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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
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