Question

The following table gives the distribution of the life time of $400$ neon lamps:

Lifetime(in hours)	Number of lamps
$1500 - 2000$	$14$
$2000 - 2500$	$56$
$2500 - 3000$	$60$
$3000 - 3500$	$86$
$3500 - 4000$	$74$
$4000 - 4500$	$62$
$4500 - 5000$	$48$

Find the median lifetime of a lamp.

Answer 1

Hint: The formula used to find the median of a given data is as follows: $Median = l + \left( {\dfrac{{\dfrac{n}{2} - cf}}{f}} \right) \times h$
Where $l$ is the lower limit of median class, $n$ is the sum of all frequencies, $cf$ is the cumulative frequency before the median class, $f$ is the frequency of median class and $h$ is the size of median class.

Complete step-by-step answer:
The class intervals with respective cumulative frequencies can be represented as follows:

Life time(in hours)	Number of lamps	Cumulative frequency$\left( {cf} \right)$
$1500 - 2000$	$14$	$14$
$2000 - 2500$	$56$	$70$
$2500 - 3000$	$60$	$130$
$3000 - 3500$	$86$	$216$
$3500 - 4000$	$74$	$290$
$4000 - 4500$	$62$	$352$
$4500 - 5000$	$48$	$400$
	$n = \sum {f = 400} $

From the table, we obtain $n = 400 \Rightarrow \dfrac{n}{2} = 200$
Cumulative frequency $\left( {cf} \right)$just greater than $\dfrac{n}{2}$$\left( {i.e.,200} \right)$ is $216$, which lies in the interval $3000 - 3500$.
Therefore, median class=$3000 - 3500$
Lower limit of the median class, $l = 3000$
Frequency of the median class, $f = 86$
Cumulative frequency of the class preceding the median class, $cf = 130$
Class size, $h = 500$
Therefore, $Median = l + \left( {\dfrac{{\dfrac{n}{2} - cf}}{f}} \right) \times h$
$Median = 3000 + \left( {\dfrac{{200 - 130}}{{86}}} \right) \times 500$
$ \Rightarrow Median = 3000 + \dfrac{{70 \times 500}}{{86}}$
$ \Rightarrow Median = 3000 + \dfrac{{35000}}{{86}}$
$ \Rightarrow Median = 3406.976$
Therefore, the median lifetime of the lamp is $3406.976$.


Note: The cumulative frequency is calculated by adding each frequency from a frequency distribution table to the sum of its predecessors. The last value will always be equal to the sum of all frequencies, since all frequencies will already have been added to the previous total.