Two sections of Class $IX$ having $30$ students each appeared for a mathematics olympiad. The marks obtained by them are shown below:
$\begin{array}{lllllllllllllll}46 & 31 & 74 & 68 & 42 & 54 & 14 & 61 & 83 & 48 & 37 & 26 & 8 & 64 & 57 \\ 93 & 72 & 53 & 59 & 38 & 16 & 88 & 75 & 56 & 46 & 66 & 45 & 61 & 54 & 27 \\ 27 & 44 & 63 & 58 & 43 & 81 & 64 & 67 & 36 & 49 & 50 & 76 & 38 & 47 & 55 \\ 77 & 62 & 53 & 40 & 71 & 60 & 58 & 45 & 42 & 34 & 46 & 40 & 59 & 42 & 29\end{array}$
Construct a grouped frequency distribution table for the data above using the classes $0-9, 10-19$,etc.,and hence find the number of students who secured more than $49$ marks.
$\begin{array}{lllllllllllllll}46 & 31 & 74 & 68 & 42 & 54 & 14 & 61 & 83 & 48 & 37 & 26 & 8 & 64 & 57 \\ 93 & 72 & 53 & 59 & 38 & 16 & 88 & 75 & 56 & 46 & 66 & 45 & 61 & 54 & 27 \\ 27 & 44 & 63 & 58 & 43 & 81 & 64 & 67 & 36 & 49 & 50 & 76 & 38 & 47 & 55 \\ 77 & 62 & 53 & 40 & 71 & 60 & 58 & 45 & 42 & 34 & 46 & 40 & 59 & 42 & 29\end{array}$
Construct a grouped frequency distribution table for the data above using the classes $0-9, 10-19$,etc.,and hence find the number of students who secured more than $49$ marks.
(32) To construct the grouped frequency distribution,we count the number of students in each class interval:
To find the number of students who secured more than $49$ marks,we sum the frequencies of the classes $50-59, 60-69, 70-79, 80-89$,and $90-99$.
Number of students $= 12 + 10 + 6 + 3 + 1 = 32$.
| Class Interval | Frequency |
| $0-9$ | $1$ |
| $10-19$ | $2$ |
| $20-29$ | $4$ |
| $30-39$ | $6$ |
| $40-49$ | $15$ |
| $50-59$ | $12$ |
| $60-69$ | $10$ |
| $70-79$ | $6$ |
| $80-89$ | $3$ |
| $90-99$ | $1$ |
| Total | $60$ |
To find the number of students who secured more than $49$ marks,we sum the frequencies of the classes $50-59, 60-69, 70-79, 80-89$,and $90-99$.
Number of students $= 12 + 10 + 6 + 3 + 1 = 32$.
Explore More
Similar Questions
The median of the data $78, 56, 22, 34, 45, 54, 39, 54, 84$ is
Medium
View SolutionLet $m$ be the mid-point and $l$ be the upper class limit of a class in a continuous frequency distribution. The lower class limit of the class is:
Easy
View SolutionThe mean of $10$ observations is calculated as $22.7$. It was later noticed that one observation was taken as $15$ instead of $51$ by mistake. Then,the correct mean of the data is ........
Medium
View SolutionThe weights (in $kg$) of $11$ students are given below:
$37, 42, 40, 39, 41, 35, 37, 36, 40, 40, 42$
Find the median of the data.
$37, 42, 40, 39, 41, 35, 37, 36, 40, 40, 42$
Find the median of the data.
Medium
View SolutionThe width of each of five continuous classes in a frequency distribution is $5$ and the lower class-limit of the lowest class is $10$. The upper class-limit of the highest class is:
Easy
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo