Question

Calculate the mean and the median for the following continuous frequency distributions.

$\begin{array}{cccccccc}Class& 0-10& 10-20& 20-30& 30-40& 40-50& 50-60& 60-70\\ f_i& 6& 8& 20& 25& 15& 10& 4\end{array}$

• A. 34.25, 32.4
• B. 34.20, 32.4
• C. 34.20, 32.1
• D. 34, 32.1

Answer 1

The correct option is B

34.20, 32.4

$\begin{array}{cccccccc}Class& 0-10& 10-20& 20-30& 30-40& 40-50& 50-60& 60-70\\ f_i& 6& 8& 20& 25& 15& 10& 4\\ c.fi& 6& 14& 34& 59& 74& 84& 88\end{array}$

Mean: To calculate the mean, we calculate $\sum _{i=1}^{n}xifi$ for each observation and divide by $\sum _{i=1}^{n}fi$

$\sum _{i=1}^n{x}_i{f}_i=5\left(6\right)+15\left(8\right)+25\left(20\right)+35\left(25\right)+45\left(15\right)+55\left(10\right)+65\left(40\right)$

$=30+120+500+875+675+550+260$

$=3010$

$\Rightarrow \overline{x}=\frac{{\sum }_{i-1}^7{x}_i{f}_i}{{\sum }_{i=1}^7{f}_i}=\frac{3010}{88}=34.20=mean$

Median: To calculate the median for a continuous frequency lies and apply the $formul{a}_1$

Median$=l+(\frac{\left(\frac{N}{2}\right)-c}{f}\times h)$

l$\to$ lower limit of median class

N$\to$ sum of frequencies

c $\to$ cumulative frequency of class precending median class

f $\to$ frequency of median class

h $\to$ width of median class

In the given example,

Median class=30-40 $\left\{because\right(\frac{88}{2}{)}^{th}termliesin30-40class\}$

$\Rightarrow l=30$

$N={\sum }_{i=1}^{7}fi=88$

$C=34$

$f=25$

$h=10$

$\therefore Median=M=30+=\left(\frac{\left(\frac{80}{2}\right)-34}{25}\right)\times 10$

$=30+\left(\frac{6}{25}\right)10=30+24=\mathrm{32.4.}$