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Question
The following frequency distribution shows the weight of students in a class:
| Weight (in Kg) | 40 | 45 | 50 | 55 | 60 | 65 |
| Number of Students | 15 | 40 | 29 | 21 | 10 | 5 |
(a) Find the percentage of students whose weight is more than 50 kg.
(b) If the weight column provided is of mid values then find the percentage of students whose weight is more than 50 kg.
Solution
(a)
| Weight (in kg) |
Number of students (f) | Less than cumulative frequency (c.f.) |
| 40 | 15 | 15 |
| 45 | 40 | 55 |
| 50 | 29 | 84 |
| 55 | 21 | 105 |
| 60 | 10 | 115 |
| 65 | 5 | 120 |
| Total | 120 |
Let the percentage of students weighing less than 50 kg be x.
∴ Px = 50
From the table, out of 120 students, 84 students have their weight less than 50 kg.
∴ Number of students weighing more than 50 kg
= 120 – 84 = 36
∴ Percentage of students having there weight more than 50 kg = `36/120 × 100` = 30%
(b) The difference between any two consecutive mid values of weight is 5 kg. The class intervals must of width 5, with 40, 45, ... as their mid values.
∴ The class intervals will be 37.5 - 42.5, 42.5 - 47.5, etc. We construct the less than cumulative frequency table as given below:
| Weight (in kg) |
Number of students (f) | Less than cumulative frequency (c.f.) |
| 37.5 - 42.5 | 15 | 15 |
| 42.5 - 47.5 | 40 | 55 |
| 47.5 - 52.5 | 29 | 84 |
| 52.5 - 57.5 | 21 | 105 |
| 57.5 - 62.5 | 10 | 115 |
| 62.5 - 67.5 | 5 | 120 |
| Total | 120 |
Here, N = 120
Let Px = 50
The value 50 lies in the class 47.5 - 52.5.
∴ L= 47.5, f = 29, c.f. = 55, h = 5
Px = `"L"+"h"/"f"("xN"/100 - "c.f.")`
50 = `47.5 + (5)/(29)(("x"×120)/(100) - 55)`
∴ 50 − 47.5 = `5/29((6"x")/5-55)`
∴ 2.5 = `5/29((6"x")/5-55)`
∴ `(6"x")/5-55` = 14.5
∴ `(6"x")/5`= 55 + 14.5
∴ `"6x"/5` = 69.5
∴ x = `69.5xx5/6`
x = 58 (approximately)
∴ 58% of students are having weight below 50 kg.
∴ Percentage of students having weight above 50 kg is 100 – 58 = 42
∴ 42% of students are having weight above 50 kg.
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