Question

Helppp me plsssssssss

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Answer 1

The class 35 - 40 has maximum frequency. So, it is the modal class.

From the given data,

• 
  `\sf \:\:\:\:\:\:\:\:\:\:x_(k)=35`
• 
  `\sf \:\:\:\:\:\:\:\:\:\:f_(k)=50`
• 
  `\sf \:\:\:\:\:\:\:\:\:\:f_(k-1)=34`
• 
  `\sf \:\:\:\:\:\:\:\:\:\:f_(k+1)=42`
• 
  `\sf \:\:\:\:\:\:\:\:\:\:h=5`

`{\bf \:\: {By\:using\:the\: formula}} \\ \\`

`\:\dag\:{\small{\underline{\boxed{\sf {Mode,\:M_(o) =\sf\red{x_k + {\bigg(h * \: ( ( f_k - f_(k-1)))/( (2f_k - f_(k - 1) - f_(k +1)))\bigg)}}}}}}} \\ \\`

`\sf \:\:\:\:\:\:\:\:\:= 35+ {\bigg(5 * ((50 - 34))/( ( 2 * 50 - 34 - 42))\bigg)} \\ \\`

`\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= 35 +{\bigg(5 * (16)/(24)\bigg)} \\ \\`

`\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= {\bigg(35+(10)/(3)\bigg)} \\ \\`

`\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:(35 + 3.33) =.38.33 \\ \\`

`\:\:\sf {Hence,}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ \large{\underline{\mathcal{\gray{ mode\:=\:38.33}}}} \\ \\`

`{\large{\frak{\pmb{\underline{Additional\: information }}}}}`

MODE

• Most precisely, mode is that value of the variable at which the concentration of the data is maximum.

MODAL CLASS

• In a frequency distribution the class having maximum frequency is called the modal class.

`{\bf{\underline{Formula\:for\: calculating\:mode:}}} \\`

`{\underline{\boxed{\sf {Mode,\:M_(o) =\sf\red{x_k + {\bigg(h * \: ( ( f_k - f_(k-1)))/( (2f_k - f_(k - 1) - f_(k +1)))\bigg)}}}}}} \\ \\`

Where,

`\sf \small\pink{ \bigstar} \: x_(k)= lower\:limit\:of\:the\:modal\:class\:interval.`

`\small \blue{ \bigstar}`
`\sf \: f_(k)=frequency\:of\:the\:modal\:class`

`\sf \small\orange{ \bigstar}\: f_(k-1)=frequency\:of\:the\:class\: preceding\:the\;modal\:class`

`\sf \small\green{ \bigstar}\: f_(k+1)=frequency\:of\:the\:class\: succeeding\:the\;modal\:class`

`\small \purple{ \bigstar}`
`\sf \: h= width \:of\:the\:class\:interval`