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Question
Find the mean, median and mode of the following data:
| Marks obtained | 25 - 35 | 35 – 45 | 45 – 55 | 55 – 65 | 65 – 75 | 75 - 85 |
| No. of students | 7 | 31 | 33 | 17 | 11 | 1 |
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Solution
To find the mean let us put the data in the table given below:
| Marks obtained | Number of students `(f_i)` | Class mark `(x_i)` | `f_i x_i` |
| 25 – 35 | 7 | 30 | 210 |
| 35 – 45 | 31 | 40 | 1240 |
| 45 – 55 | 33 | 50 | 1650 |
| 55 – 65 | 17 | 60 | 1020 |
| 65 – 75 | 11 | 70 | 770 |
| 75 – 85 | 1 | 80 | 80 |
| Total | `Ʃ f_i` = 100 | `Ʃ f_i x_i` = 4970 |
Mean =`(sum _i f_i x_i )/(sum _i f_i )`
=`4970/100`
= 49.7
Thus, mean of the given data is 49.7.
Now, to find the median let us put the data in the table given below:
| Class | Frequency `(f_i)` | Cumulative Frequency (cf) |
| 25 – 35 | 7 | 7 |
| 35 – 45 | 31 | 38 |
| 45 – 55 | 33 | 71 |
| 55 – 65 | 17 | 88 |
| 65 – 75 | 11 | 99 |
| 75 – 85 | 1 | 100 |
| Total | `N = Σ f_i`= 100 |
Now, N = 100 ⇒`N/2 =50`
The cumulative frequency just greater than 50 is 71 and the corresponding class is 45 – 55.
Thus, the median class is 45 – 55.
∴ l = 45, h = 10, N = 100, f = 33 and cf = 38.
Now,
Median = l +`((N/2- cf)/f) xxh`
`=45+((50-38)/33) xx 10`
= 45 + 3.64
= 48.64
Thus, the median is 48.64.
We know that,
Mode = 3(median) – 2(mean)
= 3 × 48.64 – 2 × 49.70
= 145.92 – 99.4
= 46.52
Hence, Mean = 49.70, Median = 48.64 and Mode = 46.52
