Question

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.​

Answer 1

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

Answer 2

Solution :

$\begin{gathered}\boxed{\begin{array}{c|c|c} \bf Classes & \bf Frequency & \bf Cumulative\; Frequency \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 5 - 7 & 55& 55 \\ \\ \sf 7 - 9& 55 & 110\\ \\ \sf \: 9 - 11& \sf 70 & \sf 180\\ \\ 11 - 13 & 150 & 330 \\ \\ \sf 13 - 15 & \sf 86 & 416\\ \\ \sf \: 15 - 17& \sf \: 84 & \sf 500 \end{array}} \\ \end{gathered}$

ANSWER :

To find the median class:

• ➡ First add the total number of frequencies.

$\rm \implies \: 55 +55+ 70 + 150 + 86+84$

$\bf\implies \: 500$

Here , we know that

• ➡ f = 500

Now ,

$\rm \implies \: \dfrac{(Total \: number \: of \: frequency)}{2}$

• Put the given values in this and solve

$\rm \implies \dfrac{500}{2} = 250$

Henceforth,

• ➡ The value 250 lies in the class interval 11-13 from the cummulative frequency table.

• ➡ As 330 is just greater than 250,

$\rm \longrightarrow \: \therefore \: Median \: class = 11 - 13$

So:

$\rm \therefore \: The \: lower \: limit \: of \: the \: modal \: class \: is \: 11.$