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प्रश्न
A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.
| Number of plants | 0 - 2 | 2 - 4 | 4 - 6 | 6 - 8 | 8 - 10 | 10 - 12 | 12 - 14 |
| Number of houses | 1 | 2 | 1 | 5 | 6 | 2 | 3 |
Which method did you use for finding the mean, and why?
उत्तर
To find the class mark (xi) for each interval, the following relation is used.
Class mark (xi) = `("Upper class limit + Lower class limit")/2`
xi and fixi can be calculated as follows.
| Number of plant | Number of houses (fi) |
xi | fixi |
| 0 − 2 | 1 | 1 | 1 × 1 = 1 |
| 2 − 4 | 2 | 3 | 2 × 3 = 6 |
| 4 − 6 | 1 | 5 | 1 × 5 = 5 |
| 6 − 8 | 5 | 7 | 5 × 7 = 35 |
| 8 − 10 | 6 | 9 | 6 × 9 = 54 |
| 10 − 12 | 2 | 11 | 2 ×11 = 22 |
| 12 − 14 | 3 | 13 | 3 × 13 = 39 |
| Total | 20 | 162 |
From the table, it can be observed that
`sum f_i = 20`
`sumf_ix_i = 162`
Mean `barx = (sumf_ix_i)/(sumf_i)`
= `162/20 = 8.1`
Therefore, mean number of plants per house is 8.1.
Here, the direct method has been used as the values of class marks (xi) and fi are small.
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 following table gives the frequency distribution of trees planted by different Housing Societies in a particular locality:
| No. of Trees | No. of Housing Societies |
| 10-15 | 2 |
| 15-20 | 7 |
| 20-25 | 9 |
| 25-30 | 8 |
| 30-35 | 6 |
| 35-40 | 4 |
Find the mean number of trees planted by Housing Societies by using ‘Assumed Means 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 |
The following table gives the distribution of total household expenditure (in rupees) of manual workers in a city. Find the average expenditure (in rupees) per household.
| Expenditure (in rupees) (x1) |
Frequency(f1) |
| 100 - 150 | 24 |
| 150 - 200 | 40 |
| 200 - 250 | 33 |
| 250 - 300 | 28 |
| 300 - 350 | 30 |
| 350 - 400 | 22 |
| 400 - 450 | 16 |
| 450 - 500 | 7 |
The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Compute the missing frequency f1 and f2.
| Class | 0 - 20 | 20 - 40 | 40 - 60 | 60 - 80 | 80 - 100 | 100 - 120 |
| Frequency | 5 | f1 | 10 | f2 | 7 | 8 |
There are three dealers A, B and C in Maharashtra. Suppose, the trade of each of them in september 2018 was as shown in the following table.
The rate of GST on each transaction was 5%.
Read the table and answer the questions below it.
| Dealer | GST collected on the sale |
GST paid at the time of purchase |
ITC | Tax paid to the Govt. |
Taxbalance with the Govt. |
| A | Rs.5000 | Rs. 6000 | Rs. 5000 | Rs. 0 | Rs. 1000 |
| B | Rs 5000 | Rs. 4000 | Rs. 4000 | Rs. 1000 | Rs. 0 |
| C | Rs.5000 | Rs. 5000 | Rs. 5000 | Rs. 0 | Rs. 0 |
(i) How much amount did the dealer A get by sale ?
(ii) For how much amount did the dealer B buy the articles ?
(iii) How much is the balance of CGST and SGST left with the government that was paid by A ?
Define mean.
Which of the following cannot be determined graphically?
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 frequency distribution is 8.1 and Σfixi = 132 + 5k, Σfi = 20, then k =?
While computing mean of grouped data, we assume that the frequencies are ______.
In the formula
If for certain frequency distribution, Median = 156 and Mode = 180, Find the value of the Mean.
If the mean of the following distribution is 7.5, find the missing frequency ‘f’:
| Variable : | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| Frequency: | 20 | 17 | f | 10 | 8 | 6 | 7 | 6 |
Find the mean of the following distribution:
| x | 10 | 30 | 50 | 70 | 89 |
| f | 7 | 8 | 10 | 15 | 10 |
The mean weight of 150 students in a certain class is 60 kgs. The mean weight of boys in the class is 70 kg and that of girls is 55 kgs. Find the number of boys and the number of girls in the class.
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 mean of 6 distinct observations is 6.5 and their variance is 10.25. If 4 out of 6 observations are 2, 4, 5 and 7, then the remaining two observations are ______.
The following table gives the marks scored by a set of students in an examination. Calculate the mean of the distribution by using the short cut method.
| Marks | Number of Students (f) |
| 0 – 10 | 3 |
| 10 – 20 | 8 |
| 20 – 30 | 14 |
| 30 – 40 | 9 |
| 40 – 50 | 4 |
| 50 – 60 | 2 |
