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प्रश्न
Find the median wages for the following frequency distribution:
| Wages per day (in Rs) | 61 – 70 | 71 – 80 | 81 – 90 | 91 – 100 | 101 – 110 | 111 – 120 |
| No. of women workers | 5 | 15 | 20 | 30 | 20 | 8 |
उत्तर
| Class | Frequency (f) | Cumulative Frequency (cf) |
| 60.5 – 70.5 | 5 | 5 |
| 70.5 – 80.5 | 15 | 20 |
| 80.5 – 90.5 | 20 | 40 |
| 90.5 – 100.5 | 30 | 70 |
| 100.5 – 110.5 | 20 | 90 |
| 110.5 – 120.5 | 8 | 98 |
| N = Σ𝑓 = 98 |
Now, N = 98
⇒ `N/2` = 49.
The cumulative frequency just greater than 49 is 70 and the corresponding class is 90.5 – 100.5.
Thus, the median class is 90.5 – 100.5.
Now, l = 90.5, h = 10, f = 30, cf = c.f. of preceding class = 40 and `N/2` = 49.
∴ Median, `M = l + {h×((N/2−f)/f)}`
`= 90.5 + {10 × ((49 − 40)/30)}`
= 90.5 + 3
= 93.5
Hence, median wages = Rs. 93.50.
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For a certain frequency distribution, the values of Assumed mean (A) = 1300, `sumf_id_i` = 900 and `sumfi` = 100. Find the value of mean (`barx`) .
The following is the distribution of the size of certain farms from a taluka (tehasil):
| Size of Farms (in acres) |
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| 5 – 15 | 7 |
| 15 – 25 | 12 |
| 25 – 35 | 17 |
| 35 – 45 | 25 |
| 45 – 55 | 31 |
| 55 – 65 | 5 |
| 65 – 75 | 3 |
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An incomplete distribution is given below:
| Variable: | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
| Frequency: | 12 | 30 | - | 65 | - | 25 | 18 |
You are given that the median value is 46 and the total number of items is 230.
(i) Using the median formula fill up missing frequencies.
(ii) Calculate the AM of the completed distribution.
Compute the median for the following data:
| Marks | No. of students |
| Less than 10 | 0 |
| Less than 30 | 10 |
| Less than 50 | 25 |
| Less than 70 | 43 |
| Less than 90 | 65 |
| Less than 110 | 87 |
| Less than 130 | 96 |
| Less than 150 | 100 |
The following table shows the daily wages of workers in a factory:
| Daily wages in (Rs) | 0 – 100 | 100 – 200 | 200 – 300 | 300 – 400 | 400 – 500 |
| Number of workers | 40 | 32 | 48 | 22 | 8 |
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Calculate the median from the following frequency distribution table:
| Class | 5 – 10 | 10 – 15 | 15 – 20 | 20 – 25 | 25 – 30 | 30 – 35 | 35 – 40 | 40 – 45 |
| Frequency | 5 | 6 | 15 | 10 | 5 | 4 | 2 | 2 |
Calculate the median from the following data:
| Height(in cm) | 135 - 140 | 140 - 145 | 145 - 150 | 150 - 155 | 155 - 160 | 160 - 165 | 165 - 170 | 170 - 175 |
| Frequency | 6 | 10 | 18 | 22 | 20 | 15 | 6 | 3 |
Find the median from the following data:
| Class | 1 – 5 | 6 – 10 | 11 – 15 | 16 – 20 | 21 – 25 | 26 – 30 | 31 – 35 | 35 – 40 | 40 – 45 |
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The prices of different articles and demand for them is shown in the following frequency distribution table. Find the median of the prices.
|
Price (Rupees)
|
20 less than | 20 – 40 | 40 – 60 | 60 – 80 | 80 – 100 |
| No. of articles | 140 | 100 | 80 | 60 | 20 |
A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.
| Age (in years) | Number of policy holders |
| Below 20 | 2 |
| Below 25 | 6 |
| Below 30 | 24 |
| Below 35 | 45 |
| Below 40 | 78 |
| Below 45 | 89 |
| Below 50 | 92 |
| Below 55 | 98 |
| Below 60 | 100 |
