Name: Frequency Distribution Table Using Tally Marks
Uploaded: 2019-07-19
Duration: 3 min 41 s
Description: In this video, we will understand the tally marks and draw the frequency distribution table by using the tally marks in a detailed manner with an example.

Question

In the following table the heights of some Indian cricketers are given. Find the median height of the team. [ $5' 10 "$ means $5$ feet $10$ inches]

S.No.	Players Name	Heights
$1$	VVS Laxman	$5' 11 "$
$2$	Parthiv Patel	$5' 3 "$
$3$	Harbhajan Singh	$6' 0 "$
$4$	Sachin Tendulkar	$5' 5 "$
$5$	Gautam Gambhir	$5' 7 "$
$6$	Yuvraj Singh	$6' 1 "$
$7$	Robin Uthappa	$5' 9 "$
$8$	Virender Sehwag	$5' 8 "$
$9$	Zaheer Khan	$6' 0 "$
$10$	MS Dhoni	$5' 11 "$

Accepted Answer

Here, we have

S.No. Players Name Heights

$1$ VVS Laxman $5' 11 "$

$2$ Parthiv Patel $5' 3 "$

$3$ Harbhajan Singh $6' 0 "$

$4$ Sachin Tendulkar $5' 5 "$

$5$ Gautam Gambhir $5' 7 "$

$6$ Yuvraj Singh $6' 1 "$

$7$ Robin Uthappa $5' 9 "$

$8$ Virender Sehwag $5' 8 "$

$9$ Zaheer Khan $6' 0 "$

$10$ MS Dhoni $5' 11 "$

If we arrange the data in ascending order, we get

$5' 3'', 5' 5 ", 5' 7 ", 5' 8 ", 5' 9 ", 5' 11 ", 5' 11 ", 6' 0 ", 6' 0 ", 6' 1 "$ .

Now, we know that if $n$ is the number of observation arranged in ascending or descending order, then

Median when $n$ is odd, ${(\frac{n + 1}{2})}^{th}$ observation

Median when $n$ is even, arithmetic mean of ${(\frac{n}{2})}^{th}$ and ${(\frac{n}{2} + 1)}^{th}$ observation.

Here, total number of observation $(n) = 10$ , which is even.

So, median of the data $= arithmetic mean of (\frac{10}{2}) and (\frac{10}{2} + 1)$ observation

$=$ arithmetic mean of $5^{th}$ and $6^{th}$ observations

$= \frac{5' 9 " + 5' 11 "}{2}$

$= \frac{140 "}{2}$ [Since $1 ft = 12 inc$ ]

$= 70 "$

$= 5' 8 "$ [Correct up to one decimal]

Hence, the median of the given data is $5' 8 "$ .

Important Questions on Frequency Distribution Tables and Graphs

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