Advertisements
Advertisements
Question
The number of students absent in a class were recorded every day for 120 days and the information is given in the following frequency table:
| No. of students absent (x) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| No. of days (f) | 1 | 4 | 10 | 50 | 34 | 15 | 4 | 2 |
Find the mean number of students absent per day.
Solution
Let the assumed mean (A) = 3
| No. of students absent x1 |
No. of days f1 |
u1 = x1 - A x1 - 3 |
f1u1 |
| 0 | 1 | -3 | -3 |
| 1 | 4 | -2 | -8 |
| 2 | 10 | -1 | -10 |
| 3 | 50 | 0 | 0 |
| 4 | 34 | 1 | 34 |
| 5 | 15 | 2 | 30 |
| 6 | 4 | 3 | 12 |
| 7 | 2 | 4 | 8 |
| N = 120 | `sumf_1"u"_1=63` |
Mean number of students absent per day `=A+(sumf_1"u"_1)/N`
`=3+63/120`
`=(360+63)/120`
`=423/120`
= 3.525
= 3.53(Approx)
APPEARS IN
RELATED QUESTIONS
Consider the following distribution of daily wages of 50 worker of a factory.
|
Daily wages (in Rs) |
100 − 120 |
120 − 140 |
140 −1 60 |
160 − 180 |
180 − 200 |
|
Number of workers |
12 |
14 |
8 |
6 |
10 |
Find the mean daily wages of the workers of the factory by using an appropriate method.
Candidates of four schools appear in a mathematics test. The data were as follows:-
| Schools | No. of Candidates | Average Score |
| I | 60 | 75 |
| II | 48 | 80 |
| III | NA | 55 |
| IV | 40 | 50 |
If the average score of the candidates of all the four schools is 66, find the number of candidates that appeared from school III.
Find the mean from the following frequency distribution of marks at a test in statistics:
| Marks(x) | 5 | 10 | 15 | 20 | 25 | 30 | 35 | 40 | 45 | 50 |
| No. of students (f) | 15 | 50 | 80 | 76 | 72 | 45 | 39 | 9 | 8 | 6 |
Find the mean of each of the following frequency distributions
| Class interval | 0 - 8 | 8 - 16 | 16 - 24 | 24 - 32 | 32 - 40 |
| Frequency | 5 | 6 | 4 | 3 | 2 |
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.
If the mean of the following distribution is 2.6, then the value of y is:
| Variable (x) | 1 | 2 | 3 | 4 | 5 |
| Frequency | 4 | 5 | y | 1 | 2 |
If the mean of first n natural numbers is \[\frac{5n}{9}\], then n =
The median from the table is?
| Value | Frequency |
| 7 | 2 |
| 8 | 1 |
| 9 | 4 |
| 10 | 5 |
| 11 | 6 |
| 12 | 1 |
| 13 | 3 |
Given that the mean of the following frequency distribution is 30, find the missing frequency ‘f’.
| Class Interval | 0 - 10 | 10 - 20 | 20 - 30 | 30 - 40 | 40 - 50 | 50 - 60 |
| Frequency | 4 | 6 | 10 | f | 6 | 4 |
Find the mean of the following frequency distribution:
| Class | 1 – 5 | 5 – 9 | 9 – 13 | 13 – 17 |
| Frequency | 4 | 8 | 7 | 6 |
