Find the medianclass interval frequency0-1000 2501000-2000 1902000-300

1.	Find the medianclass interval frequency0-1000 2501000-2000 1902000-3000 1003000-4000 404000-5000 155000-6000 5
Answer» $\large\underline{\sf{Solution-}}$ $\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c\|c\|c}\sf Class\: interval&\sf Frequency\: (f)&\sf \: <klux>CUMULATIVE</klux> \: frequency\\\frac{\qquad \qquad \qquad}{}&\frac{\qquad \qquad}{}\\\sf 0 - <klux>1000</klux>&\sf 250&\sf250\\\\\sf 1000 - 2000 &\sf 190&\sf440\\\\\sf 2000-3000 &\sf 100&\sf540\\\\\sf 3000 - 4000&\sf 40&\sf580\\\\\sf 4000-5000&\sf 15&\sf595\\\\\sf 5000-6000&\sf 5&\sf600\\\frac{\qquad}{}&\frac{\qquad}{}\\\sf & \sf & \end{array}}\end{gathered}\end{gathered}\end{gathered}$ we know that N = 60 and N/2 = 300 The cumulative frequency just greater than 300 is 440 and the corresponding class is 1000 - 2000 Median formula : $\boxed{ \sf M= l + \Bigg \{h \times \dfrac{ \bigg( \dfrac{N}{2} - cf \bigg)}{f} \Bigg \}}$ Here, l denotes lower limit of median class h denotes width of median class f denotes frequency of median class cf denotes cumulative frequency of the class preceding the median class N denotes sum of frequency ACCORDING to the question, median class is 1000 - 2000 so, l = 1000, h = 1000, f = 190, cf = cf of preceding class = 250 N/2 = 300 By substituting all the given values in the formula, $\sf \: \: \: \: M = \: \: 1000 + \bigg(\dfrac{300 - 250}{190} \bigg) \times 1000$ $\sf \: \: \: \: M = \: \: 1000 + \dfrac{5000}{19}$ $\sf \: \: \: \: M = \: \: 1000 + 263.16$ $\sf \: \: \: \: M = \: \: 1263.16$ ─━─━─━─━─━─━─━─━─━─━─━─━─ Additional Information :- Mean :- Using Direct Method $\dashrightarrow\sf Mean = \dfrac{ \sum f_i x_i}{ \sum f_i}$ Using Short Cut Method $\dashrightarrow\sf Mean = \: A \: + \dfrac{ \sum f_i d_i}{ \sum f_i}$ ASSUMED Mean Method :- $\dashrightarrow\sf Mean = \: A \: + \dfrac{ \sum f_i u_i}{ \sum f_i} \times h$

Find the medianclass interval frequency0-1000 2501000-2000 1902000-3000 1003000-4000 404000-5000 155000-6000 5

Answer»

$\large\underline{\sf{Solution-}}$

$\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c|c}\sf Class\: interval&\sf Frequency\: (f)&\sf \: <klux>CUMULATIVE</klux> \: frequency\\\frac{\qquad \qquad \qquad}{}&\frac{\qquad \qquad}{}\\\sf 0 - <klux>1000</klux>&\sf 250&\sf250\\\\\sf 1000 - 2000 &\sf 190&\sf440\\\\\sf 2000-3000 &\sf 100&\sf540\\\\\sf 3000 - 4000&\sf 40&\sf580\\\\\sf 4000-5000&\sf 15&\sf595\\\\\sf 5000-6000&\sf 5&\sf600\\\frac{\qquad}{}&\frac{\qquad}{}\\\sf & \sf & \end{array}}\end{gathered}\end{gathered}\end{gathered}$

we know that

N = 60 and N/2 = 300

The cumulative frequency just greater than 300 is 440 and the corresponding class is 1000 - 2000

Median formula :

$\boxed{ \sf M= l + \Bigg \{h \times \dfrac{ \bigg( \dfrac{N}{2} - cf \bigg)}{f} \Bigg \}}$

Here,

l denotes lower limit of median class

h denotes width of median class

f denotes frequency of median class

cf denotes cumulative frequency of the class preceding the median class

N denotes sum of frequency

ACCORDING to the question,

median class is 1000 - 2000

so,

l = 1000,

h = 1000,

f = 190,

cf = cf of preceding class = 250

N/2 = 300

By substituting all the given values in the formula,

$\sf \: \: \: \: M = \: \: 1000 + \bigg(\dfrac{300 - 250}{190} \bigg) \times 1000$

$\sf \: \: \: \: M = \: \: 1000 + \dfrac{5000}{19}$

$\sf \: \: \: \: M = \: \: 1000 + 263.16$

$\sf \: \: \: \: M = \: \: 1263.16$

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Additional Information :-

Mean :-

Using Direct Method

$\dashrightarrow\sf Mean = \dfrac{ \sum f_i x_i}{ \sum f_i}$

Using Short Cut Method

$\dashrightarrow\sf Mean = \: A \: + \dfrac{ \sum f_i d_i}{ \sum f_i}$

ASSUMED Mean Method :-

$\dashrightarrow\sf Mean = \: A \: + \dfrac{ \sum f_i u_i}{ \sum f_i} \times h$