Advertisements
Advertisements
प्रश्न
Prepare a continuous grouped frequency distribution from the following data:
| Mid-point | Frequency |
| 5 | 4 |
| 15 | 8 |
| 25 | 13 |
| 35 | 12 |
| 45 | 6 |
Also find the size of class intervals.
उत्तर
The common difference of mid-point = 15 – 5 = 10
Let lower limit is a. So, upper limit will be = a + 10
As we know that:
Mid-value = `("Lower limit" + "Upper limit")/2`
`5 = (a + a + 10)/2`
2a + 10 = 10
2a = 10 – 10
a = 0
Therefore, the first interval is 0 – 10.
Now, a frequency distribution table for the given data is given below:
| Mid-point | Class interval | Frequency |
| 5 | 0 – 10 | 4 |
| 15 | 10 – 20 | 8 |
| 25 | 20 – 30 | 13 |
| 35 | 30 – 40 | 12 |
| 45 | 40 – 50 | 6 |
APPEARS IN
संबंधित प्रश्न
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?
A company manufactures car batteries of a particular type. The lives (in years) of 40 such batteries were recorded as follows:-
| 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 |
Construct a grouped frequency distribution table for this data, using class intervals of size 0.5 starting from the intervals 2 − 2.5.
Following data gives the number of children in 40 families:
1,2,6,5,1,5, 1,3,2,6,2,3,4,2,0,0,4,4,3,2,2,0,0,1,2,2,4,3, 2,1,0,5,1,2,4,3,4,1,6,2,2.
Represent it in the form of a frequency distribution.
The following cumulative frequency distribution table shows the daily electricity consumption (in kW) of 40 factories in an industrial state:
| Consumption (in kW) | No. of Factories |
| Below 240 | 1 |
| Below 270 | 4 |
| Below 300 | 8 |
| Below 330 | 24 |
| Below 360 | 33 |
| Below 390 | 38 |
| Below 420 | 40 |
(i) Represent this as a frequency distribution table.
(ii) Prepare a cumulative frequency table.
Given below is a cumulative frequency distribution table showing the ages of people living in a locality:
| Ace in years | No. of persons |
| Above 108 | 0 |
| Above 96 | 1 |
| Above 84 | 3 |
| Above 72 | 5 |
| Above 60 | 20 |
| Above 48 | 158 |
| Above 36 | 427 |
| Above 24 | 809 |
| Above 12 | 1026 |
| Above 0 | 1124 |
Prepare a frequency distribution table
The difference between the highest and lowest values of the observations is called
In a frequency distribution, the mid-value of a class is 15 and the class intervals is 4. The lower limit of the class is
Tallys are usually marked in a bunch of
The blood groups of 30 students are recorded as follows:
A, B, O, A, AB, O, A, O, B, A, O, B, A, AB, B, A, AB, B, A, A, O, A, AB, B, A, O, B, A, B, A
Prepare a frequency distribution table for the data.
The scores (out of 100) obtained by 33 students in a mathematics test are as follows:
69, 48, 84, 58, 48, 73, 83, 48, 66, 58, 84, 66, 64, 71, 64, 66, 69, 66, 83, 66, 69, 71, 81, 71, 73, 69, 66, 66, 64, 58, 64, 69, 69
Represent this data in the form of a frequency distribution.
