Question

An analysis for more efficiency in a factory , indicating the distribution of ages of workers was as follows:

Age (in years)	16-19	20-29	30-39	40-49	50-59	60-64
Frequency	15	46	49	32	28	14

(a) Calculate the mean and median of the above data.
(b) Draw a histogram and indicate mode therein.

Answer 1

(a)

Age	Frequency (f)	Midpoint (m)	fm
16 – 19	15	17.5	262.5
20 – 29	46	24.5	1127
30 – 39	49	34.5	1690.5
40 – 49	32	44.5	1424
50 – 59	28	54.5	1526
60 – 64	14	62	868
	$\sum f=184$		$\sum fm=6898$

Mean

$$\begin{gathered} \overline{X}=\frac{\mathrm{\Sigma }fm}{\mathrm{\Sigma }f} \\ =\frac{6898}{184} \\ =37.48 \end{gathered}$$

Hence, the mean of the given distribution is 37.48

Median

Note that the given distribution is in the form of inclusive class intervals. For the calculation of median, first the class intervals must be converted into exclusive form using the following formula.
$\mathrm{Value}\mathrm{of}\mathrm{Adjustment}=\frac{\mathrm{Lower}\mathrm{limit}\mathrm{of}\mathrm{one}\mathrm{class}-\mathrm{Upper}\mathrm{limit}\mathrm{of}\mathrm{the}\mathrm{preceeding}\mathrm{class}}{2}$ Value of lower limit of one class − Value of upper limit of the preceeding class2
The value of adjustment as calculated is then added to the upper limit of each class and subtracted from the lower limit of each class. In this manner, we get the following distribution.

Age	Frequency (f)	Cumulative Frequency (c.f)
15.5 – 19.5 19.5 – 29.5 L1→29.5 – 39.5 39.5 – 49.5 49.5 – 59.5 59.5 – 64.5	15 46 49→ f1 32 28 14	15 61→c.f 110 142 170 184

Median class is given by the
Size of ${\left(\frac{N}{2}\right)}^{\mathrm{th}}$ item = ${\left(\frac{184}{2}\right)}^{\mathrm{th}}$ item = 92^th item.
This corresponds to the class interval of (29.5 − 39.5), so this is the median class.
$$\begin{gathered} \mathrm{Median}=L_1+\frac{{\displaystyle \frac{N}{2}-c.f}}{f}\times i \\ \mathrm{so},\mathrm{Median}=29.5+\frac{{\displaystyle \frac{184}{2}-61}}{49}\times 10 \\ \mathrm{or},\mathrm{Median}=29.5+\frac{31}{49}\times 10 \\ \therefore \mathrm{Median}=29.5+6.32=35.826 \end{gathered}$$

Hence, the median is 35.826

(b)
Here the data is in the form of unequal class interval. So, we will first make appropriate adjustment in the frequencies to make the class intervals equal.

Age	Frequency (f)	Adjusted Frequency
15.5 – 19.5 19.5 – 29.5 29.5 – 39.5 39.5 – 49.5 49.5 – 59.5 59.5 – 64.5	15 46 49 32 28 14	$\frac{15\times 10}{4}=37.5$ $-$ $-$ $-$ $-$ $\frac{14\times 10}{5}=28$

From the above graph, it can be seen that mode is 30.5. However, mean can not be determined graphically.