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50 students enter for a school javelin throw competition. The distance (in metres) thrown are recorded below :

Distance (in m) 0-20 20-40 40-60 60-80 80-100
Number of students 6 11 17 12 4

(i) Construct a cumulative frequency table.
(ii) Draw a cumulative frequency curve (less than type) and calculate the median distance thrown by using this curve. 

(iii) Calculate the median distance by using the formula for median.

(iv) Are the median distance calculated in (ii) and (iii) same ? 

(i) Solution. Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.

Distance fi CF
0-20 6 6
20-40 11 6+11 = 17
40-60 17 17+17 = 34
60-80 12 34 + 12 = 46
80-100 4 46 + 4 =50

(ii) Answer.  [49.41]

Solution. Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables

Distance Cumulative frequency (C.F)
0
less than 20
less than 40
less than 60
less than 80
less than 100		 0
6
17
34
46
50

n = 50
\frac{n}{2}= \frac{50}{2}= 25

median = 49.41

(iii)Answer.  [49.41]

Solution.  n = 50
\frac{n}{2}= \frac{50}{2}= 25

which lies in interval 40 – 60
l = 40, h = 20, CF = 17 and f = 17
median = l+\frac{\left ( \frac{n}{2}-cf \right )}{f}\times h
=40+\frac{\left ( 25-17 \right )}{17}\times 20
=40+\frac{8\times 20}{17}
=40 + 9.41
= 49.41

(iv) Yes, the median distance calculated in (ii) and (iii) are same.

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The following is the frequency distribution of duration for100 calls made on a mobile phone :

Duration (in seconds) Number of calls
95-125
125-155
155-185
185-215
215-245		 14
22
28
21
15

Calculate the average duration (in sec) of a call and also find the median from cumulative frequency curve.
 

Answer. [170]

Solution.

Duration fi xi u_{i}\frac{\left ( x_{i}-a \right )}{h} fiui
95-125 14 110 -2 -28
125-155 22 140 -1 -22
155=185 28 170 = a 0 0
185-215 21 200 1 21
215-245 21 230 2 30
  \sum f_{i}= 100     \sum f_{i}u_{i}= 1

a = 170, h = 30
Average = a+\frac{\sum f_{i}u_{i}}{\sum f_{i}}\times h= 170+\frac{1}{100}\times 30= 170+0\cdot 3= 170\cdot 3

less than type
Duration Number of calls
less than 95
less than 125
less than 155
less than 185
less than 215
less than 245		 0
0+14 = 14
14+22 =36
36 + 28 = 64
64 + 21 = 85
85 + 15 =100

n = 100

\frac{n}{2}= \frac{100}{2}= 50
median is 170

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The annual rainfall record of a city for 66 days is given in the following table.

Rainfall (in cm) 0-10 10-20 20-30 30-40 40-50 50-60
Number of days 22 10 8 15 5 6

Calculate the median rainfall using ogives (of more than type and of less than type)

Answer. [20]
Solution.

(i) less than type (ii) more than type
Rain fall No.of days Rain fall Number of days
less than 0 0 more than or equal to 0 66
less than 10 0+22 = 22 more than or equal to 10 66-22 = 44
less than 20 22+10 = 32 more than or equal to 20 44-10 = 34
less than 30 32+8 = 40 more than or equal to 30 34-8 = 26
less than 40 40+15 = 55 more than or equal to 40 26-15 = 11
less than 50 55+5 =60 more than or equal to 50 11 - 5 =6
less than 60 60+6 =66 more than or equal to 60 6-6 =0

Now let us draw ogives of more than type and of less than type then find the median

N= 66,\frac{N}{2}= \frac{66}{2}= 33

Here median is 20

 

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Size of agricultural holdings in a survey of 200 families is given in the following table:

Size of agricultural holdings (in ha) Number of families
0-5
5-10
10-15
15-20
20-25
25-30
30-35		 10
15
30
80
40
20
5

Compute median and mode size of the holdings

Answer. [17.77]

Solution.  
  mode= \iota +\left [ \frac{f_{m}-f_{1}}{2f_{m}-f_{1}-f_{2}} \right ]\times h

Size of agricultural holdings fi cf
0-5 10 10
5-10 15 25
10-15 30 55
15-20 80 135
20-25 40 175
25-30 20 195
30-35 5 200

(i) Here n = 200

 \frac{n}{2}= \frac{200}{2}= 100 which lies in interval  (15 – 20)

l = 15, h = 5, f = 80 and cf = 55

 median = l+\frac{\left ( \frac{n}{2}-cf \right )}{4}\times h= 15+\frac{\left ( 100-55 \right )\times 5}{80} 
=15+\frac{45}{16}= 15+2\cdot 81= 17\cdot 81
l = 15, fm = 80, f1 = 30, f2 = 40 and h = 5
mode= \iota +\left [ \frac{f_{m}-f_{1}}{2f_{m}-f_{1}-f_{2}} \right ]\times h

= 15+\left [ \frac{80-30}{2\times 80-30-40} \right ]\times 5
= 15+\left [ \frac{50}{160-70} \right ]\times 5
= 15+\left [ \frac{50}{90} \right ]\times 5

=15+\frac{25}{9}
=15 + 2.77 = 17.77

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The distribution of heights (in cm) of 96 children is given below :               

Height (in cm) Number of children
124-128
128-132
132-136
136-140
140-144
144-148
148-152
152-156
156-160
160-164		 5
8
17
24
16
12
6
4
3
1

Draw a less than type cumulative frequency curve for this data and use it to compute median height of the children.

Answer.          [139]
Solution.            

Height Number of children
less than 124
less than 128
less than  132
less than 136
less than  140
less than  144
less than  148
less than  152
less than  156
less than  160
less than  164		 0
5
13
30
54
70
82
88
92
95
96

\frac{n}{2}= \frac{96}{2}= 48
Hence the median is = 139

 

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The median of the following data is 50. Find the values of p and q, if the sum of all the frequencies is 90.

Marks Frequency
20-30
30-40
40-50
50-60
60-70
70-80
80-90		 p
15
25
20
q
8
10

 

Solution.     

marks Frequency Cummulative frequency
20-30 1 p
30-40 15 15+p
40-50 25 40+p = cf
50-60 20=f 60+p
60-70 q 68+p+q
70-80 8 68+p+q
80-90 10 78+p+q

n = 90, \frac{n}{2}= 45
l = 50, f = 20, cf = 40 + p, h = 10
median = l+\frac{\left ( \frac{n}{2}-cf \right )}{f}\times h

 50= 50+\frac{\left ( 45-40-p \right )}{20}\times 10

0= \frac{5-p}{2}
5 – p = 0
p = 5
78 + 5 + q = 90
q = 90 – 83
q = 7

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The mean of the following frequency distribution is 50, but the frequencies f1 and f2 in classes 20-40 and 60-80, respectively are not known. Find these frequencies, if the sum of all the frequencies is 120.

Class 0-20 20-40 40-60 60-80 80-100
Frequency 17 fi 32 f2 19

            

Solution.

Class (fi) xi \mu _{c}= \frac{\left ( x_{i}-a \right )}{h} fi\mu _{i}
0-20 17 10 -2 -34
20-40 f1 30 -1 -f1
40-60 32 50=a 0 0
60-80 f2 70 1 f2
80-100 19 90 2 38
  \sum f_{i}= 68+f_{i}+f_{2}      

Sum of all frequencies = 120 
\Rightarrow68 + f1 + f2 = 120
f1 + f2 = 52                  …(1)
a = 50, h = 20
mean = a+\frac{\sum f_{i}{\mu _{i}}}{\sum f_{i}}\times h   
50= 50 +  \frac{\left ( 4+f_{2}-f_{1} \right )\times 20}{20}
0= (4 + f2 – f1)
–f2 + f1 = 4                  …(2)
add (1) and (2) we get
2f1 = 56  \Rightarrow f_{1}= 28
Put f1 = 28 in equation (1)
f2 = 52 – 28  \Rightarrow f_{2}= 24

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The table below shows the salaries of 280 persons.

Salary(in thousand (Rs)) Number of persons
5-10
10-15
15-20
20-25
25-30
30-35
35-40
40-45
45-50		 49
133
63
15
6
7
4
2
1

Calculate the median and mode of the data.

Solution.       

Salary fi cf
5-10 49 49
10-15 133 49+133=182
15-20 63 182+63=245
20-25 15 245+15 = 260
25-30 6 260+6 = 266
30-35 7 266+7 = 273
35-40 4 273+4 = 277
40-45 2 277+2 = 279
45-50 1 279+1 = 280

n= 280,\frac{n}{2}= \frac{280}{2}= 140
f1 = 49, fm= 133, f2= 63, cf = 49, f = 133
l = 10, h = 5
median = \iota +\frac{\left ( \frac{n}{2}-cf \right )}{f}\times h
=10+\frac{\left ( 140-49 \right )}{133}\times 5
=10+\frac{91\times 5}{133}
=10+\frac{455}{133}= 10+3\cdot 421
= 13\cdot 421
In thousands = 13.421 × 1000 = 13421 Rs.
Mode = \iota +\left [ \frac{f_{m}-f_{i}}{2f_{m}-f_{i}-f_{2}} \right ]\times h

=10+\left [ \frac{133-49}{2\times 133-49-63} \right ]\times 5

=10+\frac{84\times 5}{266-112}=10+\frac{84\times 5}{154}
=10 + 2.727
=12.727
In thousands = 12.727 × 1000 = 12727 Rs.
 

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The weights of tea in 70 packets are shown in the following table : 

Weight (in gram) Number of packets
200-201
201-202
202-203
203-204
204-205
205-206		 13
27
18
10
1
1

 Draw the less than type and more than type ogives for the data and use them to find the median weight.

Answer.     [201.8]
Solution.

Less than type More than type  
Weight Number of packets Number of packets Number of students
Less than 200 0 More than or equal to 200 70
Less than 201 13 More than or equal to 201 70-13 = 57
Less than 202 40 More than or equal to 202 57-27 =30
Less than 203 58 More than or equal to 203 30-18 =12
Less than 204 68 More than or equal to 204 12-10 = 2
Less than 205 69 More than or equal to 205 2-1 = 1
Less than 206 70 More than or equal to 206 1-1 = 0

Here\, \, n = 70,\frac{n}{2}= \frac{70}{2}= 35                             
Hence median = 201.8

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The weights of tea in 70 packets are shown in the following table :  

Weight (in gram) Number of packets
200-201
201-202
202-203
203-204
204-205
205-206		 13
27
18
10
1
1

Draw the less than type ogive for this data and use it to find the median weight.

Answer. [201.8]
Solution.                 

Weight cf
Less than 201 13
Less than 202 27+13=40
Less than 203 40+18=58
Less than 204 58+10=68
Less than 205 68+1 =69
Less than 206 69+1 = 70


n = 70,\frac{n}{2}= \frac{70}{2}= 35
Hence the median is 201.8

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