$A$ 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:
$\begin{array}{llllll}0.03 & 0.08 & 0.08 & 0.09 & 0.04 & 0.17 \\ 0.16 & 0.05 & 0.02 & 0.06 & 0.18 & 0.20 \\ 0.11 & 0.08 & 0.12 & 0.13 & 0.22 & 0.07 \\ 0.08 & 0.01 & 0.10 & 0.06 & 0.09 & 0.18 \\ 0.11 & 0.07 & 0.05 & 0.07 & 0.01 & 0.04\end{array}$
$(i)$ 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?
$\begin{array}{llllll}0.03 & 0.08 & 0.08 & 0.09 & 0.04 & 0.17 \\ 0.16 & 0.05 & 0.02 & 0.06 & 0.18 & 0.20 \\ 0.11 & 0.08 & 0.12 & 0.13 & 0.22 & 0.07 \\ 0.08 & 0.01 & 0.10 & 0.06 & 0.09 & 0.18 \\ 0.11 & 0.07 & 0.05 & 0.07 & 0.01 & 0.04\end{array}$
$(i)$ 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?
(8) To construct the grouped frequency distribution table,we count the number of observations falling into each class interval. Note that the upper limit of each class is excluded from that class.
$(ii)$ The number of days for which the concentration of $SO_2$ is more than $0.11$ $ppm$ corresponds to the sum of frequencies for the intervals $0.12 - 0.16, 0.16 - 0.20$,and $0.20 - 0.24$.
Required number of days $= 2 + 4 + 2 = 8$.
Therefore,for $8$ days,the concentration of $SO_2$ was more than $0.11$ $ppm$.
| Concentration of $SO_2$ (in $ppm$) | Number of days (frequency) |
| $0.00 - 0.04$ | $4$ |
| $0.04 - 0.08$ | $9$ |
| $0.08 - 0.12$ | $9$ |
| $0.12 - 0.16$ | $2$ |
| $0.16 - 0.20$ | $4$ |
| $0.20 - 0.24$ | $2$ |
| Total | $30$ |
$(ii)$ The number of days for which the concentration of $SO_2$ is more than $0.11$ $ppm$ corresponds to the sum of frequencies for the intervals $0.12 - 0.16, 0.16 - 0.20$,and $0.20 - 0.24$.
Required number of days $= 2 + 4 + 2 = 8$.
Therefore,for $8$ days,the concentration of $SO_2$ was more than $0.11$ $ppm$.
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Similar Questions
Consider the marks obtained by $10$ students in a mathematics test as given below:
$55, 36, 95, 73, 60, 42, 25, 78, 75, 62$
$55, 36, 95, 73, 60, 42, 25, 78, 75, 62$
Easy
View SolutionConsider the marks,out of $100$,obtained by $51$ students of a class in a test,given in the table below. Draw a frequency polygon corresponding to this frequency distribution table.
Marks Number of students $0-10$ $5$ $10-20$ $10$ $20-30$ $4$ $30-40$ $6$ $40-50$ $7$ $50-60$ $3$ $60-70$ $2$ $70-80$ $2$ $80-90$ $3$ $90-100$ $9$ Total $51$
| Marks | Number of students |
| $0-10$ | $5$ |
| $10-20$ | $10$ |
| $20-30$ | $4$ |
| $30-40$ | $6$ |
| $40-50$ | $7$ |
| $50-60$ | $3$ |
| $60-70$ | $2$ |
| $70-80$ | $2$ |
| $80-90$ | $3$ |
| $90-100$ | $9$ |
| Total | $51$ |
Medium
View SolutionThe following table gives the life times of $400$ neon lamps:
The time (in hours) Number of lamps $300-400$ $14$ $400-500$ $56$ $500-600$ $60$ $600-700$ $86$ $700-800$ $74$ $800-900$ $62$ $900-1000$ $48$
$(i)$ Represent the given information with the help of a histogram.
$(ii)$ How many lamps have a life time of more than $700$ hours?
| The time (in hours) | Number of lamps |
| $300-400$ | $14$ |
| $400-500$ | $56$ |
| $500-600$ | $60$ |
| $600-700$ | $86$ |
| $700-800$ | $74$ |
| $800-900$ | $62$ |
| $900-1000$ | $48$ |
$(i)$ Represent the given information with the help of a histogram.
$(ii)$ How many lamps have a life time of more than $700$ hours?
Medium
View Solution$A$ company manufactures car batteries of a particular type. The lives (in years) of $40$ such batteries were recorded as follows:
$\begin{array}{llllllll}2.6 & 3.0 & 3.7 & 3.2 & 2.2 & 4.1 & 3.5 & 4.5 \\ 3.5 & 2.3 & 3.2 & 3.4 & 3.8 & 3.2 & 4.6 & 3.7 \\ 2.5 & 4.4 & 3.4 & 3.3 & 2.9 & 3.0 & 4.3 & 2.8 \\ 3.5 & 3.2 & 3.9 & 3.2 & 3.2 & 3.1 & 3.7 & 3.4 \\ 4.6 & 3.8 & 3.2 & 2.6 & 3.5 & 4.2 & 2.9 & 3.6\end{array}$
Construct a grouped frequency distribution table for this data,using class intervals of size $0.5$ starting from the interval $2 - 2.5$.
$\begin{array}{llllllll}2.6 & 3.0 & 3.7 & 3.2 & 2.2 & 4.1 & 3.5 & 4.5 \\ 3.5 & 2.3 & 3.2 & 3.4 & 3.8 & 3.2 & 4.6 & 3.7 \\ 2.5 & 4.4 & 3.4 & 3.3 & 2.9 & 3.0 & 4.3 & 2.8 \\ 3.5 & 3.2 & 3.9 & 3.2 & 3.2 & 3.1 & 3.7 & 3.4 \\ 4.6 & 3.8 & 3.2 & 2.6 & 3.5 & 4.2 & 2.9 & 3.6\end{array}$
Construct a grouped frequency distribution table for this data,using class intervals of size $0.5$ starting from the interval $2 - 2.5$.
Medium
View SolutionThe runs scored by two teams $A$ and $B$ on the first $60$ balls in a cricket match are given below:
Number of balls Teams $A$ Teams $B$ $1-6$ $2$ $5$ $7-12$ $1$ $6$ $13-18$ $8$ $2$ $19-24$ $9$ $10$ $25-30$ $4$ $5$ $31-36$ $5$ $6$ $37-42$ $6$ $3$ $43-48$ $10$ $4$ $49-54$ $6$ $8$ $55-60$ $2$ $10$
Represent the data of both the teams on the same graph by frequency polygons.
[Hint : First make the class intervals continuous.]
| Number of balls | Teams $A$ | Teams $B$ |
| $1-6$ | $2$ | $5$ |
| $7-12$ | $1$ | $6$ |
| $13-18$ | $8$ | $2$ |
| $19-24$ | $9$ | $10$ |
| $25-30$ | $4$ | $5$ |
| $31-36$ | $5$ | $6$ |
| $37-42$ | $6$ | $3$ |
| $43-48$ | $10$ | $4$ |
| $49-54$ | $6$ | $8$ |
| $55-60$ | $2$ | $10$ |
Represent the data of both the teams on the same graph by frequency polygons.
[Hint : First make the class intervals continuous.]
Medium
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