Sarthaks Test
0 votes
398 views
in Mathematics by (29.7k points)

A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows: 

0.03 0.08 0.08 0.09 0.04 0.17 0.16 0.05 0.02 0.06 0.18 0.20 0.11 0.08 0.12 0.13 0.22 0.07 0.08 0.01 0.10 0.06 0.09 0.18 0.11 0.07 0.05 0.07 0.01 0.04 

(i) Make a grouped frequency distribution table for this data with class intervals as 0.00 - 0.04, 0.04 - 0.08, and so on. 

(ii) For how many days, was the concentration of sulphur dioxide more than 0.11 parts per million? 

1 Answer

0 votes
by (128k points)
selected by
 
Best answer

(i) Taking class intervals as 0.00, −0.04, 0.04, −0.08, and so on, a grouped frequency table can be constructed as follows. 

The number of days for which the concentration of SO2 is more than 0.11 is the number of days for which the concentration is in between 0.12 − 0.16, 0.16 − 0.20, 0.20 − 0.24. 

(ii) Required number of days = 2 + 4 + 2 = 8 Therefore, for 8 days, the concentration of SO2 is more than 0.11 ppm.  

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

...