Question

The mean of following frequency table is $50.$

If the total of frequency is $120.$ Then the missing frequencies $f_1,f_2$ are respectively
$\begin{array}{cccccc}\text{Class}& 0-20& 20-40& 40-60& 60-80& 80-100\\ \text{Frequency}& 17& f_1& 32& f_2& 19\end{array}$

• A. $24,28$
• B. $28,24$
• C. $25,27$
• D. $27,25$

Answer 1

The correct option is B $28,24$

Total frequency $=120$
Hence, $17+f_1+32+f_2+19=120$
So, $f_1+f_2=52\cdots \left(1\right)$
We know mean $\overline{x}=\sum \frac{f_i{x}_i}{N}$,
$x_1=\frac{x_i+x_f}{2}=\frac{0+20}{2}=10$
$x_2=\frac{x_i+x_f}{2}=\frac{20+40}{2}=30$
$x_3=\frac{x_i+x_f}{2}=\frac{40+60}{2}=50$
$x_4=\frac{x_i+x_f}{2}=\frac{60+80}{2}=70$
$x_5=\frac{x_i+x_f}{2}=\frac{80+100}{2}=90$
$\overline{x}=\sum \frac{f_i{x}_i}{N}$
$\overline{x}=\frac{f_1{x}_1+f_2{x}_2+f_3{x}_3+f_4{x}_4+f_5{x}_5}{N}$
$50=\frac{17\times 10+f_1\times 30+32\times 50+f_2\times 70+19\times 90}{120}$
$6000=170+30f_1+1600+70f_2+1710$
$6000=30f_1+70f_2+3480$
$2520=30f_1+70f_2$
$3f_1+7f_2=252\cdots \left(2\right)$
Solving $\left(1\right)$ and $\left(2\right)$ we get
$f_1=28,f_2=24$