Advertisements
Advertisements
प्रश्न
Find the mean, median and mode of the given data:
| Class | 65 – 85 | 85 – 105 | 105 – 125 | 125 – 145 | 145 – 165 | 165 – 185 | 185 –205 |
| Frequency | 8 | 7 | 22 | 17 | 13 | 5 | 3 |
उत्तर
| Class `(f_i)` |
Frequency `(x_i)` |
Class mark | `d_i = x_i - A` `h = 20` |
`u_i = d_i/h` | `f_iu_i` |
| 65 – 85 | 8 | 75 | – 60 | – 3 | – 24 |
| 85 – 105 | 7 | 95 | – 40 | – 2 | – 14 |
| 105 –125 | 22 | 115 | – 20 | – 1 | – 22 |
| 125 – 145 | 17 | 135 = A | 0 | 0 | 0 |
| 145 –165 | 13 | 155 | 20 | 1 | 13 |
| 165 – 185 | 5 | 175 | 40 | 2 | 10 |
| 185 –205 | 3 | 195 | 60 | 3 | 9 |
| `sumf_i = 75` | `sumf_i u_i = - 28` |
We know, Mean `bar"X" = "A" + ((sumf_i u_i)/(sumf_i)) xx h`
= `135 + ((-28)/75) xx 20`
= `135 + ((-28)/15) xx 4`
= 135 – 7.47
= 127.53
Thus, the mean of the data is 127.53.
Now, the maximum frequency is 22 belonging to class interval 105 – 125.
∴ Modal class = 105 – 125
So, L = 105, f1 = 22, f0 = 7, f2 = 17, h = 20.
Mode = `"L" + ((f_1 - f_0)/(2f_1 - f_0 - f_2)) xx h"`
= `105 + ((22 - 7)/(2(22) - 7 - 17)) xx 20`
= `105 + (15/(44 - 24)) xx 20`
= `105 + (15/20) xx 20`
= 105 + 15
= 120
Thus, the mode is 120.
| Class | Frequency `(f_i)` | Cumulative frequency |
| 65 – 85 | 8 | 8 |
| 85 – 105 | 7 | 8 + 7 = 15 |
| 105 –125 | 22 | 15 + 22 = 37 |
| 125 – 145 | 17 | 37 + 17 = 54 |
| 145 –165 | 13 | 54 + 13 = 67 |
| 165 – 185 | 5 | 67 + 5 = 72 |
| 185 –205 | 3 | 72 + 3 = 75 |
Here, N = 75
∴ `(N/2)^("th") "term" = 75/3 = (37.5)^("th") "term"`
Since, 37.5th term lies in the class interval 125 – 145.
∴ Median class =125 – 145
`"So", L = 125, f = 17, N/2 = 37.5, c.f. = 37, h = 20.`
Median = `L + (("N"/2 - c.f.)/f) xx h`
= `125 + ((37.5 - 37)/17) xx 20`
= `125 + (0.5/17) xx 20`
= `125 + (5/17) xx 2`
= 125 + 0.588
= 125.588
As a result, the median is 125.588.
As a result, the data's mean, median, and mode are 127.53, 125.588, and 120, respectively.
APPEARS IN
संबंधित प्रश्न
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.
| Number of mangoe | 50 − 52 | 53 − 55 | 56 − 58 | 59 − 61 | 62 − 64 |
| Number of boxes | 15 | 110 | 135 | 115 | 25 |
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Find the mean of the following data:-
| x | 19 | 21 | 23 | 25 | 27 | 29 | 31 |
| f | 13 | 15 | 16 | 18 | 16 | 15 | 13 |
Candidates of four schools appear in a mathematics test. The data were as follows:-
| Schools | No. of Candidates | Average Score |
| I | 60 | 75 |
| II | 48 | 80 |
| III | NA | 55 |
| IV | 40 | 50 |
If the average score of the candidates of all the four schools is 66, find the number of candidates that appeared from school III.
The number of telephone calls received at an exchange per interval for 250 successive one minute intervals are given in the following frequency table:
| No. of calls(x) | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| No. of intervals (f) | 15 | 24 | 29 | 46 | 54 | 43 | 39 |
Compute the mean number of calls per interval.
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 |
Find the mean of the following data, using direct method:
| Class | 0-100 | 100-200 | 200-300 | 300-400 | 400-500 |
| Frequency | 6 | 9 | 15 | 12 | 8 |
The daily expenditure of 100 families are given below. Calculate `f_1` and `f_2` if the mean daily expenditure is ₹ 188.
|
Expenditure (in Rs) |
140-160 | 160-180 | 180-200 | 200-220 | 220-240 |
|
Number of families |
5 | 25 | `f_1` | `f_2` | 5 |
Find the mean of the following frequency distribution table using a suitable method:
| Class | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 - 70 |
| Frequency | 25 | 40 | 42 | 33 | 10 |
Find the mean of the following data using step-deviation method:
| Class | 500 – 520 | 520 – 540 | 540 – 560 | 560 – 580 | 580 – 600 | 600 – 620 |
| Frequency | 14 | 9 | 5 | 4 | 3 | 5 |
Find the mean age from the following frequency distribution:
| Age (in years) | 25 – 29 | 30 – 34 | 35 – 39 | 40 – 44 | 45 – 49 | 50 – 54 | 55 – 59 |
| Number of persons | 4 | 14 | 22 | 16 | 6 | 5 | 3 |
The yield of soyabean per acre in the farm of Mukund for 7 years was 10,7,5,3,9,6,9 quintal. Find the mean of yield per acre.
While computing mean of grouped data, we assume that the frequencies are ______.
The mean of the following distribution is 6. Find the value at P:
| x | 2 | 4 | 6 | 10 | P + 5 |
| f | 3 | 2 | 3 | 1 | 2 |
A frequency distribution of the life times of 400 T.V., picture tubes leased in tube company is given below. Find the average life of tube:
| Life time (in hrs) | Number of tubes |
| 300 - 399 | 14 |
| 400 - 499 | 46 |
| 500 - 599 | 58 |
| 600 - 699 | 76 |
| 700 - 799 | 68 |
| 800 - 899 | 62 |
| 900 - 999 | 48 |
| 1000 - 1099 | 22 |
| 1100 - 1199 | 6 |
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.
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 |
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 |
Find the mean of the following frequency distribution:
| Class: | 10 – 15 | 15 – 20 | 20 – 25 | 25 – 30 | 30 – 35 |
| Frequency: | 4 | 10 | 5 | 6 | 5 |
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 |
