Advertisements
Advertisements
प्रश्न
Find the mean marks per student, using assumed-mean method:
| Marks | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 |
| Number of Students |
12 | 18 | 27 | 20 | 17 | 6 |
उत्तर
| Class | Frequency `(f_i)` | Mid values `(x_i)` | Deviation `(d_i) d_i = (x_i – 25)` |
`(f_i× d_i)` |
| 0 –10 | 12 | 5 | -20 | -240 |
| 10 –20 | 18 | 15 | -10 | -180 |
| 20 – 30 | 27 | 25 = A | 0 | 0 |
| 30 – 40 | 20 | 35 | 10 | 200 |
| 40 – 50 | 17 | 45 | 20 | 340 |
| 50 – 60 | 6 | 55 | 30 | 180 |
| Total | `Ʃ f_i = 100` | `Ʃ (f_i × d_i) = 300` |
Let A = 25 be the assumed mean. Then we have:
Mean, x = A + `(sum (f_i xx d_i))/(sum f_i)`
= 25+`300/100`
=28
∴ 𝑥 = 28
APPEARS IN
संबंधित प्रश्न
Find the missing value of p for the following distribution whose mean is 12.58
| x | 5 | 8 | 10 | 12 | P | 20 | 25 |
| f | 2 | 5 | 8 | 22 | 7 | 4 | 2 |
If the mean of the following frequency distribution is 24, find the value of p.
| Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
| Frequency | 3 | 4 | p | 3 | 2 |
Find the mean of the following data, using assumed-mean method:
| Class | 0 – 20 | 20 – 40 | 40 – 60 | 60 – 80 | 80 – 100 | 100 - 120 |
| Frequency | 20 | 35 | 52 | 44 | 38 | 31 |
Find the arithmetic mean of the following frequency distribution using step-deviation method:
| Age (in years) | 18 – 24 | 24 – 30 | 30 – 36 | 36 – 42 | 42 – 48 | 48 – 54 |
| Number of workers | 6 | 8 | 12 | 8 | 4 | 2 |
Consider the following distribution of daily wages of 50 workers of a factory:
| Daily wages (in ₹) |
500-520 | 520-540 | 540-560 | 560-580 | 580-600 |
| Number of workers | 12 | 14 | 8 | 6 | 10 |
Find the mean daily wages of the workers of the factory by using an appropriate method.
The mean of n observation is `overlineX` .f the first item is increased by 1, second by 2 and so on, then the new mean is
The mean of 1, 3, 4, 5, 7, 4 is m. The numbers 3, 2, 2, 4, 3, 3, p have mean m − 1 and median q. Then, p + q =
In X standard, there are three sections A, B and C with 25, 40 and 35 students respectively. The average marks of section A is 70%, section B is 65% and of section C is 50%. Find the average marks of the entire X standard.
In calculating the mean of grouped data, grouped in classes of equal width, we may use the formula `barx = a + (sumf_i d_i)/(sumf_i)` where a is the assumed mean. a must be one of the mid-points of the classes. Is the last statement correct? Justify your answer.
Find the values of x and y if the mean and total frequency of the distribution are 25 and 50 respectively.
| Class Interval | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 |
| Frequency | 7 | x | 5 | y | 4 | 2 |
