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प्रश्न
The weight of 60 boys are given in the following distribution table:
Weight (kg) | 37 | 38 | 39 | 40 | 41 |
No. of boys | 10 | 14 | 18 | 12 | 6 |
Find:
- Median
- Lower quartile
- Upper quartile
- Inter-quartile range
उत्तर
Weight (kg) x |
No. of boys f |
Cumulative frequency |
37 | 10 | 10 |
38 | 14 | 24 |
39 | 18 | 42 |
40 | 12 | 54 |
41 | 6 | 60 |
Number of terms = 60
i. Median = The mean of the 30th and 31st terms
∴ Median = `(39 + 39)/2`
= `78/2`
= 39
ii. Lower quartile (Q1) = `60^(th)/4` term
= 15th term
= 38
iii. Upper quartile (Q3) = `(3 xx 60^(th))/4` term
= 45th term
= 40
iv. Inter-quartile range = Q3 – Q1
= 40 – 38
= 2
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संबंधित प्रश्न
An incomplete distribution is given as follows:
Variable: | 0 - 10 | 10 - 20 | 20 - 30 | 30 - 40 | 40 - 50 | 50 - 60 | 60 - 70 |
Frequency: | 10 | 20 | ? | 40 | ? | 25 | 15 |
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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 |
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Class | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 |
Frequency | 4 | 9 | 15 | 14 | 8 |
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Frequency | 4 | 5 | 13 | 20 | 14 | 7 | 4 |
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0 - 200 |
200 - 400 |
400 - 600 |
600 - 800 |
800 - 1000 |
1000 - 1200 |
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0 – 100 | 2 |
100 – 200 | 5 |
200 – 300 | x |
300 – 400 | 12 |
400 – 500 | 17 |
500 – 600 | 20 |
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700 – 800 | 9 |
800 – 900 | 7 |
900 – 1000 | 4 |
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