HOLA MATES !!! PLEASE SOLVE ABOVE QUESTION.... ☺☺✌✌

1.	HOLA MATES !!! PLEASE SOLVE ABOVE QUESTION.... ☺☺✌✌
Answer» $\textbf{\huge{\red{Gud\:Morning\:Miss}}}$ $\huge \underline \bold \pink {Answer}$ $\rule{200}{20}$ $\huge \underline \bold \blue{Given}$ Frequency distribution 104 114 124 134 154 164 $\huge \underline \bold \green {To \: Find}$ Class size and First and SECOND interval $\huge \underline \bold \purple {Solution}$ The class size is 114 - 104 => 10 $\huge\pink{The\:size\:of\:class\:interval\:is\:10}$ If a is a class mark and h is size of class interval, then lower limit and upper limit of the class interval area a $=>\frac{h}{2}\: and \:a\: + \:\frac{h}{2}\: respectively$ ∴ we have h = 10 ∴ Lower limit of first class interval = 104 - $\frac{10}{2}$ = 99 Upper limit of first class interval = 104 + $\frac{10}{2}$ = 109 ∴ First class interval is 99 - 109 Hence, the class intervals are 99 - 109, 109 - 119, 119 - 129, 129 - 139, 139 - 149, 149 - 159, 159 - 169. So first interval is 99 - 109 and Second interval is 109 - 119 . $\rule{200}{20}$ $\textbf{\huge{\orange{Thank\:You}}}$

Answer»

$\textbf{\huge{\red{Gud\:Morning\:Miss}}}$