Question

HOLA MATES !!!


PLEASE SOLVE ABOVE QUESTION....

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Answer 1

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

$\huge \underline \bold \pink {Answer}$

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$\huge \underline \bold \blue{Given}$

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 .

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$\textbf{\huge{\orange{Thank\:You}}}$

Answer 2

$\huge\bold\pink{Heyaa !♡}$

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