Question
For the following frequency table obtain the relative frequencies and the cumulative frequencies of the ‘more than’ type.
| Life (in hours)
499.5 – 649.5 | Frequency
5 |
| 649.5 – 799.5 | 20 |
| 799.5 – 949.5 | 40 |
| 949.5 – 1099.5 | 81 |
| 1099.5 – 1249.5 | 57 |
| 1249.5 – 1399.5 | 34 |
| 1399.5 – 1549.5 | 13 |
| Total | 250 |
Also determine the number of bulbs having a life of 1249.5 hours or less.
Answer :
Word Count : 265
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Skip in 10
First, we calculate the relative frequencies by dividing each class frequency by the total frequency $N = 250$: $$ \text{Relative frequency} = \frac{\text{Frequency}}{\text{Total frequency}} $$ | Life (hours) | Frequency | Relative Frequency | | --------------- | --------- | ------------------ | | 499.5 – 649.5 | 5 | $5/250 = 0.02$ | | 649.5 – 799.5 | 20 | $20/250 = 0.08$ | |
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