Question 28 (OR 1 st question)

The median of the following data is 525. Find the values of x and y if the total frequency is 100.

 Class Interval Frequency 0 – 100 2 100 – 200 5 200 – 300 x 300 – 400 12 400 – 500 17 500 – 600 20 600 – 700 y 700 – 800 9 800 – 900 7 900 – 100 4     1. Class 10
2. Solutions of Sample Papers for Class 10 Boards
3. CBSE Class 10 Sample Paper for 2019 Boards

Transcript

Question 28 (OR 1st question) The median of the following data is 525. Find the values of x and y if the total frequency is 100. Given that total frequency is 100 So, Sum of frequency = 100 2 + 5 + x + 12 + 17 + 17 + 20 + y + 8 + 7 + 4 = 100 76 + x + y = 100 x + y = 100 – 76 x + y = 24 Now, to find values of x and y, We first find median Since Median is 525 500 – 600 is the median class Median = l + (𝑛/2 − 𝑐𝑓)/𝑓 × h Where l = lower limit of median class = 500 h = class-interval = 100 − 0 = 100 n = ∑▒𝑓𝑖 = 100 cf = cumulative frequency of the class before median class = 36 + x f = frequency of the median class = 20 Now, Median = l + (𝑛/2 −𝑐𝑓)/𝑓 × h 525 = 500 + (100/2 − (36 + 𝑥))/20 × 100 525 – 500 = (50 − (36 + 𝑥))/20 × 100 25 = (14 − 𝑥)/20 × 100 25 = (14 – x) × 5 25 = 70 – 5x 5x = 70 – 25 5x = 45 x = 45/9 x = 5 Now, from (1) x + y = 24 Putting x = 5 5 + y = 24 y = 24 – 5 y = 19 Thus, x = 5, y = 19

CBSE Class 10 Sample Paper for 2019 Boards

Class 10
Solutions of Sample Papers for Class 10 Boards 