Name: Arithmetic Mean of Grouped and Ungrouped Data
Uploaded: 2019-07-19
Duration: 8 min 59 s
Description: The arithmetic mean is used to measure the central tendency of data. Watch this video and learn to find the arithmetic mean of grouped and ungrouped data wit...

Question

Find the median, the two quartiles and the mode of the following distribution.

Marks below	Number of students
$10$	$15$
$20$	$35$
$30$	$60$
$40$	$84$
$50$	$96$
$60$	$127$
$70$	$198$
$80$	$250$

Accepted Answer

We have to find the median, the two quartiles and the mode of the following distribution.

For finding median we need to write the class interval and the frequency.

Marks below Class interval Number of students (frequency) Cumulative frequency

$10$ $0 - 10$ $15$ $15$

$20$ $10 - 20$ $50$ $96$

$60$ $127$ $25$ $60$

$40$ $30 - 40$ $24$ $84$

$50$ $40 - 50$ $12$ $96$

$60$ $50 - 60$ $31$ $127$

$70$ $60 - 70$ $71$ $198$

$80$ $70 - 80$ $52$ $250$

$N = 250$ , $\frac{N}{2} = \frac{250}{2} = 125$

Thus, cumulative frequency near to $\frac{N}{2}$ is $127$ , so the median class is $50 - 60$ .

We use the formula, $250$ where,

$l =$ lower limit of the median class

$N =$ size of the sample or population

$c =$ cumulative frequency of the class just before the median class

$f =$ frequency of the median class

$h =$ width of the median class

$l = 50, N = 250, c = 96, f = 31, h = 10$

$\Rightarrow M = 50 + \frac{\frac{250}{2} - 96}{31} \times 10 = 50 + \frac{125 - 96}{31} \times 10 = 50 + \frac{29}{31} \times 10$

$= 50 + \frac{142.5}{127} \times 10 = 50 + 9.35 = 59.35$

Hence, the median is $59.35$ .

The first or lower Quartile $Q_{1} = l + \frac{\frac{N}{4} - c}{f} \times h$

$\Rightarrow \frac{N}{4} = \frac{250}{4} = 62.5$ , so first Quartile class is $(19 - 29)$ .

$l = 20, N = 250, c = 35, f = 25, h = 10$

$Q_{1} = 20 + \frac{\frac{250}{4} - 35}{25} \times 10 = 20 + \frac{62.5 - 35}{25} \times 10$

$= 20 + \frac{27.5}{25} \times 10 = 20 + 11 = 30$

Third or upper Quartile $Q_{3} = l + \frac{\frac{3}{4} N - c}{f} \times h$

$\Rightarrow \frac{3}{4} N = \frac{3 \times 250}{4} = 187.5$

Third Quartile class is $(60 - 70)$ .

$Q_{3} = 60 + \frac{\frac{3}{4} \times 250 - 127}{71} \times 10 = 60 + \frac{187.5 - 127}{198} \times 10$

$Q_{3} = 60 + \frac{60.5}{71} \times 10 = 60 + 8.52 = 68.52$

From the table we can state that $(60 - 70)$ is the modal class as it has the highest frequency.

For finding the mode, we use the formula

Mode $= l + \frac{f_{m} - f_{1}}{2 f_{m} - f_{1} - f_{2}} \times h$

$\Rightarrow 60 + \frac{71 - 31}{2 \times 71 - 31 - 52} \times 10 = 60 + \frac{40}{59} \times 10 = 60 + 6.78 = 66.78$

Daily savings (in ₹)	Number of workers
$1 - 2.99$	$6$
$3 - 4.99$	$53$
$5 - 6.99$	$85$
$7 - 8.99$	$56$
$9 - 10.99$	$21$
$11 - 12.99$	$16$
$13 - 14.99$	$4$
$15 - 16.99$	$4$

Find the median, the two quartiles and the mode of the following distribution.

Marks below Number of students

$10$ $15$

$20$ $35$

$30$ $60$

$40$ $84$

$50$ $96$

$60$ $127$

$70$ $198$

$80$ $250$

Important Questions on Statistics

Learn and Prepare for any exam you want !

Find the median, the two quartiles and the mode of the following distribution. Marks below Number of students 10 15 20 35 30 60 40 84 50 96 60 127 70 198 80 250

Important Questions on Statistics

Learn and Prepare for any exam you want !

Find the median, the two quartiles and the mode of the following distribution.

Marks below Number of students

$10$ $15$

$20$ $35$

$30$ $60$

$40$ $84$

$50$ $96$

$60$ $127$

$70$ $198$

$80$ $250$