Statistics For Class 10th
Class 10Statistics for class 10th students Welcome to the world of statistics! As a student in 10th grade, you may be wondering why statistics is even relevant to your studies. The truth is, statistics is an essential branch of mathematics that helps us make sense of the world around us. By analyzing data and drawing conclusions, we can better understand patterns and trends in everything from social issues to scientific experiments. In this article, we will explore the fundamentals of statistics for class 10th students, including concepts such as mean, median, and mode, as well as how to represent data using graphs and charts. So, let's dive in and explore the exciting world of statistics!
Answers to Maths NCERT Class 10 Chapter 14 – Statistics
Exercise 14.1 Page: 270  Statistics for class 10th students
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:
To find the mean value, we will use the direct method because the numerical value of f_{i} and x_{i} are small.
Find the midpoint of the given interval using the formula.
Midpoint (x_{i}) = (upper limit + lower limit)/2
No. of plants (Class interval) 
No. of houses Frequency (f_{i}) 
Midpoint (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 f_{i }= 20 
Sum f_{i}x_{i} = 162 
The formula to find the mean is:
Mean = x̄ = ∑f_{i }x_{i} /∑f_{i}
= 162/20
= 8.1
Therefore, the mean number of plants per house is 8.1.  Statistics for class 10th students
2. Consider the following distribution of daily wages of 50 workers of a factory.
Daily wages (in Rs.) 
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:
Find the midpoint of the given interval using the formula.
Midpoint (x_{i}) = (upper limit + lower limit)/2
In this case, the value of midpoint (x_{i}) is very large, so let us assume the mean value, a = 550.
Class interval (h) = 20
So, u_{i }= (x_{i} – a)/h
u_{i }= (x_{i} – 550)/20
Substitute and find the values as follows:
Daily wages (Class interval) 
Number of workers frequency (f_{i}) 
Midpoint (x_{i}) 
u_{i }= (x_{i} – 550)/20 
f_{i}u_{i} 
500520 
12 
510 
2 
24 
520540 
14 
530 
1 
14 
540560 
8 
550 = a 
0 
0 
560580 
6 
570 
1 
6 
580600 
10 
590 
2 
20 
Total 
Sum f_{i }= 50 
Sum f_{i}u_{i} = 12 
So, the formula to find out the mean is:
Mean = x̄ = a + h(∑f_{i}u_{i} /∑f_{i }) = 550 + [20 × (12/50)] = 550 – 4.8 = 545.20
Thus, mean daily wage of the workers = Rs. 545.20
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. Statistics for class 10th students
Daily Pocket Allowance(in c) 
1113 
1315 
1517 
1719 
1921 
2123 
2335 
Number of children 
7 
6 
9 
13 
f 
5 
4 
Solution:
To find out the missing frequency, use the mean formula.
Given, mean x̄ = 18
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 
234 
1921 
f 
20 
20f 
2123 
5 
22 
110 
2325 
4 
24 
96 
Total 
f_{i} = 44+f 
Sum f_{i}x_{i} = 752+20f 
The mean formula is
Mean = x̄ = ∑f_{i}x_{i} /∑f_{i }= (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.
4. Thirty women were examined in a hospital by a doctor, and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.
Number of heart beats per minute 
6568 
6871 
7174 
7477 
7780 
8083 
8386 
Number of women 
2 
4 
3 
8 
7 
4 
2 
Solution:
From the given data, let us assume the mean as a = 75.5
x_{i }= (Upper limit + Lower limit)/2
Class size (h) = 3
Now, find the u_{i }and f_{i}u_{i} as follows:
Class Interval 
Number of women (f_{i}) 
Midpoint (x_{i}) 
u_{i} = (x_{i} – 75.5)/h 
f_{i}u_{i} 
6568 
2 
66.5 
3 
6 
6871 
4 
69.5 
2 
8 
7174 
3 
72.5 
1 
3 
7477 
8 
75.5 = a 
0 
0 
7780 
7 
78.5 
1 
7 
8083 
4 
81.5 
2 
8 
8386 
2 
84.5 
3 
6 
Sum f_{i}= 30 
Sum f_{i}u_{i }= 4 
Mean = x̄ = a + h(∑f_{i}u_{i} /∑f_{i })
= 75.5 + 3 × (4/30)
= 75.5 + (4/10)
= 75.5 + 0.4
= 75.9
Therefore, the mean heart beats per minute for these women is 75.9
5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number 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? Statistics for class 10th students
Solution:
The given data is not continuous, so we add 0.5 to the upper limit and subtract 0.5 from the lower limit as the gap between two intervals is 1.
Here, assumed mean (a) = 57
Class size (h) = 3
Here, the step deviation is used because the frequency values are big.
Class Interval 
Number of boxes (f_{i}) 
Midpoint (x_{i}) 
u_{i} = (x_{i} – 57)/h 
f_{i}u_{i} 
49.552.5 
15 
51 
2 
30 
52.555.5 
110 
54 
1 
110 
55.558.5 
135 
57 = a 
0 
0 
58.561.5 
115 
60 
1 
115 
61.564.5 
25 
63 
2 
50 
Sum f_{i} = 400 
Sum f_{i}u_{i} = 25 
The formula to find out the Mean is:
Mean = x̄ = a + h(∑f_{i}u_{i} /∑f_{i })
= 57 + 3(25/400)
= 57 + 0.1875
= 57.19
Therefore, the mean number of mangoes kept in a packing box is 57.19
6. The table below shows the daily expenditure on food of 25 households in a locality.
Daily expenditure(in c) 
100150 
150200 
200250 
250300 
300350 
Number of households 
4 
5 
12 
2 
2 
Find the mean daily expenditure on food by a suitable method.
Solution:
Find the midpoint of the given interval using the formula.
Midpoint (x_{i}) = (upper limit + lower limit)/2
Let us assume the mean (a) = 225
Class size (h) = 50
Class Interval 
Number of households (f_{i}) 
Midpoint (x_{i}) 
d_{i} = x_{i} – 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 f_{i} = 25 
Sum f_{i}u_{i} = 7 
Mean = x̄ = a + h(∑f_{i}u_{i} /∑f_{i })
= 225 + 50(7/25)
= 225 – 14
= 211
Therefore, the mean daily expenditure on food is 211.
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 
0.00 – 0.04 
4 
0.04 – 0.08 
9 
0.08 – 0.12 
9 
0.12 – 0.16 
2 
0.16 – 0.20 
4 
0.20 – 0.24 
2 
Find the mean concentration of SO_{2} in the air.
Solution:
To find out the mean, first find the midpoint of the given frequencies as follows:
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 
Total 
Sum f_{i} = 30 
Sum (f_{i}x_{i}) = 2.96 
The formula to find out the mean is
Mean = x̄ = ∑f_{i}x_{i} /∑f_{i}
= 2.96/30
= 0.099 ppm
Therefore, the mean concentration of SO_{2} in the air is 0.099 ppm.  Statistics for class 10th students
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:
Find the midpoint of the given interval using the formula.
Midpoint (x_{i}) = (upper limit + lower limit)/2
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 f_{i} = 40 
Sum f_{i}x_{i} = 499 
The mean formula is,
Mean = x̄ = ∑f_{i}x_{i} /∑f_{i}
= 499/40
= 12.48 days
Therefore, the mean number of days a student was absent = 12.48.
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 
8598 
Number of cities 
3 
10 
11 
8 
3 
Solution:
Find the midpoint of the given interval using the formula.
Midpoint (x_{i}) = (upper limit + lower limit)/2
In this case, the value of midpoint (x_{i}) is very large, so let us assume the mean value, a = 70.
Class interval (h) = 10
So, u_{i }= (x_{i }– a)/h
u_{i }= (x_{i }– 70)/10
Substitute and find the values as follows:
Class Interval 
Frequency (f_{i}) 
(x_{i}) 
u_{i }= (x_{i }– 70)/10 
f_{i}u_{i} 
4555 
3 
50 
2 
6 
5565 
10 
60 
1 
10 
6575 
11 
70 = a 
0 
0 
7585 
8 
80 
1 
8 
8595 
3 
90 
2 
6 
Sum f_{i} = 35 
Sum f_{i}u_{i} = 2 
So, Mean = x̄ = a + (∑f_{i}u_{i} /∑f_{i}) × h
= 70 + (2/35) × 10
= 69.43
Therefore, the mean literacy part = 69.43%
Exercise 14.2 Page: 275
1. The following table shows the ages of the patients admitted to 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.  Statistics for class 10th students
Solution:
To find out the modal class, let us the consider the class interval with high frequency.
Here, the greatest frequency = 23, so the modal class = 35 – 45,
Lower limit of modal class = l = 35,
class width (h) = 10,
f_{m} = 23,
f_{1} = 21 and f_{2} = 14
The formula to find the mode is
Mode = l + [(f_{m }– f_{1})/ (2f_{m }– f_{1 }– f_{2})] × h
Substitute the values in the formula, we get
Mode = 35+[(2321)/(462114)]×10
= 35 + (20/11)
= 35 + 1.8
= 36.8 years
So the mode of the given data = 36.8 years
Calculation of Mean:
First find the midpoint using the formula, 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 
The mean formula is
Mean = x̄ = ∑f_{i}x_{i} /∑f_{i}
= 2830/80
= 35.375 years
Therefore, the mean of the given data = 35.375 years
2. The following data gives the 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.  Statistics for class 10th students
Solution:
From the given data the modal class is 60–80.
Lower limit of modal class = l = 60,
The frequencies are:
f_{m} = 61, f_{1} = 52, f_{2} = 38 and h = 20
The formula to find the mode is
Mode = l+ [(f_{m }– f_{1})/(2f_{m }– f_{1 }– f_{2})] × h
Substitute the values in the formula, we get
Mode = 60 + [(61 – 52)/ (122 – 52 – 38)] × 20
Mode = 60 + [(9 × 20)/32]
Mode = 60 + (45/8) = 60 + 5.625
Therefore, modal lifetime of the components = 65.625 hours.
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 (in Rs.) 
Number of families 
10001500 
24 
15002000 
40 
20002500 
33 
25003000 
28 
30003500 
30 
35004000 
22 
40004500 
16 
45005000 
7 
Solution:
Given data:
Modal class = 15002000,
l = 1500,
Frequencies:
f_{m} = 40 f_{1} = 24, f_{2} = 33 and
h = 500
Mode formula:
Mode = l + [(f_{m }– f_{1})/ (2f_{m }– f_{1 }– f_{2})] × h
Substitute the values in the formula, we get
Mode = 1500 + [(40 – 24)/ (80 – 24 – 33)] × 500
Mode = 1500 + [(16 × 500)/23]
Mode = 1500 + (8000/23) = 1500 + 347.83
Therefore, modal monthly expenditure of the families = Rupees 1847.83
Calculation for mean:
First find the midpoint using the formula, x_{i }=(upper limit +lower limit)/2
Let us assume a mean, (a) be 2750.
Class Interval 
f_{i} 
x_{i} 
d_{i} = x_{i} – 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 = a 
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 
The formula to calculate the mean,
Mean = x̄ = a +(∑f_{i}u_{i} /∑f_{i}) × h
Substitute the values in the given formula
= 2750 + (35/200) × 500
= 2750 – 87.50
= 2662.50
So, the mean monthly expenditure of the families = Rs. 2662.50
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  Statistics for class 10th students
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:
Given data:
Modal class = 30 – 35,
l = 30,
Class width (h) = 5,
f_{m} = 10, f_{1} = 9 and f_{2} = 3
Mode Formula:
Mode = l + [(f_{m }– f_{1})/ (2f_{m }– f_{1 }– f_{2})] × h
Substitute the values in the given formula
Mode = 30 + [(10 – 9)/ (20 – 9 – 3)] × 5
= 30 + (5/8)
= 30 + 0.625
= 30.625
Therefore, the mode of the given data = 30.625
Calculation of mean:
Find the midpoint using the formula, 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 
105.0 
Sum f_{i} = 35 
Sum f_{i}x_{i} = 1022.5 
Mean = x̄ = ∑f_{i}x_{i} /∑f_{i}
= 1022.5/35
= 29.2 (approx)
Therefore, mean = 29.2  Statistics for class 10th students
5. The given distribution shows the number of runs scored by some top batsmen of the world in one day 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:
Given data:
Modal class = 4000 – 5000,
l = 4000,
class width (h) = 1000,
f_{m} = 18, f_{1} = 4 and f_{2} = 9
Mode Formula:
Mode = l + [(f_{m }– f_{1})/ (2f_{m }– f_{1 }– f_{2})] × h
Substitute the values
Mode = 4000 + [(18 – 4)/ (36 – 4 – 9)] × 1000
= 4000 + (14000/23)
= 4000 + 608.695
= 4608.695
= 4608.7 (approximately)
Thus, the mode of the given data is 4608.7 runs.
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:
Given Data:
Modal class = 40 – 50, l = 40,
Class width (h) = 10, f_{m} = 20, f_{1} = 12 and f_{2} = 11
Mode = l + [(f_{m }– f_{1})/(2f_{m }– f_{1 }– f_{2})] × h
Substitute the values
Mode = 40 + [(20 – 12)/ (40 – 12 – 11)] × 10
= 40 + (80/17)
= 40 + 4.7
= 44.7
Thus, the mode of the given data is 44.7 cars.
Exercise 14.3 Page: 287
1. The following frequency distribution gives the monthly consumption of an 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:
Find the cumulative frequency of the given data as follows:
Class Interval 
Frequency 
Cumulative frequency 
6585 
4 
4 
85105 
5 
9 
105125 
13 
22 
125145 
20 
42 
145165 
14 
56 
165185 
8 
64 
185205 
4 
68 
N = 68 
From the table, it is observed that, N = 68 and hence N/2=34
Hence, the median class is 125145 with cumulative frequency = 42
Where, l = 125, N = 68, cf = 22, f = 20, h = 20
Median is calculated as follows:
= 125 + [(34 − 22)/20] × 20
= 125 + 12
= 137
Therefore, median = 137
To calculate the mode:
Modal class = 125145,
f_{m} or f_{1 }= 20, f_{0 }= 13, f_{2 }= 14 & h = 20
Mode formula:
Mode = l+ [(f_{1 }– f_{0})/(2f_{1 }– f_{0 }– f_{2})] × h
Mode = 125 + [(20 – 13)/ (40 – 13 – 14)] × 20
= 125 + (140/13)
= 125 + 10.77
= 135.77
Therefore, mode = 135.77
Calculate 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 = a 
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 
x̄ = a + h (∑f_{i}u_{i}/∑f_{i}) = 135 + 20 (7/68)
Mean = 137.05
In this case, mean, median and mode are more/less equal in this distribution.
2. If the median of a distribution given below is 28.5, find the value of x & y.
Class Interval 
Frequency 
010 
5 
1020 
x 
2030 
20 
3040 
15 
4050 
y 
5060 
5 
Total 
60 
Solution:
Given data, n = 60
Median of the given data = 28.5
CI 
010 
1020 
2030 
3040 
4050 
5060 
Frequency 
5 
x 
20 
15 
y 
5 
Cumulative frequency 
5 
5+x 
25+x 
40+x 
40+x+y 
45+x+y 
Where, N/2 = 30
Median class is 20 – 30 with a cumulative frequency = 25 + x
Lower limit of median class, l = 20,
cf = 5 + x,
f = 20 & h = 10
Substitute the values
28.5 = 20 + [(30 − 5 − x)/20] × 10
8.5 = (25 – x)/2
17 = 25 – x
Therefore, x = 8.
Now, from cumulative frequency, we can identify the value of x + y as follows:
Since,
60 = 45 + x + y
Now, substitute the value of x, to find y
60 = 45 + 8 + y
y = 60 – 53
y = 7
Therefore, the value of x = 8 and y = 7.
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.  Statistics for class 10th students
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:
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 = 3540
Then, l = 35, cf = 45, f = 33 & h = 5
Median = 35 + [(50 – 45)/33] × 5
= 35 + (25/33)
= 35.76
Therefore, the median age = 35.76 years.
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:  Statistics for class 10th students
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 the leaves.
(Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 – 126.5, 126.5 – 135.5, . . ., 171.5 – 180.5.)
Solution:
Since the data are not continuous reduce 0.5 in the lower limit and add 0.5 in the upper limit.
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 
So, the data obtained are:
N = 40 and N/2 = 20
Median class = 144.5153.5
then, l = 144.5,
cf = 17, f = 12 & h = 9
Median = 144.5 + [(20 – 17)/ 12] × 9
= 144.5 + (9/4)
= 146.75 mm
Therefore, the median length of the leaves = 146.75 mm.
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:
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 
Data:
N = 400 & N/2 = 200
Median class = 3000 – 3500
Therefore, l = 3000, cf = 130,
f = 86 & h = 500
Median = 3000 + [(200 – 130)/86] × 500
= 3000 + (35000/86)
= 3000 + 406.98
= 3406.98
Therefore, the median lifetime of the lamps = 3406.98 hours
6. 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 
14 
47 
710 
1013 
1316 
1619 
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.  Statistics for class 10th students
Solution:
To calculate median:
Class Interval 
Frequency 
Cumulative Frequency 
14 
6 
6 
47 
30 
36 
710 
40 
76 
1013 
16 
92 
1316 
4 
96 
1619 
4 
100 
Given:
N = 100 & N/2 = 50
Median class = 710
Therefore, l = 7, cf = 36, f = 40 & h = 3
Median = 7 + [(50 – 36)/40] × 3
Median = 7 + (42/40)
Median = 8.05
Calculate the Mode:
Modal class = 710,
Where, l = 7, f_{1} = 40, f_{0} = 30, f_{2} = 16 & h = 3
Mode = 7 + [(40 – 30)/(2 × 40 – 30 – 16)] × 3
= 7 + (30/34)
= 7.88
Therefore mode = 7.88
Calculate 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 
58 
1619 
4 
17.5 
70 
Sum f_{i} = 100 
Sum f_{i}x_{i} = 832 
Mean = x̄ = ∑f_{i }x_{i} /∑f_{i}
Mean = 832/100 = 8.32
Therefore, mean = 8.32
7. The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
Weight(in kg) 
4045 
4550 
5055 
5560 
6065 
6570 
7075 
Number of students 
2 
3 
8 
6 
6 
3 
2 
Solution:
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 
Given: N = 30 and N/2= 15
Median class = 5560
l = 55, C_{f} = 13, f = 6 & h = 5
Median = 55 + [(15 – 13)/6] × 5
= 55 + (10/6)
= 55 + 1.666
= 56.67
Therefore, the median weight of the students = 56.67
Exercise 14.4 Page: 293
1. The following distribution gives the daily income of 50 workers in a factory.
Daily income (in Rs.) 
100120 
120140 
140160 
160180 
180200 
Number of workers 
12 
14 
8 
6 
10 
Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive.  Statistics for class 10th students
Solution
Convert the given distribution table to a less than type cumulative frequency distribution, and we get
Daily Income 
Cumulative Frequency (or) Number of workers 
Less than 120 
12 
Less than 140 
26 
Less than 160 
34 
Less than 180 
40 
Less than 200 
50 
From the table plot the points corresponding to the ordered pairs such as (120, 12), (140, 26), (160, 34), (180, 40) and (200, 50) on graph paper and the plotted points are joined to get a smooth curve and the obtained curve is known as less than type ogive curve
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:
From the given data, to represent the table in the form of graph, choose the upper limits of the class intervals are in xaxis and frequencies on yaxis by choosing the convenient scale. Now plot the points corresponding to the ordered pairs given by (38, 0), (40, 3), (42, 5), (44, 9),(46, 14), (48, 28), (50, 32) and (52, 35) on a graph paper and join them to get a smooth curve. The curve obtained is known as less than type ogive.
Locate the point 17.5 on the yaxis and draw a line parallel to the xaxis cutting the curve at a point. From the point, draw a perpendicular line to the xaxis. The intersection point perpendicular to xaxis is the median of the given data. Now, to find the median by making a table.
Class interval 
Number of students(Frequency) 
Cumulative Frequency 

Less than 38 
0 – 38 
0 
0 
Less than 40 
38 – 40 
3 – 0 = 3 
3 
Less than 42 
40 – 42 
5 – 3 = 2 
5 
Less than 44 
42 – 44 
9 – 5 = 4 
9 
Less than 46 
44 – 46 
14 – 9 = 5 
14 
Less than 48 
46 – 48 
28 – 14 = 14 
28 
Less than 50 
48 – 50 
32 – 28 = 4 
32 
Less than 52 
50 – 52 
35 – 22 = 3 
35 
Here, N = 35 and N/2 = 35/2 = 17.5
Median class = 46 – 48
Here, l = 46, h = 2, cf= 14, f = 14
The mode formula is given as:
= 46 + [(17.5 – 14)/ 14] × 2
= 46 + 0.5
= 46 + 0.5 = 46.5
Thus, median is verified.
3. The following table gives production yield per hectare of wheat of 100 farms of a village.
Production Yield (in kg/ha) 
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:
Converting the given distribution to a more than type distribution, we get
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 
From the table obtained draw the ogive by plotting the corresponding points where the upper limits in xaxis and the frequencies obtained in the yaxis are (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) on the graph paper. The graph obtained is known as more than type ogive curve.
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