om the given data0-20 20-40 40-60 60-8015р18 10eservations of the data are5of mode is 52. Then find the value of p’. About the author Hadley
Answer: Class = 0-20, 20-40, 40-60, 60-80, 80-100 Frequency = 15, 6, 18, 10, 5 The maximum class frequency is 18 and the class corresponding to this frequency is 40-60. So, the modal class is 40-60. Here, lower limit of modal class l= 40 class size h= 20 Frequency of the modal class f_{1}=18f 1 =18 Frequency of class preceding the modal class f_{0}=6f 0 =6 Frequency of class succeeding the modal class f_{2}=10f 2 =10 We need to calculate the mode Using formula of mode Mode =l+(\dfrac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}})\times hMode=l+( 2f 1 −f 0 −f 2 f 1 −f 0 )×h Put the value into the formula Mode=40+(\dfrac{18-6}{2\times18-6-10})\times20Mode=40+( 2×18−6−10 18−6 )×20 Mode=52Mode=52 Hence, The mode is 52. Reply
Answer:
Class = 0-20, 20-40, 40-60, 60-80, 80-100
Frequency = 15, 6, 18, 10, 5
The maximum class frequency is 18 and the class corresponding to this frequency is 40-60.
So, the modal class is 40-60.
Here, lower limit of modal class l= 40
class size h= 20
Frequency of the modal class f_{1}=18f
1
=18
Frequency of class preceding the modal class f_{0}=6f
0
=6
Frequency of class succeeding the modal class f_{2}=10f
2
=10
We need to calculate the mode
Using formula of mode
Mode =l+(\dfrac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}})\times hMode=l+(
2f
1
−f
0
−f
2
f
1
−f
0
)×h
Put the value into the formula
Mode=40+(\dfrac{18-6}{2\times18-6-10})\times20Mode=40+(
2×18−6−10
18−6
)×20
Mode=52Mode=52
Hence, The mode is 52.