Question

find the missing frequencies f1 and f2 if the mean of 50 observations is 38.2
ci f

0-10 4

10-20 4

20-30 f1

30-40 10

40-50 f2

50-60 8

60-70 5

Answer 1

Solution:

Let us make a table for easy calculation

$\begin{table}[] \begin{tabular}{|l|l|l|l|} \cline{1-4} Class & \begin{tabular}[c]{@{}l@{}}Frequency\\ fi\end{tabular} & (xi) & xi fi \\ \cline{1-4} 0-10 & 4 & 5 & 20 \\ \cline{1-4} 10-20 & 4 & 15 & 60 \\ \cline{1-4} 20-30 & f1 & 25 & 25 f1 \\ \cline{1-4} 30-40 & 10 & 35 & 350 \\ \cline{1-4} 40-50 & f2 & 45 & 45f2 \\ \cline{1-4} 50-60 & 8 & 55 & 440 \\ \cline{1-4} 60-70 & 5 & 65 & 325 \\ \cline{1-4} Total & 31+f1+f2 & & \begin{tabular}[c]{@{}l@{}}1195+25f1+\\ 45f2\end{tabular} \\ \cline{1-4} \end{tabular} \end{table}$

It is given that Mean is 38.2

$mean = \frac{\Sigma xifi}{\Sigma fi} \\ \\ 38.2 = \frac{1195 + 25f1 + 45f2}{50} \\ \\ 1910 = 1195 + 25f1 + 45f2 \\ \\ 715 = 25f1 + 45f2 \\ \\ 5f1 + 9f2 = 143 \: \: \: \: \: eq1 \\ \\ 31 + f1 + f2 = 50 \\ \\ f1 + f2 = 19 \: \: \: \: eq2$
now solve eq1 and eq2 to find the value of f1 and f2
$5f1 + 9f2 = 143 \\ \\ 5f1 + 5f2 = 95 \\ \\ - \: \: \: \: \: \: - \: \: \: \: \: - \\ \\ 4f2 = 48 \\ \\ f2 = 12 \\ \\ f1 + 12 = 19 \\ \\ f1 = 7$

Thus, f1 =7

f2= 12

Hope it helps you.

Answer 2

Answer:

Table:

$\boxed{\begin{tabular}{ c|c|c|c }CI & f_i & x_i & f_ix_i\\0 - 10 & 4 & 5 & 20\\10 - 20 & 4 & 15 & 60 \\20 - 30 & f_1 & 25 & 25f_1\\30 - 40 & 10 & 35 & 350 \\40 - 50 & f_2 & 45 & 45f_2 \\50 - 60 & 8 & 55 & 440 \\60 - 70 & 5 & 65 & 520\\& \boxed{\Sigma f_i=50} & & \boxed{\Sigma f_ix_i=1195+25f_1+45f_2}\end{tabular}}$

Now, use the formula:

$\large \underline{\boxed{\bf\purple{Mean=\dfrac{\Sigma f_ix_i}{\Sigma f_i} }}}$

$\bf \to 38.2=\dfrac{1195+25f_1+45f_2}{50}$

$\bf \to 25f_1+45f_2=715$

$\bf \to 5f_i+9f_2=143$ - Equation 1

Now, total observations are 50.

$\to \bf 4+4+f_1+10+f_2+8+5=50$

$\to \bf 31+f_1+f_2=50$

$\to \bf f_1+f_2=19$ - Equation 2

Solve equation 1 and 2, we get:

• f₁ = 7
• f₂ = 12

Graph for the lines is in the file attached.