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Question
The following table shows the marks scored by 140 students in an examination of a certain paper:
| Marks: | 0 - 10 | 10 - 20 | 20 - 30 | 30 - 40 | 40 - 50 |
| Number of students: | 20 | 24 | 40 | 36 | 20 |
Calculate the average marks by using all the three methods: direct method, assumed mean deviation and shortcut method.
Solution
Direct method
| Class interval | Mid value(x1) | Frequency(f1) | f1x1 |
| 0 - 10 | 5 | 20 | 100 |
| 10 - 20 | 15 | 24 | 360 |
| 20 - 30 | 25 | 40 | 1000 |
| 30 - 40 | 35 | 36 | 1260 |
| 40 - 50 | 45 | 20 | 900 |
| N = 140 | 3620 |
Mean `=(sumf_1"u"_1)/N`
`=3620/140=25.857`
Assume mean method : Let the assumed mean = 25
| Class interval | Mid value(x1) | u1 = x1 - 25 | f1 | f1u1 |
| 0 - 10 | 5 | -20 | 20 | -400 |
| 10 - 20 | 15 | -10 | 24 | -240 |
| 20 - 30 | 25 | 0 | 40 | 0 |
| 30 - 40 | 35 | 10 | 36 | 360 |
| 40 - 50 | 45 | 20 | 20 | 400 |
| N = 140 | 120 |
Mean `=A+(sumf_1"u"_1)/N`
`=25+120/140`
= 25 + 0.857
= 25.857
Step deviation method
Let the assumed mean A = 25
| Class interval | Mid value(x1) | d1 = x1 - 25 | `"u"_1=(x_1-25)/10` | f1 | f1u1 |
| 0 - 10 | 5 | -20 | -2 | 20 | -40 |
| 10 - 20 | 15 | -10 | -1 | 24 | -24 |
| 20 - 30 | 25 | 0 | 0 | 40 | 0 |
| 30 - 40 | 35 | 10 | 1 | 36 | 36 |
| 40 - 50 | 45 | 20 | 2 | 20 | 40 |
| N = 140 | 12 |
Mean `=A+hxx(sumf_1"u"_1)/N`
`=25+10xx12/140`
`=25+120/140`
= 25 + 0.857
= 25.857
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