class interval: 5-7 7-9 9-11 11-13 13-15 15-17 frequency: 55 55 70 150 86 84 find the lower limit of the median class of the distribution.
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Subject:
CBSE BOARD X -
Author:
braelynhouston -
Created:
1 year ago
Answers 2
Explanation:
Solution :-Classes :-
5-7 , 7-9 , 9-11 , 11-13 , 13-15 , 15-17
Frequency :-
55, 55, 70, 150, 86, 84
Cumulative frequencies :-
55, 110, 180, 330, 416 , 500
Sum of all frequencies (N)
= 55+55+70+150+86+84
= 500
Therefore, N = 500
N/2 = 500/2 = 250
250 lies in the cumulative frequency 330
Frequency of Median class = 150
Class interval of Median class = 11-13
Lower boundary of the median class (l) = 11
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Author:
derekgppl
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3
[tex]\begin{gathered}\boxed{\begin{array}{c|c|c} \bf Classes & \bf Frequency & \bf Cumulative\; Frequency \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 5 - 7 & 55& 55 \\ \\ \sf 7 - 9& 55 & 110\\ \\ \sf \: 9 - 11& \sf 70 & \sf 180\\ \\ 11 - 13 & 150 & 330 \\ \\ \sf 13 - 15 & \sf 86 & 416\\ \\ \sf \: 15 - 17& \sf \: 84 & \sf 500 \end{array}} \\ \end{gathered}[/tex]
ANSWER :To find the median class:
- ➡ First add the total number of frequencies.
[tex]\rm \implies \: 55 +55+ 70 + 150 + 86+84 [/tex]
[tex]\bf\implies \: 500[/tex]
Here , we know that
- ➡ f = 500
Now ,
[tex]\rm \implies \: \dfrac{(Total \: number \: of \: frequency)}{2} [/tex]
- Put the given values in this and solve
[tex]\rm \implies \dfrac{500}{2} = 250[/tex]
Henceforth,
- ➡ The value 250 lies in the class interval 11-13 from the cummulative frequency table.
- ➡ As 330 is just greater than 250,
[tex]\rm \longrightarrow \: \therefore \: Median \: class = 11 - 13[/tex]
So:[tex] \rm \therefore \: The \: lower \: limit \: of \: the \: modal \: class \: is \: 11.[/tex]
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Author:
staruiov
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