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प्रश्न
In calculating the mean of grouped data, grouped in classes of equal width, we may use the formula `barx = a + (sumf_i d_i)/(sumf_i)` where a is the assumed mean. a must be one of the mid-points of the classes. Is the last statement correct? Justify your answer.
उत्तर
No, it is not necessary that assumed mean consider as the mid-point of the class interval. It is considered as any value which is easy to simplify it.
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संबंधित प्रश्न
The following table gives the number of boys of a particular age in a class of 40 students. Calculate the mean age of the students
| Age (in years) | 15 | 16 | 17 | 18 | 19 | 20 |
| No. of students | 3 | 8 | 10 | 10 | 5 | 4 |
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The following distribution gives the number of accidents met by 160 workers in a factory during a month.
| No. of accidents(x) | 0 | 1 | 2 | 3 | 4 |
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Find the mean of each of the following frequency distributions
| Class interval | 10 - 30 | 30 - 50 | 50 - 70 | 70 - 90 | 90 - 110 | 110 - 130 |
| Frequency | 5 | 8 | 12 | 20 | 3 | 2 |
Find the mean marks per student, using assumed-mean method:
| Marks | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 |
| Number of Students |
12 | 18 | 27 | 20 | 17 | 6 |
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Write the empirical relation between mean, mode and median.
A frequency distribution of the life times of 400 T.V., picture tubes leased in tube company is given below. Find the average life of tube:
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| 300 - 399 | 14 |
| 400 - 499 | 46 |
| 500 - 599 | 58 |
| 600 - 699 | 76 |
| 700 - 799 | 68 |
| 800 - 899 | 62 |
| 900 - 999 | 48 |
| 1000 - 1099 | 22 |
| 1100 - 1199 | 6 |
The mean of the following distribution is
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| fi | 2 | 5 | 7 | 6 |
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| 0 – 10 | 3 |
| 10 – 20 | 8 |
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| 40 – 50 | 4 |
| 50 – 60 | 2 |
