Advertisements
Advertisements
प्रश्न
Mean of a certain number of observation is `overlineX`. If each observation is divided by m(m ≠ 0) and increased by n, then the mean of new observation is
पर्याय
`overlineX/m +n`
`overlineX/n+m`
`overlineX +n/m`
`overlineX +m/n`
उत्तर
Let
\[y_1 , y_2 , y_3 , . . . , y_k\]
be k observations.
Mean of the observations = `overlineX`
\[\Rightarrow \frac{y_1 + y_2 + y_3 + . . . + y_k}{k} = x\]
\[ \Rightarrow y_1 + y_2 + y_3 + . . . + y_k = kx . . . . . \left( 1 \right)\]
If each observation is divided by m and increased by n, then the new observations are
\[\frac{y_1}{m} + n, \frac{y_2}{m} + n, \frac{y_3}{m} + n, . . . , \frac{y_k}{m} + n\]
∴ Mean of new observations
\[= \frac{\left( \frac{y_1}{m} + n \right) + \left( \frac{y_2}{m} + n \right) + . . . + \left( \frac{y_k}{m} + n \right)}{k}\]
\[ = \frac{\left( \frac{y_1}{m} + \frac{y_2}{m} + . . . + \frac{y_k}{m} \right) + \left( n + n + . . . + n \right)}{k}\]
\[ = \frac{y_1 + y_2 + . . . + y_k}{mk} + \frac{nk}{k}\]
`(koverlineX)/mk + (nk)/k`
\[ = \fr
`overlineX/m +n`
APPEARS IN
संबंधित प्रश्न
The table below shows the daily expenditure on food of 25 households in a locality.
| Daily expenditure (in Rs) | 100 − 150 | 150 − 200 | 200 − 250 | 250 − 300 | 300 − 350 |
| Number of households | 4 | 5 | 12 | 2 | 2 |
Find the mean daily expenditure on food by a suitable method.
The following table gives the number of children of 150 families in a village. Find the average number of children per family.
| No. of children (x) | 0 | 1 | 2 | 3 | 4 | 5 |
| No. of families (f) | 10 | 21 | 55 | 42 | 15 | 7 |
Find the mean of each of the following frequency distributions
| Class interval | 50 - 70 | 70 - 90 | 90 - 110 | 110 - 130 | 130 - 150 | 150 - 170 |
| Frequency | 18 | 12 | 13 | 27 | 8 | 22 |
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 |
The measurements (in mm) of the diameters of the head of the screws are given below :
| Diameter (in mm) | no. of screws |
| 33 - 35 | 9 |
| 36 - 38 | 21 |
| 39 - 41 | 30 |
| 42 - 44 | 22 |
| 45 - 47 | 18 |
Calculate the mean diameter of the head of a screw by the ' Assumed Mean Method'.
A study of the yield of 150 tomato plants, resulted in the record:
| Tomatoes per Plant | 1 - 5 | 6 - 10 | 11 - 15 | 16 - 20 | 21 - 25 |
| Number of Plants | 20 | 50 | 46 | 22 | 12 |
Calculate the mean of the number of tomatoes per plant.
If the mean of n observation ax1, ax2, ax3,....,axn is a`bar"X"`, show that `(ax_1 - abar"X") + (ax_2 - abar"X") + ...(ax_"n" - abar"X")` = 0.
In the formula `barx = a + (f_i d_i)/f_i`, for finding the mean of grouped data di’s are deviations from a of ______.
Calculate the mean of the following data:
| Class | 4 – 7 | 8 – 11 | 12 – 15 | 16 – 19 |
| Frequency | 5 | 4 | 9 | 10 |
An aircraft has 120 passenger seats. The number of seats occupied during 100 flights is given in the following table:
| Number of seats | 100 – 104 | 104 – 108 | 108 – 112 | 112 – 116 | 116 – 120 |
| Frequency | 15 | 20 | 32 | 18 | 15 |
