In a study of number of accidents per day, the observations for 30 days were obtained as follows:

6, 3, 5, 6, 4, 3, 2, 5, 4, 2, 4, 2, 1, 2, 2,

0, 5, 4, 6, 1, 6, 0, 5, 3, 6, 1, 5, 5, 2, 6

Prepare a frequency distribution table.

Open in App

Solution

The frequency table for the number of accidents per day for a period of 30 days is given below:

Number of Accidents Tally Marks Frequency

0 || 2

1 ||| 3

2 |||| | 6

3 ||| 3

4 |||| 4

5 |||| | 6

6 |||| | 6

Suggest Corrections

Similar questions

Q. In a study of number of accidents per day, the observations for 30 days were obtained as follows:

6	3	5	6	4	3	2	5	4	2
4	2	1	2	2	0	5	4	6	1
6	0	5	3	6	1	5	5	2	6

Prepare a frequency distribution table.

Q. 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{matrix} 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{matrix}

(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?

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?

Question 5 (i)
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 isas follows:
$\begin{matrix} 0.03 & 0.08 & 0.08 & 0.09 & 0.04 & 0.17 0.16 & 0.05 & 0.02 & 0.06 & 0.18 & 0.20 0.11 & 0.08 & 0.12 & 0.13 & 0.22 & 0.07 0.08 & 0.01 & 0.10 & 0.06 & 0.09 & 0.18 0.11 & 0.07 & 0.05 & 0.07 & 0.01 & 0.04 \end{matrix}$
(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.

Q. The table is a frequency table for number of days and temperature obtained in a month. Find the median day of temperature.

Number of days	$3$	$4$	$5$	$6$	$7$
Temperature $^{o} C$	$20$	$30$	$10$	$15$	$5$

Join BYJU'S Learning Program

Number of Accidents	Tally Marks	Frequency
0	\|\|	2
1	\|\|\|	3
2	~~\|\|\|\|~~ \|	6
3	\|\|\|	3
4	\|\|\|\|	4
5	~~\|\|\|\|~~ \|	6
6	~~\|\|\|\|~~ \|	6

In a study of number of accidents per day, the observations for 30 days were obtained as follows: 6, 3, 5, 6, 4, 3, 2, 5, 4, 2, 4, 2, 1, 2, 2, 0, 5, 4, 6, 1, 6, 0, 5, 3, 6, 1, 5, 5, 2, 6 Prepare a frequency distribution table.

The frequency table for the number of accidents per day for a period of 30 days is given below: Number of Accidents Tally Marks Frequency 0 || 2 1 ||| 3 2 |||| | 6 3 ||| 3 4 |||| 4 5 |||| | 6 6 |||| | 6

In a study of number of accidents per day, the observations for 30 days were obtained as follows:

6, 3, 5, 6, 4, 3, 2, 5, 4, 2, 4, 2, 1, 2, 2,

0, 5, 4, 6, 1, 6, 0, 5, 3, 6, 1, 5, 5, 2, 6

Prepare a frequency distribution table.

The frequency table for the number of accidents per day for a period of 30 days is given below:

Number of Accidents Tally Marks Frequency

0 || 2

1 ||| 3

2 |||| | 6

3 ||| 3

4 |||| 4

5 |||| | 6

6 |||| | 6