Question

Calculate the median of the data 10,20,30,40,50,60,70,80, and the no of student 15,35,60,84,96,127,198,250

Answer 1

Step-by-step explanation:

CUMULATIVE FREQUENCY TABLE is in the attachment.  

From the table,  Here, n = 250

n/2 = 125

Since, the Cumulative frequency just greater than 125 is 127 and the corresponding class is 50 - 60 .  Therefore 50 - 60  is the median class.

Here, l = 50 , f = 31 , c.f = 96 ,  h = 10

MEDIAN = l + [(n/2 - cf )/f ] ×h

= 50 + [125 - 96)/31] × 10

= 50 + (29/31)×10

= 50 + ( 290/31)

= 50 + 9.35

= 59.35

Hence, the Median is 59.35.

MEDIAN: Median is defined as the middle most or the Central observation , when the observations are arranged either in ascending or descending order of their magnitudes.

★★Median is that value of the given observation which divides it into exactly two parts.i.e 50% of the observations lie below the median and the remaining are above the median.

MEDIAN for the GROUPED data :

For this we find the Cumulative frequency(cf) of all the classes and n/2 , where n =  number of observations.

Now find the class whose Cumulative frequency is greater than and nearest to n/2 and this class is called median class,then use  the following formula calculating the median.

MEDIAN = l + [(n/2 - cf )/f ] ×h

Where,

l = lower limit of the median class

n = number of observations  

cf = cumulative frequency  of class interval preceding the  median class

f = frequency  of median class

h = class  size

★★ CUMULATIVE FREQUENCY:

Cumulative frequency is defined as a consecutive sum of frequencies.

**The Cumulative frequency of first observation is the same as its frequency since there is no frequency before it.

HOPE THIS ANSWER WILL HELP YOU.  

Answer 2

Step-by-step explanation:

Given: The data and the number of students.

To find: The median of the data.

For calculation of median,

We know that $n=250$

∴ $\frac{n}{2}=\frac{250}{2}$

⇒ $n=125$

The corresponding class is $50-60.$ Thus,

Here, $l = 50 , f = 31 , c.f = 96 , h = 10$.

We know that $median= l + (\frac{\frac{n}{2}-cf}{f} )$ × $h$

⇒ $50+\frac{125-96}{31}$ × $10$

⇒ $50+ \frac{29}{31}$ × $10$

⇒ $50 + \frac{290}{31}$

⇒ $50 + 9.35$

⇒ $59.35$

The median of the data is $59.35$.