Question

Median of 2/5,5/3,1/3,5/6,1/6

Answer 1

GIVEN :

Median of $\frac{2}{5}$, $\frac{5}{3}$, $\frac{1}{3}$, $\frac{5}{6}$, $\frac{1}{6}$

TO FIND :

Median of $\frac{2}{5}$, $\frac{5}{3}$, $\frac{1}{3}$, $\frac{5}{6}$, $\frac{1}{6}$

SOLUTION :

Given data set is  $\frac{2}{5}$, $\frac{5}{3}$, $\frac{1}{3}$, $\frac{5}{6}$, $\frac{1}{6}$

Now we have to find the median for the given data set :

First arrange the given data set in ascending order as below,

⇒ $\frac{1}{6}$, $\frac{1}{3}$, $\frac{2}{5}$, $\frac{5}{6}$,  $\frac{5}{3}$

Since the number of data set is 5

ie., n is odd ( n=5 )

The median is the middle value of the ascending order arranged data set

∴ Median for the given data set is $\frac{2}{5}$

∴ Median=$\frac{2}{5}$

Answer 2

Given :- Median of 2/5, 5/3, 1/3, 5/6, 1/6 . ?

concept used :-

• The median of a set of data is the middlemost number in the set. The median is also the number that is halfway into the set. To find the median, the data should first be arranged in order from least to greatest.

• If there is an odd number of terms, the median is the center term. = (n+1)/2 .

• If there is an even number of terms, add the two middle terms and divide by 2 = {(n/2 term)+(n/2+1) term}/2 .

Solution :-

First we have to arrange given fraction from least to greatest .

By taking LCM of denominator , we get,

→ (12, 50, 10, 25, 5)/30

So,

→ fraction from least to greatest = (5/30), (10/30) , (12/30) , (25/30) and (50/30).

Or,

→ fraction from least to greatest = (1/6), (1/3), (2/5) , (5/6) and (5/3).

Now,

→ Total terms are = n = 5 = odd number.

Therefore,

→ Median = center term = (n+1)/2 .

→ Median = (5 + 1)/2

→ Median = (6/2)

→ Median = 3rd term.

→ Median = (2/5) (Ans.)

Hence, Median of given terms is (2/5).

Learn more :-

what is the mean deviation of the data:8,9,12,15,16,20,24,30,32,34?

https://brainly.in/question/26616981?

The mean and standard deviation of a data which is comprised of a set of ten positive numbers are 8 and 2 respectively ....

https://brainly.in/question/26639131