Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
191 views
in Statistics by (30 points)
closed by

2 Answers

+1 vote
by (62.5k points)
selected by
 
Best answer
Classes fi c.f.
0 - 100 15 15
100 - 200 17 32
200 - 300 f 32 + f
300 - 400 12 44 + f
400 - 500 9 53 + f
500 - 600 5 58 + f
600 - 700 2 60 + f

From table,

N = 60 + f

⇒ \(\frac N2 =\frac{60 + f}2\)

Median = 240 which lines between class 200 - 300

Median class = 200 - 300

Median = \(l + \left[\frac{\frac N2- c.f.}{f}\right] \times h\)

\(240 = 200 + \left[\frac{\frac {60+f}2- 32}{f}\right] \times 100\)

\(40 = \left[\frac{60 + f - 64}{2f}\right] \times 100\)

\(8f = 10f - 40\)

\(2f = 40\)

\(f =20\)

0 votes
by (50.3k points)
By multiplying both sides by f/5 , we get the required result.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...