Advertisements
Advertisements
प्रश्न
If the mean of the following data is 20.6. Find the value of p.
| x | 10 | 15 | P | 25 | 35 |
| f | 3 | 10 | 25 | 7 | 5 |
उत्तर
| x | f | fx |
| 10 | 3 | 30 |
| 5 | 10 | 150 |
| P | 25 | 25P |
| 25 | 7 | 175 |
| 35 | 5 | 175 |
| N = 50 | `sum`fx = 530 + 25P |
Given
⇒ Mean = 20.6
`rArr(sumfx)/N=20.6`
`rArr(530+25P)/50=20.6`
⇒ 25P = 20.6(50) - 530
`rArrP=500/25`
⇒ P = 20
APPEARS IN
संबंधित प्रश्न
The measurements (in mm) of the diameters of the head of the screws are given below:
| Diameter (in mm) | No. of Screws |
| 33 — 35 | 10 |
| 36 — 38 | 19 |
| 39 — 41 | 23 |
| 42 — 44 | 21 |
| 45 — 47 | 27 |
Calculate mean diameter of head of a screw by ‘Assumed Mean Method’.
The arithmetic mean of the following data is 14. Find the value of k
| x1 | 5 | 10 | 15 | 20 | 25 |
| f1 | 7 | k | 8 | 4 | 5 |
Find the mean of each of the following frequency distributions
| Class interval | 0 - 8 | 8 - 16 | 16 - 24 | 24 - 32 | 32 - 40 |
| Frequency | 6 | 7 | 10 | 8 | 9 |
The following distribution shows the daily pocket allowance given to the children of a multistorey building. The average pocket allowance is Rs 18.00. Find out the missing frequency.
| Class interval | 11 - 13 | 13 - 15 | 15 - 17 | 17 - 19 | 19 - 21 | 21 - 23 | 23 - 25 |
| Frequency | 7 | 6 | 9 | 13 | - | 5 | 4 |
If the mean of the following distribution is 27, find the value of p.
| Class | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 |
| Frequency | 8 | p | 12 | 13 | 10 |
If Σfi = 25 and Σfixi = 100, then find the mean (`bar"x"`)
Consider the following distribution of SO2 concentration in the air (in ppm = parts per million) in 30 localities. Find the mean SO2 concentration using assumed mean method. Also find the values of A, B and C.
| Class interval | Frequency (fi) | Class mark (xi) | di = xi - a |
| 0.00 - 0.04 | 4 | 0.02 | -0.08 |
| 0.04 - 0.08 | 9 | 0.06 | A |
| 0.08 - 0.12 | 9 | 0.10 | B |
| 0.12 - 0.16 | 2 | 0.14 | 0.04 |
| 0.16 - 0.20 | 4 | 0.18 | C |
| 0.20 - 0.24 | 2 | 0.22 | 0.12 |
| Total | `sumf_i=30` |
There is a grouped data distribution for which mean is to be found by step deviation method.
| Class interval | Number of Frequency (fi) | Class mark (xi) | di = xi - a | `u_i=d_i/h` |
| 0 - 100 | 40 | 50 | -200 | D |
| 100 - 200 | 39 | 150 | B | E |
| 200 - 300 | 34 | 250 | 0 | 0 |
| 300 - 400 | 30 | 350 | 100 | 1 |
| 400 - 500 | 45 | 450 | C | F |
| Total | `A=sumf_i=....` |
Find the value of A, B, C, D, E and F respectively.
The following table gives the number of pages written by Sarika for completing her own book for 30 days:
| Number of pages written per day |
16 – 18 | 19 – 21 | 22 – 24 | 25 – 27 | 28 – 30 |
| Number of days | 1 | 3 | 4 | 9 | 13 |
Find the mean number of pages written per day.
If the mean of 9, 8, 10, x, 14 is 11, find the value of x.
