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 202324 and guidelines, which are as follows:
Class 10 Maths NCERT Solutions Chapter 14 Statistics Exercises 





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.
NCERT Solutions for Class 10 Maths Chapter 14 Statistics Exercise 14.1
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.
Number of Plants  02  24  46  68  810  1012  1214 
Number of houses  1  2  1  5  6  2  3 
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:
No.of Plants (Class Interval)  No. of Houses (Frequency) (f_{i})  Class Mark (x_{i})  f_{i }* x_{i} 

02  1  1  1 
24  2  3  6 
46  1  5  5 
68  5  7  35 
810  6  9  54 
1012  2  11  22 
1214  3  13  39 
 Sum: âˆ‘ f_{i} = 20 
 Sum: âˆ‘ f_{i}x_{i} = 162 
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.
Daily Wages (in â‚¹)  500520  520540  540560  560580  580600 
Number of Workers  12  14  8  6  10 
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 midpoint (xi) is very large, so let us assume the mean value, A = 150, and the class interval is h = 20.
u_{i} = (x_{i} â€“ A)/h
=> u_{i }= (xi â€“ 150)/20
Step 3: Now we will apply the Assumed Mean Formula to calculate the mean
Now, Let’s see the detailed solution:
Daily wages (Class interval)  Number of workers frequency (fi)  Midpoint (x_{i})  u_{i} = (x_{i} â€“ 150)/20  f_{i}u_{i} 

100120  12  110  2  24 
120140  14  130  1  14 
140160  8  150  0  0 
160180  6  170  1  6 
180200  10  190  2  20 
Total  Sum âˆ‘f_{i} = 50 

 Sum âˆ‘f_{i}u_{i} = 12 
So, the formula to find out the mean is:
Mean =
= 150 + (20 Ã— 12/50)
= 150 â€“ 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.
Daily pocket allowance (in â‚¹)  1113  1315  1517  1719  1921  2123  2325 
Number of children  7  6  9  13  f  5  4 
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 middleClass 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:
Class interval  Number of children (f_{i})  Midpoint (x_{i})  f_{i}x_{i} 

1113  7  12  84 
1315  6  14  84 
1517  9  16  144 
1719  13  18 = A  234 
1921  f  20  20f 
2123  5  22  110 
2325  4  24  96 
Total  âˆ‘ f_{i} = 44 + f 
 Sum âˆ‘f_{i}x_{i} = 752 + 20f 
The mean formula is
Mean =
= (752 + 20f)/(44 + f)
Now substitute the values and equate to find the missing frequency (f)
â‡’ 18 = (752 + 20f)/(44 + f)
â‡’ 18(44 + f) = (752 + 20f)
â‡’ 792 + 18f = 752 + 20f
â‡’ 792 + 18f = 752 + 20f
â‡’ 792 â€“ 752 = 20f â€“ 18f
â‡’ 40 = 2f
â‡’ 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.
Number of heartbeats per minute  6568  6871  7174  7477  7780  8083  8386 
Number of Women  2  4  3  8  7  4  2 
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 midpoint (xi) is very large, so let us assume the mean value, A = 75.5 and class size is h = 3.
d_{i} = (x_{i} â€“ A)
=> d_{i} = (x_{i} â€“ 75.5)
Step 3: Now we will apply the Assumed Mean Formula to calculate the mean
Now, Let’s see the detailed solution:
Class Interval  Number of women (f_{i})  Midpoint (x_{i})  d_{i} = (x_{i} â€“ 75.5)  f_{i}d_{i} 

6568  2  66.5  9  18 
6871  4  69.5  6  24 
7174  3  72.5  3  9 
7477  8  75.5 = A  0  0 
7780  7  78.5  3  21 
8083  4  81.5  6  24 
8386  2  84.5  9  18 
 Sum âˆ‘f_{i} = 30 

 Sum âˆ‘f_{i}u_{i} = 12 
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.
Number of Mangoes  5052  5355  5658  5961  6264 
Number of Boxes  15  110  135  115  25 
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 STEPDEVIATION METHOD.
Now, Lets see the detailed solution:
Class Interval  Number of boxes (f_{i})  Midpoint (x_{i})  d_{i} = x_{i} â€“ A  u_{i}=(x_{i} â€“ A)/h  f_{i}u_{i} 

49.552.5  15  51  6  2  30 
52.555.5  110  54  3  1  110 
55.558.5  135  57 =A  0  0  0 
58.561.5  115  60  3  1  115 
61.564.5  25  63  6  2  50 
 Sum âˆ‘f_{i} = 400 


 Sum âˆ‘f_{i}u_{i} = 25 
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.
Daily Expenditure (in â‚¹)  100150  150200  200250  250300  300350 
Number of Households  4  5  12  2  2 
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 midpoint (xi) is very large, so let us assume the mean value, A = 225 and class size is h = 50.
d_{i} = (x_{i} â€“ A)
=> d_{i} = (x_{i} â€“ 225)
u_{i} = (x_{i} â€“ A)/h
=> u_{i} = (x_{i} â€“ 225)/50
Step 3: Now we will apply the Step Deviation Formula to calculate the mean
Now, Let’s see the detailed solution:
Class Interval  Number of households (f_{i})  Midpoint (x_{i})  d_{i} = x_{i} â€“ A  u_{i} = d_{i}/50  f_{i}u_{i} 

100150  4  125  100  2  8 
150200  5  175  50  1  5 
200250  12  225 = A  0  0  0 
250300  2  275  50  1  2 
300350  2  325  100  2  4 
 Sum âˆ‘f_{i} = 25 


 Sum âˆ‘f_{i}u_{i} = 7 
Mean =
= 225 + 50 (7/25)
= 225 – 14
= 211
Therefore, the mean daily expenditure on food is â‚¹211
Question 7. To find out the concentration of SO_{2} 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:
Concentration of SO_{2} (in ppm)  Frequency 

0.000.04  4 
0.040.08  9 
0.080.12  9 
0.120.16  2 
0.160.20  4 
0.200.24  2 
Find the mean concentration of SO_{2} 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:
Concentration of SO_{2 }(in ppm)  Frequency (f_{i})  Midpoint (x_{i})  f_{i}x_{i} 

0.000.04  4  0.02  0.08 
0.040.08  9  0.06  0.54 
0.080.12  9  0.10  0.90 
0.120.16  2  0.14  0.28 
0.160.20  4  0.18  0.72 
0.200.24  2  0.22  0.44 
 Sum âˆ‘f_{i} = 30 
 Sum âˆ‘f_{i}x_{i} = 2.96 
The formula to find out the mean is
Mean =
= 2.96/30
= 0.099 ppm
Therefore, the mean concentration of SO_{2} 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.
Number of Days  06  610  1014  1420  2028  2838  3840 
Number of Students  11  10  7  4  4  3  1 
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:
Class Interval  Frequency (f_{i})  Midpoint (x_{i})  f_{i}x_{i} 

06  11  3  33 
610  10  8  80 
1014  7  12  84 
1420  4  17  68 
2028  4  24  96 
2838  3  33  99 
3840  1  39  39 
 Sum âˆ‘f_{i} = 40 
 Sum âˆ‘f_{i}x_{i} = 499 
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.
Literacy rate (in %)  4555  5565  6575  7585  8595 
Number of Cities  3  10  11  8  3 
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 midpoint (xi) is very large, so let us assume the mean value, A = 70 and class size is h = 10.
d_{i} = (x_{i} â€“ A)
=> d_{i} = (x_{i} â€“ 70)
u_{i} = (x_{i} â€“ A)/h
=> u_{i} = (x_{i} â€“ 70)/10
Step 3: Now we will apply the Step Deviation Formula to calculate the mean
Now, Let’s see the detailed solution:
Class Interval  Frequency (f_{i})  Class Mark(x_{i})  di = x_{i} â€“ a  u_{i} = d_{i}/h  f_{i}u_{i} 

4555  3  50  20  2  6 
5565  10  60  10  1  10 
6575  11  70 = A  0  0  0 
7585  8  80  10  1  8 
8595  3  90  20  2  6 
 Sum âˆ‘f_{i} = 35 


 Sum âˆ‘f_{i}u_{i }= 2 
So,
Mean =
= 70 + (2/35) Ã— 10
= 69.42
Therefore, the mean literacy rate = 69.42%.
NCERT Solutions for Class 10 Maths Chapter 14 Statistics Exercise 14.2
Question 1. The following table shows the ages of the patients admitted in a hospital during a year:
Age (in years)  515  1525  2535  3545  4555  5565 
Number of patients  6  11  21  23  14  5 
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 â€“ 45,
l = 35,
Class width = 10, and the frequencies are
f_{m} = 23, f_{1} = 21 and f_{2} = 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 (f_{i}) Midpoint (x_{i}) f_{i}x_{i} 515 6 10 60 1525 11 20 220 2535 21 30 630 3545 23 40 920 4555 14 50 700 5565 5 60 300 Sum f_{i} = 80 Sum f_{i}x_{i} = 2830 Mean = = âˆ‘f_{i}x_{i} /âˆ‘f_{i}
= 2830/80
= 35.37 years
Question 2. The following data gives information on the observed lifetimes (in hours) of 225 electrical components:
Lifetime (in hours)  020  2040  4060  6080  80100  100120 
Frequency  10  35  52  61  38  29 
Determine the modal lifetimes of the components.
Solution:
According to the given question
The modal class is 60 â€“ 80
l = 60, and the frequencies are
f_{m} = 61, f_{1} = 52, f_{2} = 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:
Expenditure  Number of families 
10001500  24 
15002000  40 
20002500  33 
25003000  28 
30003500  30 
35004000  22 
40004500  16 
45005000  7 
Solution:
According to the question
Modal class = 15002000,
l = 1500,and the frequencies are
f_{m} = 40 f_{1} = 24, f_{2} = 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 f_{i} x_{i} d_{i} = x_{i} â€“ a u_{i} = d_{i}/h f_{i}u_{i} 10001500 24 1250 1500 3 72 15002000 40 1750 1000 2 80 20002500 33 2250 500 1 33 25003000 28 2750 0 0 0 30003500 30 3250 500 1 30 35004000 22 3750 1000 2 44 40004500 16 4250 1500 3 48 45005000 7 4750 2000 4 28 f_{i} = 200 f_{i}u_{i} = 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 statewise teacherstudent ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures
No of Students per teacher  Number of states / U.T 
1520  3 
2025  8 
2530  9 
3035  10 
3540  3 
4045  0 
4550  0 
5055  2 
Solution:
According to the question
Modal class = 30 â€“ 35,
l = 30,
Class width (h) = 5, and the frequencies are
f_{m} = 10, f_{1} = 9 and f_{2} = 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 (f_{i}) Midpoint (x_{i}) f_{i}x_{i} 1520 3 17.5 52.5 2025 8 22.5 180.0 2530 9 27.5 247.5 3035 10 32.5 325.0 3540 3 37.5 112.5 4045 0 42.5 0 4550 0 47.5 0 5055 2 52.5 105.5 Sum f_{i} = 35 Sum f_{i}x_{i} = 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 oneday international cricket matches.
Run Scored  Number of Batsman 

30004000  4 
40005000  18 
50006000  9 
60007000  7 
70008000  6 
80009000  3 
900010000  1 
1000011000  1 
Find the mode of the data.
Solution:
According to the question
Modal class = 4000 â€“ 5000,
l = 4000,
class width (h) = 1000, and the frequencies are
f_{m} = 18, f_{1} = 4 and f_{2} = 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:
Number of cars  Frequency 

010  7 
1020  14 
2030  13 
3040  12 
4050  20 
5060  11 
6070  15 
7080  8 
Solution:
According to the question
Modal class = 40 â€“ 50, l = 40,
Class width (h) = 10, and the frequencies are
f_{m} = 20, f_{1} = 12 and f_{2} = 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
NCERT Solutions for Class 10 Maths 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.
Monthly consumption(in units)  No. of customers 

6585  4 
85105  5 
105125  13 
125145  20 
145165  14 
165185  8 
185205  4 
Solution:
Total number of consumer n = 68
n/2 =34
So, the median class is 125145 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
f_{1 }= 20, f_{0 }= 13, f_{2 }= 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 f_{i} x_{i} d_{i }= x_{i }– a u_{i }= d_{i}/h f_{i}u_{i} 6585 4 75 60 3 12 85105 5 95 40 2 10 105125 13 115 20 1 13 125145 20 135 0 0 0 145165 14 155 20 1 14 165185 8 175 40 2 16 185205 4 195 60 3 12 Sum f_{i }= 68 Sum f_{i}u_{i }= 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.
Class Interval  Frequency 

010  5 
1020  x 
2030  20 
3040  15 
4050  y 
5060  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 â€“ 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.
Age (in years)  Number of policyholder 

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 1520 2 2 2025 4 6 2530 18 24 3035 21 45 3540 33 78 4045 11 89 4550 3 92 5055 6 98 5560 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:
Length (in mm)  Number of leaves 

118126  3 
127135  5 
136144  9 
145153  12 
154162  5 
163171  4 
172180  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 117.5126.5 3 3 126.5135.5 5 8 135.5144.5 9 17 144.5153.5 12 29 153.5162.5 5 34 162.5171.5 4 38 171.5180.5 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.
Lifetime (in hours)  Number of lamps 

15002000  14 
20002500  56 
25003000  60 
30003500  86 
35004000  74 
40004500  62 
45005000  48 
Find the median lifetime of a lamp.
Solution:
According to the question
Class Interval Frequency Cumulative 15002000 14 14 20002500 56 70 25003000 60 130 30003500 86 216 35004000 74 290 40004500 62 352 45005000 48 400 n = 400 and n/2 = 200
Median class = 3000 â€“ 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:
Number of letters  14  47  710  1013  1316  1619 
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 14 6 6 47 30 36 710 40 76 1013 16 92 1316 4 96 1619 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, f_{1} = 40, f_{0} = 30, f_{2} = 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 f_{i} x_{i} f_{i}x_{i} 14 6 2.5 15 47 30 5.5 165 710 40 8.5 340 1013 16 11.5 184 1316 4 14.5 51 1619 4 17.5 70 Sum f_{i} = 100 Sum f_{i}x_{i} = 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.
Weight(in kg)  4045  4550  5055  5560  6065  6570  7075 
Number of students  2  3  8  6  6  3  2 
Solution:
According to the question
Class Interval Frequency Cumulative frequency 4045 2 2 4550 3 5 5055 8 13 5560 6 19 6065 6 25 6570 3 28 7075 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
NCERT Solutions for Class 10 Maths Chapter 14 Statistics Exercise 14.4
Question 1. The following distribution gives the daily income of 50 workers if a factory. Convert the distribution above to a lessthantype cumulative frequency distribution and draw its ogive.
Daily income in Rupees  100120  120140  140160  160180  180200 
Number of workers  12  14  8  6  10 
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 xaxis represents the upper limit and yaxis represent the frequency. The curve obtained from the graph is known as less than type ogive curve.
Question 2. During the medical checkup of 35 students of a class, their weights were recorded as follows:
Weight in kg  Number of students 

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 
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 xaxis represents the upper limit and yaxis represents the frequency. The curve obtained is known as less than type ogive.
Now, locate the point 17.5 on the yaxis and draw a line parallel to the xaxis cutting the curve at a point. From this point, now we draw a perpendicular line to the xaxis and the intersection point which is perpendicular to xaxis 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 â€“ 48 has the maximum frequency, hence, this is the modal class
l = 46, h = 2, f_{1 }= 14, f_{0 }= 5 and f_{2} = 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.
Production Yield  5055  5560  6065  6570  7075  7580 
Number of farms  2  8  12  24  38  16 
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 aaxis represents the upper limit and yaxis 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.
Key Features of NCERT Solutions for Class 10 Maths Chapter 14 Statistics:
 These NCERT solutions are developed by the GFG team, with a focus on studentsâ€™ benefit.
 These solutions are entirely accurate and can be used by students to prepare for their board exams.
 Each solution is presented in a stepbystep 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
NCERT Solutions for Class 10 Maths Chapter 14 Statistics – FAQs
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 subtopics.
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 GeeksforGeeks.