Question

lesson statistics pls find the answer it's a three mark question​

Answer 1

GiveN :

$\boxed{\begin{array}{c|c|c|c}\bf \: Classes&\bf \: Class \: Mark(y_i) \: &\bf \: Frequency(f_i)& \bf f_iy_i\\\dfrac{\qquad\qquad}{}&\dfrac{\qquad\qquad}{}&\dfrac{\qquad\qquad\qquad}{}& \dfrac{\qquad \qquad}{}\\\sf0 - 6&\sf 3\: &\sf7 &\sf21\\\sf6 - 12&\sf9&\sf5& \sf \:45\\\sf12 - 18&\sf15&\sf10& \sf150\\\sf18 - 24&\sf21&\sf12& \sf \:252\\\sf24 \: - \: 30&\sf27&\sf6& \sf \: 162\\\ \: \dfrac{\qquad\qquad}{ \bf \: Total}& \dfrac{\qquad\qquad}{} & \dfrac{\qquad\qquad}{ \bf \:40}& \dfrac{\qquad\qquad}{ \bf630} \end{array}}$

To FinD :

The mean.

SolutioN :

Analysis :

Since the given data is having grouped frequency therefore we have to use direct method to find the mean.

Required Formula :

Mean using Direct Method,

$\normalsize{\pink{\underline{\boxed{\sf{Mean=\dfrac{\sum f_iy_i}{\sum f_i}}}}}}$

where,

• fᵢ = Total Frequency
• yᵢ = Class Mark

ExplanatioN :

Using the formula,

$\\ :\normalsize\implies{\sf{Mean=\dfrac{\sum f_iy_i}{\sum f_i}}}$

where,

• fᵢ = 40
• fᵢyᵢ = 630

Substituting the values,

$\\ :\normalsize\implies{\sf{Mean=\dfrac{630}{40}}}$

$\\ :\normalsize\implies{\sf{Mean=\dfrac{63\cancel{0}}{4\cancel{0}}}}$

$\\ :\normalsize\implies{\sf{Mean=\dfrac{63}{4}}}$

$\\ \normalsize\therefore\boxed{\bf{\pink{Mean=15.75.}}}$

The mean is 15.75.

Explore More :

Mean by ShortCut Method :

$\normalsize{\sf{Mean=a+\dfrac{\sum f_id_i}{\sum f_i}}}$

Mean by Step - Deviation Method :

$\normalsize{\sf{Mean=a+c\times\dfrac{\sum f_iu_i}{\sum f_i}}}$

where,

• a = Assumed Mean

• fᵢ = Total Frequency

• dᵢ = Class Mark - a

• c = Width of the class

• uᵢ = ${\bf{\dfrac{Class\:Mark-a}{c}}}$