc. Any issue where relevant data can be gathered, examined, analysed, and presented so as to work towards an efficient resolution can benefit from the knowledge and assistance provided by Statistics. The possibilities are endless.

Q2: What topics are covered in NCERT Solutions for Class 10 Maths Chapter 14 Statistics?

NCERT Solutions for Class 10 Maths Chapter 14 Statistics covers topics such as calculation of mean, median, mode, frequency distribution and standard deviation and variance etc.

Q3: How can NCERT Solutions for Class 10 Maths Chapter 14 Statistics help me?

NCERT Solutions for Class 10 Maths Chapter 14 Statistics can help you solve the NCERT exercises in an easy mannner. If you are stuck on a problem you can find its solution here and free yourself from the frustration of being stuck on some question.

Q4: How many exercises are there in Class 10 Maths Chapter 14 Statistics?

There are 4 exercises in the NCERT Class 10 Maths Chapter 14 Statistics which covers all the important topics and sub-topics.

Q5: Where can I find NCERT Solutions for Class 10 Maths Chapter 14 Statistics?

You can find the NCERT Solutions for Class 10 Maths Chapter 14 Statistics in this article created by our team of experts at w3wiki.