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Maths 5 Marks Questions

Statistics · Jharkhand Board (JAC) STD 10 (Jharkhand - English Medium). 50 questions.

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Question 15 Marks
An incomplete distribution is given as follows:
Variable
0-10
10-20
20-30
30-40
40-50
50-60
60-70
Frequency
10
20
?
40
?
25
15
You are given that the median value is 35 and the sum of all the frequencies is 170. Using the median formula, fill up the missing frequencies.
Answer
$\text{Median}=25$ and $\sum\text{f}=\text{N}=170$
Let $p_1$ and $p_2$ be two missing frequencies
Variable
Frequency
c.f.
0-10
10
10
10-20
20
30
20-30
$p_1​​​​​​​$​​​​​​​
$30 + p_1​​​​​​​$
30-40
40
$70 + p_1​​​​​​​$
40-50
$p_2​​​​​​​$
$70 + p_1 + p_2​​​​​​​$
50-60
25
$95 + p_1 + p_2​​​​​​​$
60-70
15
$110 + p_1 + p_2​​​​​​​$
$\therefore110+\text{p}_1+\text{p}_2=170$
$\Rightarrow\text{p}_1+\text{p}_2=170-110=60$
Here $\text{N}=170,\frac{\text{N}}{2}=\frac{170}{2}=85$
$\therefore$ Median = 35 which lies in the class 30 - 40
Here$\text{I}=30,\text{f}=40,\text{F}=30+\text{p},$ and $\text{h}=10$
$\text{Median}=\text{l}+\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\times\text{h}$
$\Rightarrow35=30+\frac{85-(30+\text{p}_1)}{40}\times10$
$\Rightarrow35-30=\frac{85-30-\text{p}_1}{4}$
$\Rightarrow5=\frac{85-30-\text{p}_1}{4}$
$20=55-\text{p}_1$
$\Rightarrow\text{p}_1=55-20=35$
But $\text{p}_1+\text{p}_2=60 $
$\therefore\text{ p}_2=60-\text{p}_1=60-35=25 $
Hence missing frequencies are 35 and 25
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Question 25 Marks
The table below shows the daily expenditure on food of 25 households in a locality:
Daily expenditure (in Rs.)
100-150
150-200
200-250
250-.300
300-350
Number of households
4
5
12
2
3
Answer
We may calcute class marks ($x_i$​​​​​​​) for each interval by using the relatio
$\text{x}_\text{i}=\frac{\text{Upper class limit+lover class limit}}{2}$
Class size = 50
Now, taking 225 as assumed mean can we may calculated $d_i, u_i, f_i, u_i$ as follows
Daily expenditure (in Rs.)
$f_i$
$x_i$
$d_i = x_i - 225$
$\text{u}_\text{i}=\frac{\text{x}_\text{i}-225}{\text{h}}$
$f_iu_i$
100-150
4
125
-100
-2
-8
150-200
5
15
-50
-1
-5
200-250
12
255
0
0
0
250-300
2
275
50
1
2
300-350
2
325
100
2
4
 
 
 
 
 
-7
Now we may observe that
$\sum\text{f}_\text{i}=25$
$\sum\text{f}_\text{i}\text{x}_\text{i}=-7$
Mean $\bar{\text{(x)}}=\text{a}+\Big(\frac{{\sum\text{f}_\text{i}\text{u}_\text{i}}}{\sum\text{f}}\Big)\times\text{h}$
$=225+\Big(\frac{-7}{25}\Big)\times50$
$=225-14=211$
So, mean daily expenditure on food is Rs. 211
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Question 35 Marks
The following table gives the frequency distribution of married women by age at marriage:
Age (in years)
Frequency
15-19
53
20-24
140
25-29
98
30-34
32
35-39
12
40-44
9
45-49
5
50-54
3
55-59
3
60 and above
2
Calculate the median and interpret the results.
Answer
Writing classes in exclusive form,
Age (in years)
Class intervals in exclusive from
Frequency (f)
e.f.
15-19
14.5-19.5
53
53
20-24
19.5-24.5
140
193
25-29
24.5-29.5
98
291
30-34
29.5-34.5
32
323
35-39
34.5-39.5
12
335
40-44
39.5-44.5
9
344
45-49
44.5-49.5
5
349
50-54
49.5-54.5
3
352
55-59
54.5-59.5
3
355
60 and above
59.5-above
2
357
Here N = 357
$\therefore \frac{\text{N}}{2}=\frac{357}{2}=178.5=179 $ which lies in the class 20-24
$\therefore \text{l}=20,\text{F}=53,\text{f}=140,\text{h}=5$
$\therefore \text{Median}=\text{l}+\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\times\text{h}=20+\frac{179-53}{140}\times5$
$=20+\frac{126}{140}\times5=20+\frac{126}{28}$
$=20+4.5=24.5$
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Question 45 Marks
A survey regarding the height (in cm) of 51 girls of class X of a school was conducted and the following data was obtained.
Hight in cm
Number of girls
Less than 140
4
Less than 145
11
Less than 150
29
Less than 155
40
Less than 160
46
Less than 165
51
Find the median height.
Answer
Height (in cm)
No. of girls (c.f.)
F
135-140
4
4
140-145
11
7
145-150
29
18
150-155
40
11
155-160
46
6
160-165
51
5
 
 
51
Here $\frac{\text{N}}{2}=\frac{51}{2}$ = 25.5 or 26 which lies in the class 145-150
$\text{I}=145,\text{F}=11,\text{f}=18,\text{h}=5$
$\text{Median}=\text{l}+\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\times\text{h}$
$=145+\frac{25.5-11}{18} \times5$
$=145+\frac{14.5}{18} \times5=145+\frac{72.5}{18}$
$=145+4.03=149.03$
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Question 55 Marks
Find the missing frequency (p) for the following distribution whose mean is 7.68.
x 3 5 7 9 11 13
f 6 8 15 P 8 4
Answer
x f fx
3 6 18
5 8 40
7 15 105
9 P 9P
11 8 88
13 4 52
  N = P + 41 $\sum\text{fx}=9\text{P}+303$
Given
$\Rightarrow\text{Mean}=7.68$
$\Rightarrow\frac{\sum\text{fx}}{\text{N}}=68$
$\Rightarrow\frac{9\text{P}+303}{\text{P}+41}=7.68$
$\Rightarrow9\text{P}+303=\text{P}(7.68)+314.88$
$\Rightarrow9\text{P}-7.68\text{P}=314.88-303$
$\Rightarrow1.32\text{P}=11.88$
$\Rightarrow\text{P}=\frac{11.88}{1.32}$
$\Rightarrow\text{P}=9.$
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Question 65 Marks
If the mean of the following data is 20.6. Find the value of p.
x 10 15 p 25 35
y 3 10 25 7 5
Answer
Mean = 20.6
x f fx
10 3 30
15 10 150
p 25 25p
25 7 175
35 5 175
Total 50 $\sum\text{fx}=530 + 25\text{p}$
$\bar{\text{X}}=20.6$
Mean $\bar{\text{X}}=\frac{\sum\text{fx}}{\sum\text{f}}=\frac{530+25p}{50}$
$\Rightarrow\frac{530+25\text{p}}{50}=20.6$
$\Rightarrow530+25\text{p}=20.6\times50$
$\Rightarrow530+25\text{p}=1030$
$\Rightarrow25\text{p}=1030-530=500$
$\therefore\text{p}=\frac{500}{25}=20$
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Question 75 Marks
The following tables gives the distribution of total household expenditure (in rupees) of manual workers in a city.
Expenditure (in rupees) ($x_i$)
Frequency($f_i$)
Expenditure (in ruppes)($x_i​​​​​​​$)
Frequency($f_i​​​​​​​​​​​​​​$)
100-150
24
300-350
30
150-200
40
350-400
22
200-250
33
400-450
16
250-300
28
450-500
7
Find the average expenditure (in rupees) per household.
Answer
Let the assumed mean (A) = 275.
Class interval
Mid value ($x_i$​​​​​​​)
$d_i = x_i- 275$
$\text{u}_\text{i}=\frac{\text{x}_\text{i}-275}{50}$
Frequency $f_i$
$f_iu_i$
100-150
125
-150
-3
24
-72
150-200
175
-100
-2
40
-80
200-250
225
-50
-1
33
-33
250-300
275
0
0
28
0
300-350
325
50
1
30
30
350-400
375
100
2
22
44
400-450
425
150
3
16
48
450-500
475
200
4
7
28
 
 
 
 
N = 200
$\sum\text{f}_\text{i}\text{u}_\text{i}=-35$
We have
$\text{A}=275,\text{h}=50$
$\text{Mean}=\text{A}+\text{h}\times \frac{\sum\text{f}_1\text{u}_1}{\text{N}}$
$= 275+50\times\frac{-35}{200}$
$=275-8.75$
$=266.25$
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Question 85 Marks
The weekly observation on cost of living index in a certain city for the year 2004 – 2005 are given below. Compute the weekly cost of living index.
Cost of living Index
Number of Students
Cost of living Index
Number of Students
1400-1500
5
1700-1800
9
1500-1600
10
1800-1900
6
1600-1700
20
1900-2000
2
Answer
Let the assume mean (A) = 1650
Class interval
Mid-value $x_i$
$d_i = x_i= - A =x_i - 1650$
$\text{u}_\text{i}=\frac{\text{x}_\text{i}-15}{6}$
Frequency $f_i$
$f_iu_i$
1400-1500
1450
-200
-2
5
-10
1500-1600
1550
-100
-1
10
-10
1600-1700
1650
0
0
20
0
1700-1800
1750
100
1
9
9
1800-1900 1850 200 2 6 12
1900-2000
1950
300
3
2
6
 
 
 
 
N = 52
$\sum\text{f}_\text{i}\text{u}_\text{i}=7$
We have
A = 1650, h = 100
$\text{Mean}=\text{A}+\text{h}\times\frac{\sum\text{f}_\text{i}\text{u}_\text{i}}{\text{N}}$
$= 1650+100\times\frac{7}{52}$
$=1650+\frac{700}{52}$
$=1650+13.46$
$=1663.46$
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Question 95 Marks
Find the missing frequencies in the following distribution, if the sum of the frequencies is 120 and the mean is 50.
Class:
0-20
20-40
40-60
60-80
80-100
Frequency:
17
$f_1$​​​​​​​
32
$f_2​​​​​​​$
19
Answer
it is given that mean of the data is 50.
Let the assumed mean A = 50 and h = 10.
Marks
Mid-value($x_i$)
Frequency
$\text{u}_\text{i}=\frac{\text{x}_\text{i}-50}{20}$
$f_iu_i$
0-20
10
17
-2
-34
20-40
30
$f_1​​​​​​​$
-1
-f1
40-60
50
32
0
0
60-80
70
$f_2​​​​​​​$
1
$f_2​​​​​​​$
80-100
90
19
2
38
 
 
$N = 68 + f_1 +f_2​​​​​​​$​​​​​​​
 
$\sum\text{f}_\text{i}\text{u}_\text{i}=4-\text{f}_1+\text{f}_2$
We know that mean, $\bar{\text{x}}=\text{a}+\text{h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)$
Now, we have $\text{N}=68+\text{f}_1+ \text{f}_2, \ \text{h}=20,\text{A=50,}$
$\sum\text{f}_\text{i}\text{u}_\text{i}=4-\text{f}_1+\text{f}_2$
$50=50+20\Big(\frac{4-\text{f}_1+\text{f}_2}{68+\text{f}_1+\text{f}_2}\Big)$
$\Rightarrow \text{f}_1-\text{f}_2=4$
$\Rightarrow \text{f}_1=4+\text{f}_2\ \dots(\text{i})$
Now, $\text{N}=68+\text{f}_1+\text{f}_2=120$
$\text{f}_1+\text{f}_2=120-68=52$
$\Rightarrow4+\text{f}_2+\text{f}_2=52$ $\big[\text{Using(i)}\big]$
$\Rightarrow\text{f}_2=24$
So,
$\text{f}_1=4+24=28$
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Question 105 Marks
The following is the distribution of height of students of a certain class in a certain city:
Height (in cms)
160-162
163-165
166-168
16-171
172-174
No. of students
15
118
142
127
18
Find the median height.
Answer
Class interval (inclusive)
Class interval (inclusive)
Class interval Frequency
Cumulative frequency
160-162
159.2-162.5
15
15
163-164
162.5-165.5
118
133 (f)
166-168
165.5-168.5
142 (f)
275
169-171
168.5-168.5
127
402
172-174
171.5-174.5
18
420
 
 
N = 420
  
We have
N = 420
$\frac{\text{N}}{2}=\frac{420}{2}=210$
The cumulativefrequency just greater than $\frac{\text{N}}{2}$ is 275 then 165.5-168.5 is the
median class such, that
$\text{l}=165.5,\text{f}=142,\text{F}=133$ and $\text{h}=168.5-105.5=3$
$\text{Mean}= \text{l}+\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\times\text{h}$
$=165.5+\frac{210-133}{142}\times3$
$=165.5+1.626$
$=167.126$
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Question 115 Marks
Calculate the median from the following data:
Marks below
10
20
30
40
50
60
70
80
No. of students
15
35
60
84
96
127
198
250
Answer
Marks below
No. of students
Class interval
Frequency
Cumulative frequency
10
15
0-10
15
15
20
35
10-20
20
35
30
60
20-30
25
84
40
84
30-40
24
84
50
96
40-50
12
96(f)
60
127
50-60
37(f)
127
70
198
60-70
52
198
80
250
70-80
52
250
 
 
 
N = 250
  
We have N = 250
$\frac{\text{N}}{2}=\frac{250}{2}=125$
The cumulative frequency just greater than $\frac{\text{N}}{2}$ is 127 then mediam class is 50-60 such that
$\text{l}=50,\text{f}=31,\text{F}=96,\text{h}=60-50=10$
$\text{Mediam}= \text{L}+\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\times\text{h}$
$=50+\frac{125-96}{31}\times10$
$=50+\frac{29\times10}{31}$
$=\frac{155+290}{31}$
$=\frac{445}{31}$
$=59.35$
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Question 125 Marks
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
Literacy rate (in %)
45-55
55-65
65-75
75-85
85-95
Number of cities
3
10
11
8
3
Answer
We may find class marks by using the relation
$\text{x}_\text{i}=\frac{\text{Upper class limit+Lower class limit}}{2}$
Class size (h) for this data = 10
Now taking 70 as assumed mean (a) wrong Calculate $d_i, u_i,$ and $f_i, u_i$ as follows,
Library rate (in $r_i$)
Number of cities ($f_i​​​​​​​$)
$x_i$
$d_i = x_i - 70$
$\text{x}_\text{i}=\frac{\text{d}_\text{i}}{10}$
$f_iu_i$
45-55
3
50
-20
-2
-6
55-65
10
60
-10
-1
-10
65-75
11
70
0
0
0
75-85
8
80
10
1
8
85-95
3
90
20
2
6
Total
35
 
 
 
-2
Now we may observe that
$\sum\text{f}_\text{i}=35$
$\sum\text{f}_\text{i}\text{u}_\text{i}=-2$
$\text{Mean}\ \bar{\text{(x)}}=\text{a}+\Big(\frac{{{\sum\text{f}_\text{i}\text{u}_\text{i}}}}{{\sum\text{u}_\text{i}}}\Big)\times\text{h}$
$=70+\Big(\frac{-2}{35}\Big)10$
$=70\frac{-4}{7}$
$=70-0.57=69.43$
So, mean literacy rate is 69.437.
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Question 135 Marks
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:
Number of students per Teacher
Number of States / U.T.
15-20
3
20-25
8
25-30
9
30-35
10
35-40
3
40-45
0
45-50
0
50-55
2
Answer
Let assumed mean (A) = 32.5
Number of students per Teacher
Number of States
Class Mark (f)
$d_1 = x - A A = 32.5$
$f_1d_1$
15-20
3
17.5
-15
-45
20-25
8
22.5
-10
-80
25-30
9
27.5
-5
-45
30-35
10
32.5 - A
0
0
35-40
3
37.5
5
15
40-45
0
42.5
15
0
45-50
0
47.5
15
0
50-55
2
52.5
20
40
Total
35
 
 
-115
  1. We see that the class 30-35 has the maximum frequency
$\therefore$ It is the modal class

Here I = 30 $f = 10, f_1 = 9, f_2 = 3, h = 5$

$\therefore\text{Mode}=\text{l}+\frac{\text{f}-\text{f}_1}{2\text{f}-\text{f}_1-\text{f}_2}\times\text{h}$

$=30+\frac{10-9}{2\times10-9-3}\times5=30+\frac{1}{20-12}\times5$

$=30+\frac{5}{8}=30+0.625=30.625$
  1. $\text{Mean}=\text{A}+\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}_\text{i}}$
$=32.5-\frac{115}{35}=32.5-\frac{23}{7}$

$= 32.5 - 3.28 = 29.22 = 29.2$
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Question 145 Marks
Candidate of four schools appear in a mathematics test. The data were as follows:
Schools No. of Candidates Average Score
i 60 75
ii 48 80
iii Not available 55
iv 40 50
If the average score of the candidates of all the four schools is 66, find the number of candidates that appeared from school III.
Answer
Let unknown quantity = p and mean = 66
Schools No. of Candidates (f) Average Score (x) Total Score (fx)
i 60 75 4500
ii 48 80 3840
iii p 55 55p
iv 40 50 2000
Total 148 + p   10340 + 55p
$\therefore$ Average $=\frac{\sum\text{fx}}{\sum\text{f}}$
$\Rightarrow66=\frac{10340+55\text{P}}{148+\text{P}}$
$\Rightarrow66(148+\text{p})=10340+55\text{p}$
$\Rightarrow9768+66\text{p}=10340+55\text{p}$
$\Rightarrow66\text{p}-55\text{p}=10340-9768$
$\Rightarrow11\text{p}=572$
$\Rightarrow\text{p}=\frac{572}{11}=52$
$\therefore$ Number of candidates for school III = 52
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Question 155 Marks
Find the missing frequencies and the median for the following distribution if the mean is 1.46.
No. of accidents
0
1
2
3
4
5
Total
Frequency (No. of days)
46
?
?
25
10
5
200
Answer
Mean = 1.46, N = 200
Let $p_1$ and $p_2$ be the missing frequencies
No. of accidents (x)
Frequency No. of days (f)
c.f.
f
0
46
46
0
1
$p_1$
$46 + p_1$
$p_1$​​​​​​​
2
$p_2​​​​​​​$
$46 + p_1 + p_2​​​​​​​$
$2p_2​​​​​​​$
3
25
$71 + p_1+ p_2​​​​​​​$
75
4
10
$81 + p_1 + p_2​​​​​​​$
40
5
5
$86 + p_1 + p_2​​​​​​​$
25
Total
200
 
$140 + p_1 + p_2​​​​​​​$
$\therefore86+\text{p}_1+\text{p}=200$
$\Rightarrow \text{p}_1+\text{p}_2=200-86=114$
$\therefore \text{p}_1=114-\text{p}_2\ ......(1)$
Now, $\text{Mean}=\frac{\sum\text{f}\ \text{x}}{\sum\text{f}}=\frac{140+\text{p}_1+2\text{p}_2}{200}$
$\Rightarrow1.46=\frac{140+\text{p}_1+2\text{p}_2}{200}$
$\Rightarrow292=140+\text{p}_1+2\text{p}_2$
$\Rightarrow\text{p}_1+2\text{p}_2=292-140=152$
$114-\text{p}_2+2\text{p}_2=152$
$\Rightarrow\text{p}_2=152-114=38$
$\therefore \text{p}_1=114- \text{p}_2=114-38=76$
Hence missing frequencies are 76 and 38
$\text{Median}=\text{Here}\frac{\text{N}}{2}=\frac{200}{2}=100$
$\therefore$ cf of 2 nd class is 46 + 76 = 122
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Question 165 Marks
The distribution below gives the weight of 30 students in a class. Find the median weight of students:
Weight (in kg)
40-45
45-50
50-55
55-60
60-65
65-70
70-75
No. of students
2
3
8
6
6
3
2
Answer
We prepare the cumulative frequency table, as given below.
weight (in kg):
No. of students: ($f_i$​​​​​​​)
cumulative frequency (e.f.)
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
  
We have, N = 30
So, $\frac{\text{N}}{2}=15$
Now, the cumulative frequency just greater than 15 is 19 and the
corresponding class is 55-65
Therefore, 55-60 is the median class.
Here, $\text{l}=55,\text{f}=6,\text{F}=13$ and $\text{h}=5$
We know that
$\text{Median}=\text{l}+\bigg\{\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\bigg\}\times\text{h}$
$=55+\Big\{ \frac{15-13}{6}\Big\}\times5$
$=55+\frac{2\times5}{6}$
$=55+1.667$
$=56.667$
Hence, the median weight of students is 56.67kg.
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Question 175 Marks
If median of the following data is 50. Find the values of p and q, if the sum of all the frequencies is 90.
Marks:
20-30
30-40
40-50
50-60
60-70
70-80
80-90
Frequency:
p
15
25
20
q
8
10
Answer
The given series is in inclusive form. Converting it to exclusive form and preparing the cumulative frequency table, we have
Class interval
Frequency ($f_i$​​​​​​​)
Cumulative Frequency (c.f.)
20-30
p
p
30-40
15
p + 15
40-50
25
p + 40
50-60
20
p + 60
60-70
q
p + q + 60
70-80
8
p + q + 68
80-90
10
p + q + 78
 
78 + p + q = 90
  
Median = 50
It lies in the inrerval 50 - 60, so the median class is 50 - 60. now, we have
$\text{l}=50, \text{h}=10, \text{f}=20, \text{F}= \text{p}+40,\text{N}=90$
We know that
$\text{Median}=\text{l}\ +\bigg\{\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\bigg\}\times\text{h}$
$50=50+\frac{45-(\text{p}+40)}{20}\times10$
$\Rightarrow 0=\frac{5-\text{p}}{2}$
$\Rightarrow\text{p}=5$
And,
$\text{p}+\text{q}+78=90$
$\Rightarrow\text{p}+\text{q}=12$
$\Rightarrow \text{q}=12-5=7$
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Question 185 Marks
Find the mean of each of the following frequency distributions:
Class interval
0-8
 8-16 16-24 24-32 32-40
Frequency
 5 6 4 3 2
Answer
Let the assumed mean be A = 20 and h = 8.
Class interval:
Mid value($x_i$​​​​​​​):
Frequency($f_i​​​​​​​$):
$d_i = x_i - A = x_i- 20$
$\text{u}_\text{i}=\frac{1}{\text{h}}(\text{d}_\text{i})\\=\frac{1}{8}(\text{d}_\text{i})$
$f_iu_i$
0-8
4
5
-16
-2
-10
8-16
12
6
-8
-1
-6
16-24
20
4
0
0
0
24-32
28
3
8
1
3
32-40
36
2
16
2
4
 
 
$\sum\text{f}_\text{i}=20$
 
 
$\sum\text{f}_\text{i}\text{u}_\text{i}=-9$
We know that mean, $\overline{\text{X}}=\text{A}+\text{h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)$
Now, we have $\text{N}= \sum\text{f}_\text{i}=20,\sum\text{f}_\text{i}\text{u}_\text{i}=-9,\text{h}=8$ and A = 20
Putting the values in the above formula, we get
$\overline{\text{X}}=\text{A}+\text{h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)$
$=20+8\Big(\frac{1}{20}\times(-9)\Big)$
$=20-\frac{72}{20}$
$=20-3.6$
$=16.4$
Hence, the mean is 16.4.
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Question 195 Marks
The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.
Runs scored
Number of batsman
Runs scored
Number of batsman
3000-4000
4000-5000
5000-6000
6000-7000
4
18
9
7
7000-8000
8000-9000
9000-10000
10000-11000
6
3
1
1
Find the mode of the data.
Answer
The given data is shown below.
Runs scored
Number of batsmen
3000-4000
4
4000-5000
18
5000-6000
9
6000-7000
7
7000-8000
6
8000-9000
3
9000-10000
1
10000-11000
1
Here, the maximum frequency is 18 so the modal class is 4000-5000.
Therefore
l = 4000
h = 1000
$f_1 = 18$
$f_0 = 4$
$f_2 = 9$
$\therefore\text{Modal}=\text{l}+\frac{\text{f}_1-\text{f}_0}{2\text{f}_1-\text{f}_0-\text{f}_2}\times\text{h}$
$=4000+\frac{18-4}{2\times18-4-9}\times1000$
$=4000+\frac{14}{23}\times1000$
$=4000+608.7$
$=4608.7$
Thus, the mode of the data(or runs scored) is 4608.7.
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Question 205 Marks
If the mean of the following data is 20.6. Find the value of p.
x 10 15 p 25 35
y 3 10 25 7 5
Answer
x
f
fx
5
6
30
10
p
10p
15
6
90
20
10
200
25
5
125
 
N = P127
$\sum \text{fx} = 10\text{P} + 445$
Given
$\Rightarrow \text{Mean} = 15$
$\Rightarrow\frac{\sum\text{Px}}{\text{N}}=5$
$\Rightarrow\frac{109+445}{\text{P}+127}=15$
$\Rightarrow10\text{P}+445=15\text{P}+405$
$\Rightarrow15\text{P}-10\text{P}=445-405$
$\Rightarrow5\text{P}=40$
$\Rightarrow\text{P}=\frac{40}{5}$
$\Rightarrow\text{P}=8$
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Question 215 Marks
To find out the concentration of $S0_2$​​​​​​​ 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.
Answer
Let the assumed mean A = 0.1 and h = 0.04.
Concentration of $SO_2$ (in ppm):
Midvalue($x_i$):
Frequency($f_i​​​​​​​$)
$d_i = x_i - A = x_i - 0.10$
$\text{u}_\text{i}=\frac{1}{\text{h}}(\text{d}_\text{i})\\=\frac{1}{0.04}(\text{d}_\text{i})$
$f_iu_i$
0.00-0.04
0.02
4
-0.08
-2
-8
0.04-0.08
0.06
9
-0.04
-1
-9
0.08-0.12
0.10
9
0
0
0
0.12-0.16
0.14
2
0.04
1
2
0.16-0.20
0.18
4
0.08
2
8
0.20-0.24
0.22
2
0.12
3
6
 
 
$\sum\text{f}_\text{i}=30$
 
 
$\sum\text{f}_\text{i}\text{u}_\text{i}=-1$
We know that mean, $\bar{\text{x}}=\text{A}+\text{h}\Big(\frac{\sum\text{f}_\text{i}\text{u}_\text{i}}{\text{N}}\Big)$
Now, we have $\text{N}=\sum\text{f}_\text{i}=30,\sum\text{f}_\text{i}\text{u}_\text{i}=-1$
h = 0.04 and A = 0.10
Putting the values in the above formula, we have
$\bar{\text{x}}=\text{A}+\text{h}\Big(\frac{1}{\text{N}}{\sum\text{f}_\text{i}\text{u}_\text{i}}\Big)$
$=0.10+0.04\Big[\frac{1}{30}\times(-1)\Big]$
$=0.10-\frac{0.04}{30}$
$=0.10-0.001$
$=0.099$
Hence, the mean concentration of $SO_2​​​​​​​$​​​​​​​ in the air is 1.099 ppm.
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Question 225 Marks
If the mean of the following distribution is 27, find the value of p.
Class
0-10
10-20
20-30
30-40
40-50
Frequency
8
p
12
13
10
Answer
Given: Mean = 27
Let the assumed mean A = 25 and h = 10.
Class
Mid-value($x_i$​​​​​​​):
Frequency($f_i​​​​​​​$):
$d_i = x_i - A = x_i - 25$
$\text{u}_\text{i}=\frac{1}{\text{h}}(\text{d}_\text{i})\\=\frac{1}{10}(\text{d}_\text{i})$
$f_iu_i$
0-10
5
8
-20
-2
-16
10-20
15
p
-10
-1
-p
20-30
25
12
0
0
0
30-40
35
13
10
1
13
40-50
45
10
20
2
20
 
 
$\sum\text{f}_\text{i}=43+\text{p}$
 
 
$\sum\text{f}_\text{i}\text{u}_\text{i}=17-\text{p}$
We know that mean,$\bar{\text{X}}=\text{A}+\text{h}\bigg(\frac{1}{\text{N}}{\sum\text{f}_\text{i}\text{u}_\text{i}}\bigg)$
Now, we have ${\sum\text{f}_\text{i}}=43+\text{p}$
${\sum\text{f}_\text{i}\text{u}_\text{i}}=17-\text{p}$
$\text{h}=10$ and $\text{A}=25$
Putting the values in the above formula, we have
$27=25+10\Big(\frac{1}{43+\text{p}}\times(17-\text{p})\Big)$
$\frac{2}{10}=\Big(\frac{(17-\text{p)}}{43+\text{p}}\Big)$
$43+\text{p=85-5}\text{p}$
$6\text{p}=42$
$\text{p}=7$
Thus, the value of p is 7.
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Question 235 Marks
If the mean of the following frequency distribution is 18, find the missing frequency.
Class interval:
11-13
13-15
15-17
17-19
19-21
21-23
23-25
Frequency
3
6
9
13
f
5
4
Answer
Class interval
Mid-point ($x_i$​​​​​​​)
Frequency ($f_i​​​​​​​$)
$\text{u}_\text{i}=\frac{\text{x}_\text{i}-18}{2}$
$f_iu_i$
11-13
12
3
-3
-9
13-15
14
6
-2
-12
15-17
16
9
-1
-9
17-19
18
13
0
0
16-21
20
f
1
f
21-23
22
5
2
10
23-25
24
4
3
12
 
 
$\sum\text{f}_\text{i}=40+\text{f}$
 
$\sum\text{f}_\text{i}\text{u}_\text{i}=\text{f}-8$
Let us take assumed mean (a) = 18.
Here h = 2
$\text{Mean}=\bar{\text{x}}=\text{a}+\text{h}\Big(\frac{{{\sum\text{f}_\text{i}\text{u}_\text{i}}}}{{\sum\text{f}_\text{i}}}\Big)$
$= 18+2 \Big(\frac{\text{f}-8}{40+\text{f}}\Big)$
$\bar{\text{x}}=18$ (given)
So, $18=18+\frac{2(\text{f}-8)}{40+\text{f}}$
or $\text{f}=8$
Hence, the frequency of the class interval 19-21 is 8.
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Question 245 Marks
Following is the distribution of I.Q. of 100 students. Find the median I.Q.
I.Q.
55-64
65-74
75-84
85-94
95-104
105-114
115-124
125-134
135-144
No. of students
1
2
9
22
33
22
8
2
1
Answer
Here, the frequency table is given in inclusive form. Transforming the given table into exclusive form and prepare the cumulative frequency table.
IQ
Frequency ($f_i$​​​​​​​)
Cummulative Frequency (c.f.)
54.5-64.5
1
1
64.5-74.5
2
3
74.5-84.5
9
12
84.5-94.5
22
34
94.5-104.5
33
67
104.5-114.5
22
89
114.5-124.5
8
97
124.5-134.5
2
99
134.5-144.5
1
100
 
N = 100
  
Here, N = 100
$\frac{\text{N}}{2}=50$
So,
Thus, the cumulative frequency just greater than 50 is 67 and
the corresponding class is 94.5-104.5.
Therefore, 94.5-104.5 is the median class.
Here, l = 94.5, f = 33, F = 34 and h = 9
We know that,
$\text{Median} \ =\text{l} +\bigg\{\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\bigg\}\times\text{h}$
$=94.5+\Big(\frac{50-34}{33}\Big)\times10$
$=94.5+\frac{160}{33}$
$=94.5+4.85$
$=99.35$
Hence, the median is 99.35.
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Question 255 Marks
The following table shows the ages of the patients admitted in a hospital during a year :
Age (in years)
5-15
15-25
25-35
35-45
45-55
55-65
No. of students
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.
Answer
We may observe compute class marks (xi) as per the relation
$\text{x}_\text{i}=\frac{\text{Uppre class limit+ lower claas limit}}{2}$
Now taking 30 as assumed mean (a) we may calculated and $f_1d_1$ as follows
Age(in years)
No. of patients ($f_1$)
Class Marks $x_1$
$d_1 = x_1 - 30$
$f_1d_1$
5-15
6
10
-20
-120
15-25
11
20
-10
-110
25-35
21
30
0
0
35-45
23
40
10
230
45-55
14
50
20
150
55-65
5
60
30
150
Total
80
 
 
430
From the table we may observe that $\sum\text{f}_\text{i}=80$
$\sum\text{f}_\text{i}\text{d}_\text{i}=430$
$\text{Mean}=\text{a}+\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}_\text{i}}$
$=30+\big(\frac{430}{80}\big)$
$=30+5.375$
$=35.38$
Clearly mean of this data is 35.38 It represents that on an average the age of patient
admitted to hospital was 35.58 years.As we may observe that maximum class frequency 23
belonging to class interval 35-45
So, modal class = 35-45
Lower limit (L) of modal class = 35
Frequency ($f_1$​​​​​​​) of modal class = 23
Class size (h) 10
Frequency ($f_0​​​​​​​$​​​​​​​) of class Preceding the Modal = 21
Frequency ($f_2​​​​​​​$​​​​​​​) of class succeeding the modal = 14
Now, $\text{ mode}=\text{L}+\Big(\frac{\text{f}-\text{f}_0} {2\text{f}-\text{f}_0-\text{f}_2}\Big)\text{h}$ $$
$=35+\Big[\frac{23-21}{2(23)-21-14}\Big]\times6$
$=35+\frac{12}{11}$
$=36.8$
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Question 265 Marks
Compare the modal ages of two groups of students appearing for an entrance test:
Age (in years)
16-18
18-20
20-22
22-24
24-26
Group A
50
78
46
28
23
Group B
54
89
40
25
17
Answer
age (in years)
16-18
18-20
20-22
22-24
24-26
Group A
50
78
46
28
23
Group B
54
89
40
25
17
For Group A
Here the maximum frequency is 78, then the correspondingb class 18-20 is modal class
$\text{L}=18,\text{h}=20-18=2,\text{f}=48,\text{f}_1=50,\text{f}_2=46$
$\text{Model}=\text{L}+\frac{\text{f}-\text{f}_1}{2\text{f}-\text{f}_1-\text{f}_2}\times\text{h}$
$=18+\frac{78-50}{156-50-46}\times2$
$=18+\frac{56}{60}=18+0.93$
$=18.9\ \text{years}$
Here the maximum frequency is 89, then the corresponding class 18-20 is modal class
$\text{L}=18,\text{h}=20-18=2,\text{f}=89,\text{f}_1=54,\text{f}_2=40$
$\text{Model}=\text{L}+\frac{\text{f}-\text{f}_1}{2\text{f}-\text{f}_1-\text{f}_2}\times\text{h}$
$=18+\frac{78-54}{156-54-40}\times5$
$=18+0.83$
$=18.83$
Here the mode of age for the group A is higher than group B.
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Question 275 Marks
The frequency distribution table of agriculture holdings in a village is given below:
Area of land (in hactares):
1-3
3-5
5-7
7-9
9-11
11-13
Number of families:
20
45
80
55
40
12
Find the modal agriculture holdings of the village.
Answer
Here the maximum class frequency is 80,
and the class corresponding to this frequency is 5-7.
So, the modal class is 5-7.
l (lower limit of modal class) = 5
$f_1$​​​​​​​ (frequency of the modal class) = 80
$f_0 $(frequency of the class preceding the modal class) = 45
$f_2​​​​​​​$​​​​​​​ (frequency of the class succeeding the modal class) = 55
h (class size) = 2
$\text{Modal}=\text{l}+\Big(\frac{\text{f}_1-\text{f}_0}{\text{2f}_1-\text{f}_0-\text{f}_2}\Big)\text{h}$
$=5+\Big(\frac{80-45}{2(80)-45-55}\Big)\times2$
$=5+\frac{35}{60}\times2=5+\frac{35}{30}$
$=5+1.2=6.2$
Hence, the modal agricultural holdings of the village is 6.2 hectares
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Question 285 Marks
If the median of the following frequency distribution is 28.5 find the missing frequencies:
Class interval
0-10
10-20
20-30
30-40
40-50
50-60
Total
Frequency
5
$f_1$​​​​​​​
20
15
$f_2​​​​​​​$
5
60
Answer
Given: Median = 28.5
We prepare the cumulative frequency table, as given below.
Class interval:
Frequency: ($f_i$)
Cu,ulative frequency (c.f.)
0-10
5
5
10-20
$f_1​​​​​​​$
$5 + f_1​​​​​​​$
20-30
20
$25 + f_1​​​​​​​$
30-40
15
$40 + f_1​​​​​​​$
40-50
$f_2​​​​​​​$
$40 + f_{1 +}f_2​​​​​​​$
50-60
5
$45 + f_1 + f_2​​​​​​​$
 
$N = 60 = 45 + f_1 + f_2​​​​​​​$
  
Now, we have
$\text{N} =60 $
$45+\text{f}_1+\text{f}_2=60$
$\text{f}_2=15-\text{f}_1\ ....(1)$
Also,
$\frac{\text{N}}{2}=30$
,Since the median = 28.5 so the median class is 20
Here, $\text{l}=20,\text{f}=20,\text{F}=5+,\text{f}_1$ and $\text{h}=10$
We know that
$\text{Median}=\text{l}+\bigg\{\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\bigg\}\times\text{h}$
$28.5=20+\Big\{\frac{30-(5+\text{f}_1)}{20}\Big\}\times10$
$8.5=\frac{(25-\text{f}_1)\times10}{20}$
$8.5\times20=250-10\ \text{f}_1$
$10\ \text{f}_1=250-170$
$= 80$
$\text{f}_1=8$
Putting the value of $\text{f}_1 $in (1), we get
$\text{f}_2=15-8$
$= 7$
Hence, the missing frequencies are 7 and 8.
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Question 295 Marks
Find the value of p, if the mean of the following distribution is 20.
x 15 17 17 20 + p 23
f 2 3 4 5p 6
Answer
Given:
$x_i$ 15 17 19 20 + p 23
$f_i$ 2 3 4 5Pp 6
Mean = 20 First of all prepare the frequency table in such a way that its first column consist of the values of the variate($x_i$) and the second column the corresponding frequencies($f_i$). Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing($f_ix_i$). Then, sum of all entries in the column second and denoted by $\sum\text{f}_\text{i}$ and in the third column to obtain$\sum\text{f}_\text{i}\text{x}_\text{i}$.
$x_i$
$f_i$
$f_ix_i$
15
2
30
17
3
51
19
4
76
20 + p
5p
5p(20 + p)
23
6
138
 
$\sum\text{f}_\text{i}=15+5\text{p}$
$\sum\text{f}_\text{i}\text{x}_\text{i}=295+5\text{p}(20+\text{p})$
We know that mean, $\overline{\text{X}}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}$
$20=\frac{295+5\text{p}(20+\text{p})}{15+5\text{p}}$By using cross miltiplication method,
$5\text{p}^2+100\text{p}+295=300+100\text{p}$ $\Rightarrow5\text{p}^2=300-295=5$ $\Rightarrow\text{p}^2=1$ $\Rightarrow\text{p}=1$ Hence, $\text{p}=1$
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Question 305 Marks
Class interval
50-70
70-90
90-110
110-130
130-150
150-180
Frequency
18
12
13
27
8
22
Answer
Let the assumed mean be A = 100 and h = 20.
Class interval
Mid value($x_i$​​​​​​​):
Frequency($f_i​​​​​​​$)
$d_i = x_i - A = x_i - 100$
$\text{u}_\text{i}=\frac{1}{\text{h}}(\text{d}_\text{i})=\frac{1}{20}(\text{d}_\text{i})$
$f_iu_i$
50-70
60
18
-40
-2
-36
70-90
80
12
-20
-1
-12
90-110
100
13
0
0
0
110-130
120
27
20
1
27
130-150
140
8
40
2
16
150-170
160
22
60
3
66
 
 
$\sum\text{f}_\text{i}=100$
 
 
$\sum\text{f}_\text{i}\text{u}_\text{i}=61$
We know that mean $\bar{\text{X}}=\text{A}+\text{h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big) $
Now we have ${\text{N}}= \sum\text{f}_\text{i}=100, \sum\text{f}_\text{i}\text{u}_\text{i},=61,=\text{h}=20$ and A = 100
Putting the values in the above formula, we get
$\bar{\text{X}}=\text{A}+\text{h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)$
$=100+20\Big(\frac{1}{100}\times(61)\Big)$
$=100+\frac{1220}{100}$
$=100+12.20$
$=112.20$
Hence, the mean is 112.20.
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Question 315 Marks
Find the mean of each of the following frequency distributions:
Class interval
0-6
6-12
12-18
18-24
24-30
Frequency
7
5
10
12
6
Answer
Let the assumed mean be A = 15 and h = 6.
Class inteval:
Mid value($x_i$​​​​​​​):
Frequency($f_i​​​​​​​$):
$d_i = x_i - A = x_i - 15$
$\text{u}_\text{i}=\frac{1}{\text{h}}(\text{d}_\text{i}) $
$=\frac{1}{6}(\text{d}_\text{i})$
$f_iu_i$
0-6
3
7
-12
-2
-14
6-12
9
5
-6
-1
-5
12-18
15
10
0
0
0
18-24
21
12
6
1
12
24-30
27
6
12
2
12
 
 
$\sum\text{f}_\text{i}=40$
 
 
$\sum\text{f}_\text{i}\text{u}_\text{i}=5$
We know that mean, $\bar{\text{X}}=\text{A}+\text{h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)$
Now, we have $\text{N} =\sum\text{f}_\text{i}=40,\sum\text{f}_\text{i}\text{u}_\text{i}=5,\text{h}=6$ and A = 15
Putting the values in the above formula, we get
$\bar{\text{X}}=\text{A}+\text{h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)$
$=15+6\Big(\frac{1}{40}\times(5)\Big)$
$=15+\frac{30}{40}$
$=15+0.75$
$=15.75$
Hence, the mean is 15.75.
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Question 325 Marks
The marks in science of 80 students of class X are given below: Find the mode of the marks obtained by the students in science.
Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100
Frequency 3 5 16 12 13 20 5 4 1 1
Answer
Marks
0-10
10-20
20-30
30-40
40-50
50-60
602-70
70-80
80-90
90-100
Frequency
3
5
16
12
13
20
5
4
1
1
Here, the maximum frequency is 20 so the model class is 50-60.
Therefore.
$i = 50$
$h = 10$
$f = 20$
$f_1= 13$
$f_2= 5$
Now,
$\text{Model}=\text{l}+\frac{\text{f}-\text{f}_1}{2\text{f}-\text{f}_1-\text{f}_2}\times\text{h}$
$=50+\frac{20-13}{40-13-5}\times10$
$=50+\frac{7}{22}\times10$
$=50+\frac{70}{22}$
$=50+3.17$
$\text{Model}=53.17$
Thus, the mode of the marks obtained by the students in science is 53.17.
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Question 335 Marks
The median of the distribution given below is 14.4. Find the values of x and y, if the total frequency is 20.
Class interval:
0-6
6-16
12-18
18-24
24+30
Frequency:
4
x
5
y
1
Answer
Class interval
Frequency
Cumulative frequency
0-6
4
4
6-12
x
4 + x
12-18
5
9 + x
18-24
y
9 + x + y
24-30
1
10 + x + y
It is given that n = 20
So, $10+\text{x}+\text{y}=20,\text{i.e.}\ \text{x}+\text{y}=10\ ......(1)$
It is also given that median = 14.4
Which lies in the class interval 12-18
So, $\text{I}=12,\text{f}=5,\text{cf}=4+\text{x},\text{h}=6$
Using the formula,
$\text{Median} = \text{l}+\bigg(\frac{\frac{\text{n}}{2}-\text{cf}}{\text{f}}\bigg)\text{h}$
We get, $14.4=12+\Big(\frac{10-(4+{\text{x})}}{5}\Big)6$
or $14.4=12+\Big(\frac{6-{\text{x}}}{5}\Big)6$
or $\text{x}=4\ ....(2)$
Now,
$10+\text{x}+\text{y}=20$
$4+\text{y}=20-10$
$\text{y}=10-4$
$\text{y} = 6.$
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Question 345 Marks
If the median of the following data is 32.5, find the missing frequencies.
Class intertval
0-10
10-20
20-30
30-40
40-50
50-60
60-70
Total
Frequency
$f_1$​​​​​​​
5
9
12
$f_2$
3
2
40
Answer
Class interval
Frequency
Cumulative frequency
0-10
$f_1$
$f_1​​​​​​​$
10-20
5
$5 + f_1​​​​​​​$
20-30
9
$14 + f_1(f)$
30-40
12
$26 + f_1​​​​​​​$
40-50
$f_2$
$26 + f_1 + f_2​​​​​​​$
50-60
3
$29 + f_1 + f_2​​​​​​​$
60-70
2
$31 + f_1 + f_2​​​​​​​$
 
N = 40
  
Given
Median = 32.5
The median class = 90-40
$\text{l}=30 \therefore 40-32=10,$
$\text{f}=12,\text{F}= 14+\text{f}_1$
$\text{Median} = 1+\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\times\text{h}$
$\Rightarrow32.5=30+\frac{20-(14+\text{f}_1)}{12}\times10$
$\Rightarrow2.5=\frac{6-\text{f}_1}{6}\times5$
$\Rightarrow15=(6-\text{f}_1)5$
$\Rightarrow3=6-\text{f}_1$
$\Rightarrow\frac{15}{5}=6-\text{f}_1$
$\Rightarrow\text{f}_1=3$
Given sum of frequencies = 40
$\Rightarrow3+5+9+12+\text{f}_2+3+2=40$
$\Rightarrow34+\text{f}_2=40$
$\Rightarrow\text{f}_2=6$
$\therefore \text{f}_1=3; \text{f}_2=6$
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Question 355 Marks
Calculate the missing frequency from the following distribution, it being given that the median of the distribution
Age in years
0-10
10-20
20-30
30-40
40-50
No. of persons
5
25
?
18
7
Answer
Let the frequency of the class 20-30 be $f_1$​​​​​​​.It is given that median is 35 which lies
in the class 20-30. So 20-30 is the median class.
Now, lower limit of median class (l) = 20
Height of the class (h) = 10
Frequency of median class = $(f) = f_1$​​​​​​​
Cumulative frequency of preceding median class (F) = 5 + 25
Total frequency $(N) =55 + f_1​​​​​​​$​​​​​​​
Formula to be used to calculate median,
$=\text{l}+\bigg(\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\bigg)\text{(h)}$
Where,
l- Lower limit of median class
h- Height of the class
f- Frequency of median class
F- Cumulative frequency of preceding median class
N- Total frequency
Put the values in the above,
$=24=20+\Bigg(\frac{\frac{\text{(55}+\text{f}_1)}{2}-30}{\text{f}_1}\Bigg)(10)$
$\frac{4}{10}=\frac{55+\text{f}_1-60}{2\text{f}_1}$
$2\ \text{f}_1=50$
$\text{f}_1=25$
Hence, the required frequency is 25.
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Question 365 Marks
An incomplete distribution is given below:
Variable
10-20
20-30
30-40
40-50
50-60
60-70
 70-80
Frequency
12
30
-
65
-
25
 18
You are given that the median value is 46 and the total number of items is 230.
  1. Using the median formula fill up missing frequencies.
  2. Calculate the AM of the completed distribution.
Answer
  1. The median formula fill up missing frequencies.
Class interval
Frequency ($f_i$)
Cumulative frequency (c.f.)
10-20
12
12
20-30
30
42
30-40
x
$42 + f_1​​​​​​​$
40-50
65
$107 + f_1​​​​​​​$
50-60
y
$107 + f_1 + f_2​​​​​​​$
60-70
25
$132 + f_1 + f_2​​​​​​​$
70-80
18
$150 + x + y$
 
N = 230
  
Given,
Median = 46
The median class = 40-50
l = 40, h = 50 - 40 = 10, f = 65, F = 42 + x
Median $=\text{l}+\frac{\big(\frac{\text{N}}{2}\big)-\text{F}}{\text{f}}\times\text{h}$
$\Rightarrow46=40+\frac{115-(42+x)}{65}\times10$
$\Rightarrow46-40=\frac{115-42-\text{x}}{65}\times10$
$\Rightarrow6=\frac{73-\text{x}}{65}=10$
$\Rightarrow\frac{6\times65}{10}=73-\text{x}$
$\Rightarrow\frac{390}{10}=73-\text{x}$
$39=73-\text{x}$
$\text{x}=34$
Given, N = 230
$\Rightarrow12+30+\text{x}+65+\text{y}+25+18=230$
$\Rightarrow12+30+34+65+\text{y}+25+18=230$
$\Rightarrow184+\text{y}=230$
$\Rightarrow\text{y}=230-184$
$\Rightarrow\text{y}=46$
Hence, the missing frequencies are 34 and 46.
  1. The AM of the completed distribution.
Class interval
Mid value (x)
Frequency (f)
fx
10-20
15
12
180
20-30
25
30
750
30-40
35
34
1190
40-50
45
65
2925
50-60
55
46
2530
60-70
65
25
1625
70-80
75
18
1350
 
 
N = 230
$\sum\text{fx}=10550$
Mean $=\frac{\sum\text{fx}}{\text{N}}$
$=\frac{10550}{230}=45.87$
Hence, the mean is 45.87.
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Question 375 Marks
Find the mean, median and mode of the following data,
Classes
0-20
20-40
40-60
60-80
80-100
100-120
120-140
Frequency
6
8
10
12
6
5
3
Answer
Class interval
Mid value x
Frequency f
fx
Cumulative frequency
0-20
10
6
60
6
20-40
30
8
240
14
40-60
50
10
500
24
60-80
70
12
840
36
80-100
90
 06
540
42
100-120
110
5
550
47
120-140
130
3
390
50
 
 
N = 50
$\sum\text{fx}=3120$
  
$\text{Mean}=\frac{\sum\text{f}\ \text{x}}{\text{N}}=\frac{320}{50}=62.4$
We have, N = 50
Then $\frac{\text{N}}{2}=\frac{50}{2}=25$
$\text{c},>\frac{\text{N}}{2}$ is 36 then median class 60-80 such that
l = 60, h = 20, f = 12, F = 24
$\text{Median}=\text{I}+\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\times\text{h}$
$=60+\frac{25.24}{12}\times20=60+1.67$
Modal class $l = 60, h = 20, f = 10, f_2 = 6$
$\text{Mode}=\text{l}+\Big[\frac{\text{f}-\text{f}_1}{2\text{f}-\text{f}_1-\text{f}_2}\Big]\text{h}$
$=60+\Big[\frac{12-10}{24-10-6}\Big]20$
$=60+\frac{40}{8}=65$
$\text{Mode}=65$
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Question 385 Marks
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them:
Monthly consumption (in units)
65-85
85-105
105-125
125-145
145-165
165-185
185-205
No. of consumers
4
5
13
20
14
8
4
Answer
Let assumed mean (A) = 135
Here $\text{N}=68,\frac{\text{N}}{2}=\frac{68}{2}=34$
Monthly consumption (in units)
Class Marks (x)
No.of consumers (f)
c.f.
d = x - A
f.d.
65-85
75
4
4
-60
-240
85-105
95
5
9
-40
-200
105-125
115
13
22
-20
-260
125-145
135 - A
20
42
0
0
145-165
155
14
56
20
280
165-185
175
8
64
40
320
185-205
195
4
68
60
240
Total
 
68
 
 
140
  1. Medim: $\frac{\text{N}}{2}=34$ which lies in the class 125-145
  2. Here l = 125, F = 22, f = 20, h = 20
    $\therefore\text{Median}=\text{l}+\frac{\frac{\text{N}}{2}-\text{f}}{\text{f}}\times\text{h}$
    $=125 +\frac{34-22}{22}\times20=125+12=137\text{ units}$
  3. $\text{Mean}=\text{A}+\frac{\sum\text{f}_i\text{d}_i}{\sum\text{f}_i}=135+\frac{140}{68}$
  4. $=135+2.05=137.05\text{ units}$
  5. Mode; We see that the class 125 - 145 has maximum frequency, therefore it a modal class
  6. Here $l = 125, f = 20, f_1= 13, f_2 = 14, h = 20$
    $\text{Model}=\text{I}+\frac{\text{f}_-\text{f}_1}{2\text{f}-\text{f}_1-\text{f}_2}\times \text{h}$
    $=125+\frac{20-13}{20\times2-14-13}\times20$
    $=125+\frac{7\times20}{40-27}=125+\frac{140}{13}$
    $=125+10.76=135.76\text{ units}$
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Question 395 Marks
The median of the following data is 525. Find the missing frequency, if it is given that there are 100 observations in the data:
Class interval
Frequency
0-100
2
100-200
5
200-300
$f_1$
300-400
12
400-500
17
500-600
20
600-700
$f_2​​​​​​​$​​​​​​​
700-800
9
800-900
7
900-1000
4
Answer
Median = 525, N = 100
Class interval
Frequency (f)
c.f.
0-100
2
2
100-200
5
7
200-300
$f_1$
$7 + f_1​​​​​​​$
300-400
12
$19 + f_1​​​​​​​$
400-500
17
$36 + f_1​​​​​​​$
500-600
20
$56 + e_1​​​​​​​$
600-700
$f_2​​​​​​​$
$56 + f_1 + f_2​​​​​​​$
700-800
9
$65 + f_1 + f_2​​​​​​​$
800-900
7
$72 + f_1 + f_2​​​​​​​$
900-1000
4
$76 + f_1 +f_2​​​​​​​$​​​​​​​
Total
100
  
$\therefore 76+\text{f}_1+\text{f}_2=100$
$\Rightarrow\text{f}_1+\text{f}_2=100-76=24$
$\text{f}_2=24-\text{f}_1\ ......(1)$
$\therefore$ Median = 525 which lies in the alss 500-600
$\text{l}=500,\text{F}=36+\text{f}_1,\text{f}=20,\text{h=100}$
$\therefore\text{Median}= \text{l}+\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\times\text{h}$
$=500+\frac{\frac{100}{2}-(36+\text{f}_1)}{20}\times100$
$\Rightarrow525=500+\frac{50-36-\text{f}_1}{20}\times100$
$\Rightarrow 525=500+(14-\text{f}_1)\times5$
$\Rightarrow 25=70-5\text{f}_1$
$\Rightarrow5\text{f}_1=70-25=45$
$\Rightarrow\text{f}_1=\frac{45}{5}=9$
and $\text{f}_2=24-\text{f}_1=24-9=15$
Hence $\text{f}_1=9,\text{f}_2=15$
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Question 405 Marks
The following table gives the distribution of the life time of 400 neon lamps:
Life time (in hours)
Number of lamps
1500-2000
2000-2500
2500-3000
3000-3500
3500-4000
4000-4500
4500-5000
14
56
60
86
74
62
48
Find the median life.
Answer
Life time f c.f.
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  
$\frac{\text{N}}{2}=\frac{400}{2}=200$
f = 86, C.F. = 130
l = 3000
h = 500
$\text{M}=\text{l}+\bigg(\frac{\frac{\text{N}}{2}-\text{C.F.}}{\text{f}}\bigg)\text{h}$
$=3000+\Big(\frac{200-130}{86}\Big)500$
$=3000+\Big(\frac{70}{86}\Big)500$
$=3000+\frac{35000}{86}$
$=3000+406.97=3406.97$
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Question 415 Marks
The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Compute the missing frequency.
Class
0-20
20-40
40-60
60-80
80-100
100-120
Frequency
5
$f_1$
10
$f_2​​​​​​​$
7
8
Answer
Class
Class Marks (x)
Frequency (y)
fx
0-20
10
5
50
20-40
30
$f_1$
$30f_1$
40-60
50
10
500
60-80
70
$f_2​​​​​​​$​​​​​​​
$70f_2​​​​​​​$
80-100
90
7
630
100-120
110
8
880
Total
 
$30 + f_1 + f_2​​​​​​​$
$2060 + 30f_1+ 70f_2​​​​​​​$
Mean $=62.8$ and $\sum\text{f}=50$
$\therefore 30+\text{f}_1+\text{f}_2=50$
$\Rightarrow\text{f}_1+\text{f}_2=50-30$
$\Rightarrow\text{f}_1+\text{f}_2=20$
$\Rightarrow\text{f}_2=20-\text{f}_1$
Mean $=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}=\frac{2060+30\ \text{f}_1+70\ \text{f}_2}{50}$
$\Rightarrow62.8=\frac{2060+30\ \text{f}_1+70\ \text{f}_2}{50}$
$\Rightarrow 30\text{f}_1+70\ (20-\text{f}_1)=1080$
$\Rightarrow 30\text{f}_1+1400-70\text{f}_1=1080$
$\Rightarrow -40\text{f}_1=1080-1400=-320$
$\Rightarrow \text{f}_1=\frac{-320}{-40}=8$
and $ \text{f}_2=20- \text{f}_1=20-8=12$
Hence $ \text{f}_1=8,\text{f}_2=12$
$2060+30 \text{f}_1+70\text{f}_2=50\times62.8=3140$
$\Rightarrow30 \text{f}_1+70\text{f}_2=3140-2060=1080$
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Question 425 Marks
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
1-4
4-7
7-10
10-13
13-16
16-19
Number 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, and the modal size of the surnames.
Answer
onsider the following table.
Number of letters
No. of surnames
$x_i$
$f_ix_i$
c.f.
1-4
6
2.5
15
6
4-7
30
5.5
165
36
7-10
40
8.5
340
76
10-13
16
11.5
184
92
13-16
4
14.5
58
96
16-19
4
17.5
70
100
 
$\text{N}=\sum\text{f}=100$
 
$\sum\text{f}_\text{i}\text{x}_\text{i}=832$
  
Here, the maximum frequency is 40 So the modal class is 7-10.
Therefore,
$l = 7$
$h = 3$
$f = 40$
$f_1= 30$
$f_2= 16$
$\Rightarrow\text{Modal}=\text{l}+\frac{\text{f}-\text{f}_1}{2\text{f}-\text{f}_1-\text{f}_2}\times\text{h}$
$=7+\frac{10}{34}\times3$
$=7+\frac{30}{34}$
$\text{Mode}=7.88$
Thus, the modal sizes of the surnames is 7.88
$\text{Mean}=\frac{\sum\text{f}_i\ \text{x}_i}{\sum\text{f}}$
$=\frac{832}{100}$
$\text{Mean}=8.32$
Thus, the mean number of letters in the surnames is 8.32.
Median
$=\text{l}+\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\times\text{h}$
$=7+\frac{50-36}{40}\times3$
$=7+\frac{14}{40}\times3$
$=7+\frac{21}{20}$
$\text{Mean}=8.05$
Thus, the median number of letters in the surnames is 8.05.
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Question 435 Marks
Find the mean of each of the following frequency distributions:
Classes
25-29
30-34
35-39
40-44
45-49
50-54
55-59
Frequency
14
22
16
6
5
3
4
Answer
The given series is an inclusive series. Firstly, make it an exclusive series.
Classes interval
Mid-value($x_i$)
Frequency($f_i​​​​​​​$)
$\text{d}_\text{i}=\text{x}_\text{i}-\text{A}\\=\text{x}_\text{i}-42$
$\text{u}_\text{i}=\frac{\text{d}_\text{i}}{\text{h}_\text{i}}=\frac{\text{d}_\text{i}}{5}$
$f_iu_i$
24.5-29.5
27
14
-15
-3
-42
29.5-34.5
32
22
-10
-2
-44
34.5-39.5
37
16
-5
-1
-16
39.5-44.5
A = 42
6
0
0
0
44.5-49.5
47
5
5
1
5
49.5-54.5
52
3
10
2
6
54.5-59.5
57
4
15
3
12
 
 
$\sum\text{f}_\text{i}=70$
 
 
$\sum\text{f}_\text{i}\text{u}_\text{i}=-79$
Let the assumed mean be A = 42 and h = 5
We know that mean, $\overline{\text{X}}=\text{A}+\text{h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)$
Now, we have $\text{N}= \sum \text{f}_\text{i}=70, \sum \text{f}_\text{i} \text{u}_\text{i}=-79,\text{h}=5$ and A = 42
Putting the values in the above formula, we have
$\overline{\text{X}} =\text{A}+\text{h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)$
$=42+5\Big(\frac{1}{70}\times(-79)\Big)$
$=42-\frac{395}{70}$
$=42-5.6438$
$=36.3571$
Hence, the mean is 36.357.
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Question 445 Marks
Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is 50.
x 10 30 50 70 90  
f 17 $f_1$​​​​​​​ 32 $f_2​​​​​​​$ 19 Total 120
Answer
Mean = 50
x f fx
10 17 170
30 $f_1$​​​​​​​ $30f_1​​​​​​​$
50 32 1600
70 $f_2​​​​​​​$ $70f_2​​​​​​​$
90 19 1710
Total $68 + f_1 + f_2 = 120$ $3480 + 30f_1 +70f_2$
$68+\text{f}_1+\text{f}_2=120$
$\Rightarrow\text{f}_1+\text{f}_2=120-68=52$
$\Rightarrow\text{f}_2=52-\text{f}_1$
And $\overline{\text{x}}=\frac{\sum\text{f}\times\text{x}}{\sum\text{f}}$
$\Rightarrow50=\frac{3480+30\text{f}_1+70\text{f}_2}{120}$
$\Rightarrow3480+30\text{f}_1+70\text{f}_2=120\times50=6000$
$\Rightarrow30\text{f}_1+70\text{f}_2=6000-3480=2520$
$\Rightarrow3\text{f}_1+7\text{f}_2=252$
$\Rightarrow3\text{f}_1+7(52-\text{f}_1)=252$
$\Rightarrow3\text{f}_1+364-7\text{f}_1=252$
$\Rightarrow-4\text{f}_1=252-364=-112$
$\Rightarrow\text{f}_1=\frac{-112}{-4}=28$
And $\text{f}_2=52-28=24$
Hence $\text{f}_1=28,\ \text{f}_2=24$
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Question 455 Marks
Find the mean, median and mode of the following data:
Classes
0-50
50-100
100-150
150-200
200-250
250-300
300-350
Frequency
2
3
5
6
5
3
1
Answer
Classes
Frequency ($f_i$)
$x_i$
$f_ix_i$
c.f.
0-50
2
25
50
2
50-100
3
75
225
5
100-150
5
125
625
10
150-200
6
175
1050
16
200-250
5
225
1125
21
250-.300
3
275
825
24
300-350
1
325
325
25
 
$\text{N}=\sum\text{f}=25$
 
$\sum\text{f}_\text{i}\text{x}_\text{i}=4225$
  
Here, the maximum frequency is 6 so the modal class 150-200.
Therefore,
$l = 150$
$h = 50$
$f = 6$
$f_1 = 5$
$f_2 = 5$
$F = 10$
$\text{Mean}=\frac{\sum\text{f}_i\text{x}_i}{\sum\text{f}}$
$=\frac{4225}{25}$
$\text{Mean}=169$
Thus, the mean of the data is 169.
$\text{Median}=\text{l}+\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\times\text{h}$
$=150+\frac{12.5-10}{6}\times50$
$=150+\frac{2.5}{6}\times50$
$=150+\frac{125}{6}$
$\text{Median}=170.83$
Thus, the median of the data is 170.83
$\text{Mode}=\text{l}+\frac{\text{f}-\text{f}_1}{2\text{f}-\text{f}_1-\text{f}_2}\times\text{h}$
$=150+\frac{6-5}{12-5-5}\times50$
$=150+\frac{1}{2}\times50$
$=150+25$
$=175$
Thus, the mode of the data is 175.
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Question 465 Marks
A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are only given to persons having age 18 years onward but less than 60 years.
Age in years
Number of policy holders
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
Answer
Here class width is not same. There is no need to adjust the frequencies according to class intervals. Now given frequencies table is of less than type represented with upper class limits. As policies were given only to persons having age 18 years onwards but less than 60 years we can definite class intervals with their respective cumulative frequencies as below
Age (in years)
No. of policy planers
Cumulative frequency (cr)
18-20
2
2
20-25
6 - 2 = 4
 6
25-30
24 - 6 = 18
24
30-35
45 - 24 = 21
45
35-40
78 - 45 = 33
78
40-45
89 - 78 = 11
89
45-50
92 - 89 = 3
92
50-55
98 - 92 = 6
98
55-60
100 - 98 = 2
100
Total (n)
Now from the table we may observe that n = 100 cumulative frequencies (F) just greater than $\frac{\text{n}}{2}\Big(\text{i.e}.,\frac{100}{2}=50\Big)$ is 78 belonging to inteval 35-40
So median class = 35-40
Lower limit (1) o median class = 35
Class size (h) = 5
Frequlative (f) of median class = 33
Cumulative frequency (f) off class preceding median class = 45
$\text{Median}=\frac{\Big(\frac{\text{n}}{2}-\text{cf}\Big)}{\text{f}}\times\text{h}$
$=35+\Big(\frac{50-45}{33}\Big)\times5$
$=35+\frac{25}{33}$
$=35.76$
So, median age is 35.76 years.
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Question 475 Marks
The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table:
Length (in mm)
118-126
127-135
136-144
145-153
154-162
163-171
172-180
N. of leaves
 3 5 9 12 5 4 2
find the mean length of leaf.
Answer
Calculation for mean
Length (in mm)
Mid-Values($x_i$)
Number of lesves($f_i$)
$f_ix_i$
117.5-126.5
122
3
336
126.5-135.5
131
5
655
135.5-144.5
140
9
1260
144.5-153.5
149
12
1788
153.5-162.5
158
5
790
162.5-171.5
167
4
668
171.5-180.5
176
2
353
 
 
N = 40
$\sum\text{f}_\text{i}\text{x}_\text{i}=5880$
Mean length of the leaf $=\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{x}_\text{i}=\frac{1}{40}\times5880=147$
Calculation for median
The given series is in inclusive form. Converting it to exclusive form and
preparing the cumulative frequency table, we have
Length (in mm)
Number of leaves $(f_i​​​​​​​$​​​​​​​)
Cumulative Frequency (c.f.)
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
 
N = 40
  
Now, the cumulative frequency just greater than 20 is 29 and the corresponding class is
144.5–153.5.
Therefore, 144.5–153.5 is the median class.
Here, $\text{I}=145,\text{f}=12,\text{F}=17$ and $\text{h}=9$
We know that
$\text{Median}=\text{l}+\bigg\{\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\bigg\}\times\text{h}$
$=144.5+\Big(\frac{20-17}{12}\Big)\times9$
$=144.5+\frac{27}{12}$
$=144.5+2.25$
$=146.75$
Hence, the median length of leaf is 146.75mm.
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Question 485 Marks
The following table shows the marks scored by 140 students in an examination of a certain paper:
Marks
0-12
10-20
20-30
30-40
40-50
Number of students
20
24
40
36
20
Calculate the average marks by using all the three methods: direct method, assumed mean deviation and shortcut method.
Answer
We may prepare the table as shown:
Marks:
Mid value($x_i$):
No. of students($f_i$):
$d_i = x_i - A = x_i - 25$
$\text{u}_\text{i}=\frac{1}{\text{h}}(\text{d}_\text{i})\\=\frac{1}{10}(\text{d}_\text{i})$
$f_iu_i$
$f_id_i$
$f_ix_i$
0-10
5
20
-20
-2
-40
-400
100
10-20
15
24
-10
-1
124
-240
-360
20-30
25
40
0
0
0
0
1000
30-40
35
36
10
1
36
360
1260
40-50
45
20
20
2
40
400
900
 
 
$\sum\text{f}_\text{i}=140$
 
 
$\sum\text{f}_\text{i}\text{u}_\text{i}=12$
$\sum\text{f}_\text{i}\text{d}_\text{i}=120$
$\sum\text{f}\text{i}\text{x}_\text{i}=3620$
  1. Direct Method:
  2. We know that mean, $\bar{\text{X}}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}$
    = 3620140 = 25.857 = 3620140 = 25.857
    Hence, the mean is 25.857.
  3. Short-cut method:
  4. Let the assumed mean A = 25
    We know that mean, $\bar{\text{X}}=\text{A}+\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{d}_\text{i}\Big)$
    $=25+\Big(\frac{1}{140}\times(120)\Big)$
    $=25+\frac{120}{140}$
    $=25+0.857$
    $=25.857$
    Hence, the mean is 25.857.
  5. Step deviation method:
  6. Let the assumed mean
    A = 25 and h = 10.
    We know that mean,
    $\overline{\text{X}}=\text{A}+\text{h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)$
    $=25+10\Big(\frac{1}{140}\times(12)\Big)$
    $=25+\frac{120}{140}$
    $=25+0.857$
    $=25.857$
    Hence, the mean is 25.857
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Question 495 Marks
Compute the median for each of the following data:
(1)
Marks
No. of students
Less than 10
0
Less than 30
10
Less than 50
25
Less than 70
43
Less than 90
65
Less than 110
87
Less than 130
96
Less than 150
100
(2)
Marks
No. of students
More than 150
0
More than 140
12
More than 130
27
More than 120
60
More than 110
105
More than 100
124
More than 90
141
More than 80
150
Answer
(1)
We prepare the cumulative frequency table, as given below.
Marks
No. of students ($f_i$)
Cumulative frequency (c.f.)
0-10
0
0
10-30
10
10
30-50
15
25
50-70
18
43
70-90
22
65
90-110
22
87
110-130
9
96
130-150
4
100
 
N = 100
  
Now, we have
$\text{N}=100$
So,
$\frac{\text{N}}{2}=50$
Now, the cumulative frequency just greater than 50 is 65 and the
corresponding class is 70-90
Therefore, 70-90 is the median class.
Here, $\text{l}=70, \text{f}=22,\text{F}=43$ and $\text{h}=20$
We know that
$\text{Median}=\text{l}+\bigg\{\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\bigg\}\times\text{h}$
$=70+\Big\{\frac{50-43}{22}\Big\}\times20$
$=70+\frac{7\times20}{22}$
$=70+\frac{140}{22}$
$=70+6.36$
$=76.36$
Hence, the median is 76.36.
Note: The first class in the table can be omitted also.
(2)
We prepare the cumulative frequency table, as given below.
Marks
No. of students ($f_i​​​​​​​$​​​​​​​)
Cumulative frequency (c.f.)
More than 150
0
0
140-150
12
12
130-140
15
27
120-130
33
60
110-120
45
105
100-120
19
124
90-100
17
141
80-90
9
150
 
N = 150
  
Now, we have
$\text{N}=150$
So.
$\frac{\text{N}}{2}=75$
Thus, the cumulative frequency just greater than 75 is 105 and the
corresponding class is 110-120.
Therefore, 110-120 is the median class.
$\text{l}=120, \text{f}=45,\text{F}=60$ and $\text{h}=-10$
(Because class interval given in descending order)
We know that
$\text{Median}= \text{l}+\bigg\{\frac{\frac{\text{N}}{2}-\text{F}}{\text{f}}\bigg\}\times\text{h}$
$= 120+\Big\{\frac{75-60}{45}\Big\}\times(-10)$
$= 120-\frac{15\times10}{45}$
$= 120-\frac{150}{45}$
$= 120-3.333$
$=116.67$ (approx)
Hence, the median is 116.67.
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Question 505 Marks
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 ₹)
Frequency
1000-1500
24
1500-2000
40
2000-2500
33
2500-3000
28
3000-3500
30
3500-4000
22
4000-4500
16
4500-5000
7
Answer
We may observe that the given data the maximum class frequency is 40 belonging to 1500 – 2000 interval. So modal class = 1500 – 2000
L.L (L) = 1500, f. of M.C ($f_1$) = 40
Frequency of class preceeding modal class$ f_0 = 24$
Frequency of class succeeding modal class $f_2 = 33$
Class size (h) = 50
$\text{Mode}=\text{L}+\Big(\frac{\text{f}_1-\text{f}_0}{2\text{f}-\text{f}_0-\text{f}_2}\Big)\text{h}$
$=1500+\Big[\frac{40-24}{2(40)-24-33}\Big]\times500$
$=1500+\Big[\frac{16}{80-67}\times500\Big]$
$=1500+347.826$
$1847.826=1847.83$
So modal class monthly expenditure was Rs. 1847.83
Now we may find class mark as
$\text{Class mark}=\frac{\text{upper class limit+lower class limit}}{2}$
Class size (h) of given data = 500
Now taking 2750 as assumed mean (a) we may calculate d, 4 and f, 4 as follows.
Expenditure in Rs.
No. of families $f_i$
$x_i$
$\text{d}_\text{i}=\text{x}_\text{i}-2\pi0$
$4_i$
$f_iu_i$
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
28
4500-5000
7
4750
2000
4
28
Total
200
 
 
 
-35
New from table we may observe that
$\sum\text{f}_\text{i}=200$
$\sum\text{f}_\text{i}\text{u}_\text{i}=-35$
$(\bar{\text{x}})\ \text{Mean}=\text{a}+\Big(\frac{\sum\text{f}_\text{i}\text{u}_\text{i}}{\sum\text{f}_\text{i}}\Big)\times\text{h}$
$(\bar{\text{x}})=2750+\Big(\frac{-35}{200}\Big)\times500$
$=2750-87.5$
$=2662.3$
So mean, monthly expenditure was Rs. 2662.50 ps.
We may observe them the given data the maximum class frequency is 10 belonging to class
interval 30 – 35
So modal class 30 – 35
Class size (h) = 5
Lower limit (L) of modal class = 30
Frequency $(f_1$​​​​​​​) of modal class 10
Frequency ($f_0​​​​​​​$) of class preceeding modal class = 9
Frequency ($f_0​​​​​​​$​​​​​​​) of class succeeding modal class = 3
$\text{Mode}=\text{L}+\Big(\frac{\text{f}_1-\text{f}_0}{2\text{f}-\text{f}_0-\text{f}_2}\Big)\text{h}$
$=30+\Big(\frac{10-9}{30-9-3}\Big)5$
$=30+\frac{5}{8}$
$=30.625$
$\text{Mode}=30.6$
It represents that most of states 1 UT have a teacher – student ratio as 30.6
Now we may find class marks by using the relation
$\text{Class mark}=\frac{\text{upper class limit lower class limit}}{2}$
Now taking 325 as assumed mean (a) we may calculated $d_i 4_i$ and $f_i 4_i​​​​​​​$​​​​​​​as following.
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5 Marks Questions - Maths STD 10 Questions - Vidyadip