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प्रश्न
Find the mean, median and mode of the following data:
Class | 0 – 50 | 50 – 100 | 100 – 150 | 150 – 200 | 200 – 250 | 250 – 300 | 300 - 350 |
Frequency | 2 | 3 | 5 | 6 | 5 | 3 | 1 |
उत्तर
To find the mean let us put the data in the table given below:
Class | Frequency `(f_i)` | Class mark `(x_i)` | `f_i x_i` |
0 – 50 | 2 | 25 | 50 |
50 – 100 | 3 | 75 | 225 |
100 – 150 | 5 | 125 | 625 |
150 – 200 | 6 | 175 | 1050 |
200 – 250 | 5 | 225 | 1125 |
250 – 300 | 3 | 275 | 825 |
300 – 350 | 1 | 325 | 325 |
Total | `Ʃ f_i `= 25 | `Ʃ f_i x_i` = 4225 |
Mean =`( sum _i f_i x_i )/(sum _ i f_ i)`
=`4225/25`
= 169
Thus, mean of the given data is 169.
Now, to find the median let us put the data in the table given below:
Class | Frequency `(f_i)` | Cumulative Frequency (cf) |
0 – 50 | 2 | 2 |
50 – 100 | 3 | 5 |
100 – 150 | 5 | 10 |
150 – 200 | 6 | 16 |
200 – 250 | 5 | 21 |
250 – 300 | 3 | 24 |
300 – 350 | 1 | 25 |
Total | `N = Σ f_i = 25` |
Now, N = 25 ⇒`N/2 = 12.5`
The cumulative frequency just greater than 12.5 is 16 and the corresponding class is 150 – 200.
Thus, the median class is 150 – 200.
∴ l = 150, h = 50, N = 25, f = 6 and cf = 10.
Now,
Median = l +`((N/2 - cf)/f) xx h`
= 150`((122.5-10)/6) xx 50`
= 150 + 20.83
= 170.83
Thus, the median is 170.83.
We know that,
Mode = 3(median) – 2(mean)
= 3 × 170.83 – 2 × 169
= 512.49 – 338
= 174.49
Hence, Mean = 169, Median = 170.83 and Mode = 174.49
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संबंधित प्रश्न
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Explain the meaning of the term Class-mark.
Explain the meaning of the term Class limits.
Explain the meaning of the term True class limits.
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No. of students | 7 | 31 | 33 | 17 | 11 | 1 |
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Height in cm |
120 – 130 | 130 – 140 | 140 – 150 | 150 – 160 | 160 – 170 |
No. of girls |
2 | 8 | 12 | 20 | 8 |
Find the mean, median and mode of the above data.
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100 – 150 | 6 |
150 – 200 | 7 |
200 – 250 | 12 |
250 – 300 | 3 |
300 – 350 | 2 |
Find the mean and median daily expenditure on food.
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Find the class marks of classes 10−25 and 35−55.
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