Question

find the mode for the following frequescy distribution table age (in year) 8-10 = 2 student ,10-12 =7 student,12-14=21 student ,14-16=17 student ,16-18=3 student

Answer 1

$\underline\blue{\bold{Given \: data \: is:- }}$

Find the mode of the following :-

$\begin{gathered} \begin{array}{|c|c|} \bf{x_i} & \bf{f_i} \\ 8 - 10 & 2  \\10 - 12 & 7 \\12 - 14 & 21 \\14 - 16 & 17 \\16 - 18 & 3 \end{array}\end{gathered}$

$\begin{gathered}\bf\purple{Formula \: for \: mode:}\end{gathered}$

$\boxed{ \boxed{\bf{Mode = l + \bigg(\dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \bigg) \times h }}}</p><p>$

Where,

• l is lower limit of modal class.
• $\sf{f_1} is \: frequency \: of \: modal \: class$
• $\sf{f_0} \:  is \: frequency \: of \: class \: preceding \: modal \: class$
• $\sf{f_2} \: is \: frequency \: of \: class \: succeeding \: modal \: class$
• h is class height.

$\underline\blue{\bold{Solution :- }}$

☆ Here,

$\begin{gathered}\bf \:  ⟼ \bf\red{l = 12}\end{gathered}$

$\begin{gathered}\bf \:  ⟼ \bf \:\green{f_1 = 21}\end{gathered}$

$\begin{gathered}\bf \:  ⟼ \bf \:\purple{f_0 = 7}\end{gathered}$

$\begin{gathered}\bf \:  ⟼ \bf \:\pink{f_2 = 17}\end{gathered}$

$\begin{gathered}\bf \:  ⟼ \bf \:\blue{h = 2}\end{gathered}$

☆ Now, we know

${{\bf{Mode = l + \bigg(\dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \bigg) \times h }}}$

☆ On substituting the values, we get

${{\bf\bf\implies \:{Mode = 12 + \bigg(\dfrac{21 - 7}{2 \times 21 - 7 - 17} \bigg) \times2 }}}$

${{\bf\bf\implies \:{Mode = 12 + \bigg(\dfrac{14}{42 - 24} \bigg) \times2 }}}$

${{\bf\bf\implies \:{Mode = 12 + \bigg(\dfrac{14}{18} \bigg) \times2 }}}$

${{\bf\bf\implies \:{Mode = 12 + \bigg(\dfrac{7}{9} \bigg) \times2 }}}$

${{\bf\bf\implies \:{Mode = 12 + \bigg(\dfrac{14}{9} \bigg) }}}$

${{\bf\bf\implies \:{Mode = 12 + \bigg(1.55 \bigg)}}}$

${{\bf\bf\implies \:{Mode = 13.55 \: (approx)}}}$

$\large{\boxed{\boxed{\bf{Hence, \: Mode = 13.55 \: (approx.)}}}}$

Answer 2

Answer:

up answer is right hope it helso