The starting value of the model class of a distribution is $20$ . The frequency of the class is $18$ . The frequencies of the classes preceeding and succeeding are $8, 10$ width of the model class is $5$ , the mode $=$

18.5

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20.5

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21.4

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22.77

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Solution

The correct option is C $22.77$
$\begin{matrix} R = 20 f_{0} = 18 f_{(- 1)} = 8, f_{(1)} = 10 \Rightarrow M o d e (M_{o}) = l + \frac{(f_{0} - f_{(- 1)})}{(2 f_{0} - f_{(- 1)} - f_{1})} \Rightarrow e (M_{o}) = 20 + \frac{10}{18} \times 5 e (M_{o}) = 20 + \frac{50}{10} = 22.77 \end{matrix}$

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Similar questions

Q. The class marks of a frequency distribution are 15, 20, 25, 30, .. . The class corresponding to the class marks 20 is
(a) 12.5−17.5
(b) 17.5−22.5
(c) 18.5−21.5
(d) 19.5−20.5

(i) Find the class mark of the class 90-120.
(ii) In a frequency distribution, the mid-value of the class is 10 and width of the class is 6. Find the lower limit of the class.
(iii) The width of each of five continuous classes in a frequency distribution is 5 and lower class limit of the lowest class is 10. What is the upper class limit of the highest class?
(iv) The class marks of a frequency distribution are 15, 20, 25, .... Find the class corresponding to the class mark 20.
(v) In the class intervals 10-20, 20-30, find the class in which 20 is included.

Choose the correct answer in each of the following questions:
The class marks of a frequency distribution are 15, 20, 25, 30, .... The class corresponding to the class mark 20 is
(a) 12.5-17.5
(b) 17.5-22.5
(c) 18.5-21.5
(d) 19.5-20.5

Q. Question 6
The class marks of a frequency distribution are given as follows 15, 20, 25….. the class corresponding to the classmark 20 is:

A) 12.5 - 17.5
B) 17.5 - 22.5
C) 18.5 - 21.5
D) 19.5 - 20.5

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The starting value of the model class of a distribution is 20. The frequency of the class is 18. The frequencies of the classes preceeding and succeeding are 8,10 width of the model class is 5, the mode =

The correct option is C 22.77R=20f0=18f(−1)=8,f(1)=10⇒Mode(Mo)=l+(f0−f(−1))(2f0−f(−1)−f1)⇒e(Mo)=20+1018×5e(Mo)=20+5010=22.77

The starting value of the model class of a distribution is $20$ . The frequency of the class is $18$ . The frequencies of the classes preceeding and succeeding are $8, 10$ width of the model class is $5$ , the mode $=$

The correct option is C $22.77$
$\begin{matrix} R = 20 f_{0} = 18 f_{(- 1)} = 8, f_{(1)} = 10 \Rightarrow M o d e (M_{o}) = l + \frac{(f_{0} - f_{(- 1)})}{(2 f_{0} - f_{(- 1)} - f_{1})} \Rightarrow e (M_{o}) = 20 + \frac{10}{18} \times 5 e (M_{o}) = 20 + \frac{50}{10} = 22.77 \end{matrix}$