Question

32) Find the median of the following data:
CI 0-8 8-16 16-24 24-32 32-40 40-18
F 8 10 16 24 15 7
​

Answer 1

Answer:

Answer

C.I Frequency c.f

0−8 7 7

8−16 9 16

16−24 10 26

24−32 8 34

32−40 12 46

40−48 8 54

Here, N=54 then,

2

N

=

2

54

=27

We can see, 27 lies between class interval 24−32

∴ Median class =24−32

⇒ Upper limit of median class =33

hope this help you

thank you ....

Answer 2

$\begin{gathered}\boxed {\begin{array}{c|c|c|c|c|c|c}\bf {Class\:interval} &\sf 0 - 8 & \sf 8 - 16 &\sf 16 - 24 &\sf 24 - 32 & 32 - 40 &\sf 40 - 48 \\&&&&&&\\ \bf Frequency &\sf 8 &\sf 10&\sf 16 &\sf 24 &\sf 15 &\sf 7 \\&&&&&&\\ \end {array}}\end{gathered}$

$[ \star] \underline{\huge \frak{solution} : }[ \star]$

Arranging the terms in ascending order for median :

$\qquad \qquad \sf \: 8 ,10,16,24,15,7$

$\sf \: Here \qquad \: { \frak{n = 12}} \qquad \: which \: is \: even,$

$\therefore \: \sf \: \: Median \: \: M_{d}$

$= \frak{ \frac{Value \: of \: \frac{ { n}^{th} }{2} \: terms +value \: of \: {( \frac{ {n}^{th} }{2} + 1) }^{2} \: term}{2} }$

$= \sf{ \frac{Value \: of \: {3}^{th} \: terms +value \: of \: {4}^{th} \: term}{2} }$

$\sf \: = \frac{16 + 24}{2}$

$\sf = \frac{ \cancel{40}}{ \cancel2}$

$\sf = 20$

$\huge\frak{Hence, media = 20}$