Advertisements
Advertisements
प्रश्न
The following table gives the daily income of 50 workers of a factory:
Daily income (in Rs) | 100 - 120 | 120 - 140 | 140 - 160 | 160 - 180 | 180 - 200 |
Number of workers: | 12 | 14 | 8 | 6 | 10 |
Find the mean, mode and median of the above data.
उत्तर
Class interval |
Mid value(x) | Frequency(f) | fx | Cumulative frequency |
100 - 120 | 110 | 12 | 1320 | 12 |
120 - 140 | 130 | 14 |
1820 |
26 |
140 - 160 | 150 | 8 | 1200 | 34 |
160 - 180 | 170 | 6 | 1020 | 40 |
180 - 200 | 190 | 10 | 1900 | 50 |
N = 50 | `sumfx=7260` |
Mean `=(sumfx)/N=7260/50=145.2`
Thus, the mean of the given data is 145.2.
It can be seen in the data table that the maximum frequency is 14. The class corresponding to this frequency is 120−140.
∴ Modal class = 120 − 140
Lower limit of modal class (l) = 120
Class size (h) = 140 − 120 = 20
Frequency of modal class (f) = 14
Frequency of class preceding the modal class (f1) = 12
Frequency of class succeeding the modal class (f2) = 8
Mode `=l+(f-f1)/(2f-f1-f2)xxh`
`=120+(14-12)/(2xx14-12-8)xx20`
`=120+2/(28-20)xx20`
`=120+2/8xx20`
`=120+40/8`
= 120 + 5
= 125
Thus, the mode of the given data is 125.
Here, number of observations N = 50
So, N/2 = 50/2 = 25
This observation lies in class interval 120−140.
Therefore, the median class is 120−140.
Lower limit of median class (l) = 120
Cumulative frequency of class preceding the median class(c.f.) or (F) = 12
Frequency of median class(f) = 14
Median `=l+(N/2-F)/fxxh`
`=120+(25-12)/14xx20`
`=120+13/14xx20`
`=120+13/7xx10`
`=120+130/7`
= 120 + 18.57
= 138.57
Thus, the median of the given data is 138.57.
APPEARS IN
संबंधित प्रश्न
The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure.
Expenditure (in Rs) | Number of families |
1000 − 1500 | 24 |
1500 − 2000 | 40 |
2000 − 2500 | 33 |
2500 − 3000 | 28 |
3000 − 3500 | 30 |
3500 − 4000 | 22 |
4000 − 4500 | 16 |
4500 − 5000 | 7 |
The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.
Number of students per teacher |
Number of states/U.T. |
15 − 20 | 3 |
20 − 25 | 8 |
25 − 30 | 9 |
30 − 35 | 10 |
35 − 40 | 3 |
40 − 45 | 0 |
45 − 50 | 0 |
50 − 55 | 2 |
Find the mode of the following data:
3, 5, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4
Compare the modal ages of two groups of students appearing for an entrance test:
Age (in years): | 16-18 | 18-20 | 20-22 | 22-24 | 24-26 |
Group A: | 50 | 78 | 46 | 28 | 23 |
Group B: | 54 | 89 | 40 | 25 | 17 |
The monthly salary of 10 employees in a factory are given below:
₹ 5000, ₹ 7000, ₹ 5000, ₹ 7000, ₹ 8000, ₹ 7000, ₹ 7000, ₹ 8000, ₹ 7000, ₹ 5000
Find the mean, median and mode
In the formula `x-a+(sumf_i d_i)/(sumf_i),` for finding the mean of grouped data d1's are deviations from the ______.
For ‘more than ogive’ the x-axis represents ______.
For the following distribution:
Class | 0 – 5 | 5 – 10 | 10 – 15 | 15 – 20 | 20 – 25 |
Frequency | 10 | 15 | 12 | 20 | 9 |
The sum of lower limits of the median class and modal class is:
Mrs. Garg recorded the marks obtained by her students in the following table. She calculated the modal marks of the students of the class as 45. While printing the data, a blank was left. Find the missing frequency in the table given below.
Marks Obtained |
0 − 20 | 20 − 40 | 40 − 60 | 60 − 80 | 80 − 100 |
Number of Students |
5 | 10 | − | 6 | 3 |
The frequency distribution of daily working expenditure of families in a locality is as follows:
Expenditure in ₹ (x): |
0 – 50 | 50 – 100 | 100 – 150 | 150 – 200 | 200 – 250 |
No. of families (f): |
24 | 33 | 37 | b | 25 |
If the mode of the distribution is ₹ 140 then the value of b is ______.