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प्रश्न
The median of the following data is 525. If the sum of all the frequencies is 100, find the values of p and q.
| Class | Frequency |
| 0 – 100 | 2 |
| 100 – 200 | p |
| 200 – 300 | 9 |
| 300 – 400 | 12 |
| 400 – 500 | 17 |
| 500 – 600 | 20 |
| 600 – 700 | 15 |
| 700 – 800 | 9 |
| 800 – 900 | q |
| 900 – 1000 | 4 |
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उत्तर
1. Cumulative frequency table
First, let’s calculate the cumulative frequency (cf) for each class:
| Class | Frequency (f) | Cumulative Frequency (cf) |
| 0 – 100 | 2 | 2 |
| 100 – 200 | p | 2 + p |
| 200 – 300 | 9 | 11 + p |
| 300 – 400 | 12 | 23 + p |
| 400 – 500 | 17 | 40 + p |
| 500 – 600 | 20 | 60 + p |
| 600 – 700 | 15 | 75 + p |
| 700 – 800 | 9 | 84 + p |
| 800 – 900 | q | 84 + p + q |
| 900 – 1000 | 4 | 88 + p + q |
2. Using total frequency
88 + p + q = 100
p + q = 100 – 88
p + q = 12 ...(Equation 1)
3. Finding p using the median formula
The median is given as 525, which falls in the class interval 500 – 600.
Therefore, the median class is 500 – 600.
Median Formula:
Median = `l + ((N/2 - cf)/f) xx h`
Where:
l (lower limit of median class) = 500
N (total frequency) = 100
cf (cumulative frequency of the preceding class) = 40 + p
f (frequency of the median class) = 20
h (class size) = 100
Substitute the values:
`525 = 500 + ((100/2 - (40 + p))/20) xx 100`
`525 - 500 = ((50 - 40 - p)/20) xx 100`
25 = (10 – p) × 5
Divide both sides by 5:
5 = 10 – p
p = 10 – 5
p = 5
4. Finding q
Now, substitute the value of p into equation 1:
5 + q = 12
q = 12 – 5
q = 7
The value of p is 5 and the value of q is 7.
