Advertisements
Advertisements
प्रश्न
The monthly expenditure on milk in 200 families of a Housing Society is given below:
| Monthly Expenditure (in ₹) |
1000 – 1500 | 1500 – 2000 | 2000 – 2500 | 2500 – 3000 | 3000 – 3500 | 3500 – 4000 | 4000 – 4500 | 4500 – 5000 |
| Number of families | 24 | 40 | 33 | x | 30 | 22 | 16 | 7 |
Find the value of x and also, find the median and mean expenditure on milk.
The monthly expenditure on milk in 200 families of a Housing Society is given below:
| Monthly Expenditure (in Rs.) | 1000 - 1500 | 1500 - 2000 | 2000 - 2500 | 2500 - 3000 | 3000 - 3500 | 3500 - 4000 | 4000 - 4500 | 4500 - 5000 |
| Number of families | 24 | 40 | 33 | x | 30 | 22 | 16 | 7 |
Find the value of x and also find the mean expenditure.
उत्तर
Given, Total number of families = 200
∴ 24 + 40 + 33 + x + 30 + 22 + 16 + 7 = 200
`\implies` x = 200 – 172 = 28
| Monthly expenditure (in ₹) (C.I.) |
Number of families (fi) |
Mid value (xi) |
xifi | C.F. |
| 1000 – 1500 | 24 | 1250 | 30,000 | 24 |
| 1500 – 2000 | 40 | 1750 | 70,000 | 64 |
| 2000 – 2500 | 33 | 2250 | 74,250 | 97 |
| 2500 – 3000 | 28 | 2750 | 77,000 | 125 |
| 3000 – 3500 | 30 | 3250 | 97,500 | 155 |
| 3500 – 4000 | 22 | 3750 | 82,500 | 177 |
| 4000 – 4500 | 16 | 4250 | 68,000 | 193 |
| 4500 – 5000 | 7 | 4750 | 33,250 | 200 |
| `bb(N = sumf_i = 200)` | `bb(sumx_i f_i = 5,32,500)` |
Mean = `(sumx_i f_i)/(sumf_i)`
= `(5,32,500)/200`
= ₹ 2662.5
Now, `N/2` = `200/2` = 100th
100th observation lies in class 2500 – 3000, which is known as median class.
l = lower limit of median class = 2500
N = number of families = 200
f = frequency of median class = 28
C.F. = cumulative frequency of the class preceding the median class = 97
h = class size = 500
Median = `l + ((N/2 - C.F.))/f xx h`
Median = `2500 + ((200/2 - 97))/28 xx 500`
= `2500 + (100 - 97)/28 xx 500`
= `2500 + 3/7 xx 125`
= 2500 + 53.571
= 2553.571
APPEARS IN
संबंधित प्रश्न
Calculate the median from the following data:
| Marks below: | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 |
| No. of students: | 15 | 35 | 60 | 84 | 96 | 127 | 198 | 250 |
A survey regarding the height (in cm) of 51 girls of class X of a school was conducted and the following data was obtained:
| Height in cm | Number of Girls |
| Less than 140 | 4 |
| Less than 145 | 11 |
| Less than 150 | 29 |
| Less than 155 | 40 |
| Less than 160 | 46 |
| Less than 165 | 51 |
Find the median height.
A student got the following marks in 9 questions of a question paper.
3, 5, 7, 3, 8, 0, 1, 4 and 6.
Find the median of these marks.
The weights (in kg) of 10 students of a class are given below:
21, 28.5, 20.5, 24, 25.5, 22, 27.5, 28, 21 and 24.
Find the median of their weights.
The marks obtained by 19 students of a class are given below:
27, 36, 22, 31, 25, 26, 33, 24, 37, 32, 29, 28, 36, 35, 27, 26, 32, 35 and 28.
Find:
- Median
- Lower quartile
- Upper quartile
- Inter-quartile range
What is the value of the median of the data using the graph in the following figure of less than ogive and more than ogive?
If the median of the data: 6, 7, x − 2, x, 17, 20, written in ascending order, is 16. Then x=
If 35 is removed from the data, 30, 34, 35, 36, 37, 38, 39, 40 then the median increases by ______.
|
Yoga is an ancient practice which is a form of meditation and exercise. By practising yoga, we not even make our body healthy but also achieve inner peace and calmness. The International Yoga Day is celebrated on the 21st of June every year since 2015.
|
| Age Group | 15 – 25 | 25 – 35 | 35 – 45 | 45 –55 | 55 –65 | 65 –75 | 75 – 85 |
| Number of People |
8 | 10 | 15 | 25 | 40 | 24 | 18 |
Based on the above, find the following:
- Find the median age of people enrolled for the camp.
- If x more people of the age group 65 – 75 had enrolled for the camp, the mean age would have been 58. Find the value of x.
The median of the following frequency distribution is 35. Find the value of x.
| Class: | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 |
| Frequency: | 6 | 3 | x | 12 | 19 |
