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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 |
Find the median daily wage income of the workers.
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Class | Frequency (f) | Cumulative Frequency (cf) |
0 – 100 | 40 | 40 |
100 – 200 | 32 | 72 |
200 – 300 | 48 | 120 |
300 – 400 | 22 | 142 |
400 – 500 | 8 | 150 |
N = ΣЁЭСУ = 150 |
Now, N = 150
`⇒ N/2` = 75.
The cumulative frequency just greater than 75 is 120 and the corresponding class is 200 – 300.
Thus, the median class is 200 – 300.
∴ l = 200, h = 100, f = 48, cf = c.f. of preceding class = 72 and `N/2` = 75.
Now,
Median, M = `l + {h×((N/2−cf)/f)}`
`= 200 + {100 × ((75 − 72)/48)}`
= 200 + 6.25
= 206.25
Hence, the median daily wage income of the workers is Rs 206.25.
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100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:
Number of letters | Number of surnames |
1 - 4 | 6 |
4 − 7 | 30 |
7 - 10 | 40 |
10 - 13 | 6 |
13 - 16 | 4 |
16 − 19 | 4 |
Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.
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Height (in cm): | 160 - 162 | 163 - 165 | 166 - 168 | 169 - 171 | 172 - 174 |
No. of students: | 15 | 118 | 142 | 127 | 18 |
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From the following data, find:
Median
25, 10, 40, 88, 45, 60, 77, 36, 18, 95, 56, 65, 7, 0, 38 and 83
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Weight (kg) | 37 | 38 | 39 | 40 | 41 |
No. of boys | 10 | 14 | 18 | 12 | 6 |
Find:
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Rainfall (in cm ): | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
Number of days : | 22 | 10 | 8 | 15 | 5 | 6 |
Calculate the median rainfall using ogives of more than type and less than type.
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46 | 7 |
48 | 5 |
50 | 8 |
52 | 12 |
53 | 10 |
54 | 2 |
55 | 1 |
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Class | 0 – 5 | 6 – 11 | 12 – 17 | 18 – 23 | 24 – 29 |
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The upper limit of the median class is:
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Monthly Expenditure (in тВ╣) |
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.
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