Advertisements
Advertisements
Question
If the mean of the following observations is 54, find the value of p.
| Class | 0 – 20 | 20 – 40 | 40 – 60 | 60 – 80 | 80 – 100 |
| Frequency | 7 | p | 10 | 9 | 13 |
Solution
| Class | Frequency (f) | Mid value (x) | fx |
| 0 – 20 | 7 | 10 | 70 |
| 20 – 40 | p | 30 | 30p |
| 40 – 60 | 10 | 50 | 500 |
| 60 – 80 | 9 | 70 | 630 |
| 80 – 100 | 13 | 90 | 1170 |
| Total | 39 + p | 2370 + 30p |
`barx = (sumfx)/(sumf) = (2370 + 30p)/(39 + p)` ...(i)
Here mean = 54 ...(ii)
From (i) and (ii)
`(2370 + 30p)/(39 + p) = 54`
`=>` 2370 + 30p = 2106 + 54p
`=>` 54p – 30p = 2370 – 2106
`=> p = 264/24 = 11`
APPEARS IN
RELATED QUESTIONS
Find the mean of the following distribution by step deviation method:
| Class Interval | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
| Frequency | 10 | 6 | 8 | 12 | 5 | 9 |
The mean of the following distribution is 52 and the frequency of class interval 30-40 is ‘f’. Find ‘f’.
| Class Interval | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
| Frequency | 5 | 3 | f | 7 | 2 | 6 | 13 |
The mean of 1, 7, 5, 3, 4 and 4 is m. The numbers 3, 2, 4, 2, 3, 3 and p have mean m – 1 and median q. Find p and q.
Estimate the median, the lower quartile and the upper quartile of the following frequency distribution by drawing an ogive:
| Age( in yrs) | Under 10 | Under 20 | Under 30 | Under 40 | Under 50 | Under 60 |
| No. of males | 6 | 10 | 25 | 32 | 43 | 50 |
Find the median of:
241, 243, 347, 350, 327, 299, 261, 292, 271, 258 and 257
Find the median of:
63, 17, 50, 9, 25, 43, 21, 50, 14 and 34
The mean of a certain number of observations is 32. Find the resulting mean, if the observation is, divided by 0.5
Find the median of 9, 3, 20, 13, 0, 7 and 10
Find the median of 26, 33, 41, 18, 30, 22, 36, 45 and 24
The marks in a subject for 12 students are as follows:
31, 37, 35, 38, 42, 23, 17, 18, 35, 25, 35, 29
For the given data, find the median
