There are 96 students in a class and are arranged according to weight from lowest to highest. What are the median terms of the weight distribution?

Open in App

Solution

For an even number of terms,
the middle/median terms are $(\frac{n}{2})$ $^{t h}$ and $(\frac{n}{2} + 1)$ $^{t h}$ .
Here n is 96.
On substituting value of n, we get $48^{t h}$ and $49^{t h}$ terms.

Suggest Corrections

Similar questions

The weights (in kg.) of 15 students of a class are:

38, 42, 35, 37, 45, 50, 32, 43, 43, 40, 36, 38, 43, 38, 47

(i) Find the mode and median of this data.

(ii) Is there more than one mode?

Q. Median and mean weight of the students of a class are 35.83 and 37.06 respectively. Calculate the mode.

Q. There are following the weights of

30

students of class

10

of a secondary school:

34, 34, 36, 37, 38, 33, 34, 35, 36, 37, 38, 33, 34, 35, 34, 33

35, 34, 38, 36, 35, 34, 35, 37, 38, 34, 35, 35, 37

Represent the above data in the form of a frequency distribution table.

Q. Question 3(i)
The weight (in kg) of 15 students of a class are:
38, 42, 35, 37, 45, 50, 32, 43, 43, 40, 36, 38, 43, 38,47
Find the mode and median of this data.

There are 29 students in a class whose average weight is 36 kgs. Among these 29 students, 16 are boys and 13 are girls. The average weight of 16 boys is 6.75 kgs more than the average weight of 13 girls. What is the average weight of girls (in kg)?

Join BYJU'S Learning Program

There are 96 students in a class and are arranged according to weight from lowest to highest. What are the median terms of the weight distribution?

For an even number of terms, the middle/median terms are (n2)th and (n2+1)th. Here n is 96. On substituting value of n, we get 48th and 49th terms.

For an even number of terms,
the middle/median terms are $(\frac{n}{2})$ $^{t h}$ and $(\frac{n}{2} + 1)$ $^{t h}$ .
Here n is 96.
On substituting value of n, we get $48^{t h}$ and $49^{t h}$ terms.