Question

*️⃣ 50 points *️⃣

Calculate mean given in the attachment!!

Plz!! Fast!!

Answer 1

$\begin{tabular}{|c|c|} \cline{1-2}Data&Frequency \\ \cline{1-2}10-20&4 \\ \cline{1-2}20-30&16 \\\cline{1-2}30-40&56 \\\cline{1-2}40-50&97 \\ \cline{1-2}50-60&124 \\ \cline{1-2}60-70&137 \\ \cline{1-2}70-80&146 \\ \cline{1-2}80-90&150 \\ \cline{1-2} \end{tabular}$

First we will calculate its class mark which is

$\large \boxed{Class \: mark = \dfrac{upper \: limit - lower \: limit}{2}}$

$\begin{tabular}{|c|c|c|} \cline{1-3}Data&Frequency&Class \: Mark \\ \cline{1-3}10-20&4&15\\ \cline{1-3}20-30&16&25 \\\cline{1-3}30-40&56&35 \\\cline{1-3}40-50&97&45\\ \cline{1-3}50-60&124&55 \\ \cline{1-3}60-70&137&65 \\ \cline{1-3}70-80&146&75 \\ \cline{1-3}80-90&150&85 \\ \cline{1-3} \end{tabular}$

Now, we have take an assume mean let 55 be assumed mean

$\begin{tabular}{|c|c|c|} \cline{1-3}Data&Frequency&Class \: Mark \\ \cline{1-3}10-20&4&15\\ \cline{1-3}20-30&16&25 \\\cline{1-3}30-40&56&35 \\\cline{1-3}40-50&97&45\\ \cline{1-3}50-60&124&55=A \\ \cline{1-3}60-70&137&65 \\ \cline{1-3}70-80&146&75 \\ \cline{1-3}80-90&150&85 \\ \cline{1-3} \end{tabular}$

Now we have to find the value of $u_i$ which is

$\large \boxed{u_i = \dfrac{class \: mark - Assummed \: mean}{class \: width}}$

$\begin{tabular}{|c|c|c|c|} \cline{1-4}Data&Frequency&Class \: Mark&u_i=\dfrac{y-A}{c} \\ \cline{1-4}10-20&4&15&-4  \\ \cline{1-4}20-30&16&25&-3\\\cline{1-4}30-40&56&35&-2 \\\cline{1-4}40-50&97&45&-1\\ \cline{1-4}50-60&124&55=A&0 \\ \cline{1-4}60-70&137&65&1 \\ \cline{1-4}70-80&146&75&2 \\\cline{1-4}80-90&150&85&3 \\ \cline{1-4} \end{tabular}$

Now multiply   $u_i$ with frequency,

$\begin{tabular}{|c|c|c|c|c|} \cline{1-5}Data&Frequency&Class \: Mark&u_i=\dfrac{y-A}{c}&u_i \times f \\ \cline{1-5}10-20&4&15&-4&-16\\ \cline{1-5}20-30&16&25&-3&-48\\\cline{1-5}30-40&56&35&-2&-112 \\\cline{1-5}40-50&97&45&-1&-97\\ \cline{1-5}50-60&124&55=A&0&0 \\ \cline{1-5}60-70&137&65&1&137 \\\cline{1-5}70-80&146&75&2&292 \\\cline{1-5}80-90&150&85&3&450 \\ \cline{1-5} \end{tabular}$

Now add up total frequency and product of   $u_i$ with frequency,

$\begin{tabular}{|c|c|c|c|c|} \cline{1-5}Data&Frequency&Class \: Mark&u_i=\dfrac{y-A}{c}&u_i \times f \\ \cline{1-5}10-20&4&15&-4&-16\\ \cline{1-5}20-30&16&25&-3&-48\\\cline{1-5}30-40&56&35&-2&-112 \\\cline{1-5}40-50&97&45&-1&-97\\ \cline{1-5}50-60&124&55=A&0&0 \\ \cline{1-5}60-70&137&65&1&137 \\ \cline{1-5}70-80&146&75&2&292 \\\cline{1-5}80-90&150&85&3&450 \\ \cline{1-5} Total&606&&Total&730\\ \cline{1-5} \end{tabular}$

Now we need to find mean by applying values in formula,

$\large \boxed{Mean = A + c \times \dfrac{ \Sigma u_if}{\Sigma f}}$

$\mathsf{Mean = 55 + 10 \times \dfrac{606}{730}}$

$\mathsf{Mean = 55 + 10 \times 0.830}$

$\mathsf{Mean = 55 + 8.30}$

$\huge \boxed{Mean =63.3}$

Answer 2

Table of data representing the given info :

Use : $\mathsf{di=\dfrac{y-a}{c}}$

a = assumed mean .

c = class interval .

y = class mark .

Sum of frequencies ( fi ) and Sum of fidi will have to be calculated .

$\mathsf{\begin{tabular}{|c|c|c|c|c|} \cline{1-5}Class&Frequency(f_i)&Class \:Marks & u _i=\dfrac{y-a}{c}&f_iu_i \\ \cline{1-5}10-20&4&15&-4&-16\\ \cline{1-5}20-30&16&25&-3&-48\\\cline{1-5}30-40&5&35&-2&-112 \\\cline{1-5}40-50&97&45&-1&-97\\ \cline{1-5}50-60&124&55&0&0 \\ \cline{1-5}60-70&137&65&1&137 \\ \cline{1-5}70-80&146&75&2&292 \\ \cline{1-5}80-90&150&85&3&450 \\ \cline{1-5}\end{tabular}}$

Class mark is found by dividing by 2 the upper and lower class limit .

Assumed mean is the middle most element .

Here I took 55 as the assumed mean .

From the table we can calculate the total value of frequency and total value of ui .

$\mathsf{\Sigma f_i= 4+16+56+97+124+137+146+150}\\\\\implies \mathsf{\Sigma f_i=76+97+124+137+146+150}\\\\\implies \mathsf{\Sigma f_i=173+124+137+146+150}\\\\\implies \mathsf{\Sigma f_i=730}$

$\mathsf{\Sigma fiui=-16-48-112-97+0+137+292+450}\\\\\implies \mathsf{\Sigma fiui=606}$

$\boxed{\red{\mathsf{Mean=a+c\times\dfrac{\Sigma f_iu_i}{\Sigma f_i}}}}$

$\mathsf{Mean=a+c\times \dfrac{606}{730}}\\\\\implies \mathsf{Mean=55+10\times 0.83}\\\\\implies \mathsf{Mean=55+8.3}\\\\\implies \mathsf{Mean=63.3}$

$\huge{\boxed{\green{\bf{Mean=63.3}}}}$