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Question
A frequency distribution of the life times of 400 T.V., picture tubes leased in tube company is given below. Find the average life of tube:
| Life time (in hrs) | Number of tubes |
| 300 - 399 | 14 |
| 400 - 499 | 46 |
| 500 - 599 | 58 |
| 600 - 699 | 76 |
| 700 - 799 | 68 |
| 800 - 899 | 62 |
| 900 - 999 | 48 |
| 1000 - 1099 | 22 |
| 1100 - 1199 | 6 |
Solution
Here, the class-intervals are formed by exclusive method. If we make the series an inclusive one the mid-values remain same. So, there is no need to convert the series.
Let the assumed mean be A = 749·5 and h = 100.
Calculation of Mean
| Life time (in hrs) |
Frequency `(f_i)` |
Mid-values `(x_i)` |
`d_i = x_i - "A" = x_i - 749·5` |
`mu_i = (x_i - "A")/"h"` = `(x_i - 749·5)/(100)` |
`f_imu_i` |
| 300 - 399 | 14 | 349·5 | -400 | -4 | -56 |
| 400 - 499 | 46 | 449·9 | -300 | -3 | -138 |
| 500 - 599 | 58 | 549·5 | -200 | -2 | -116 |
| 600 - 699 | 76 | 649·5 | -100 | -1 | -76 |
| 700 - 799 | 68 | A = 749·5 | 0 | 0 | 0 |
| 800 - 899 | 62 | 849·5 | 100 | 1 | 62 |
| 900 - 999 | 48 | 949·5 | 200 | 2 | 96 |
| 1000 - 1099 | 22 | 1049·5 | 300 | 3 | 66 |
| 1100 - 1199 | 6 | 1149·5 | 400 | 4 | 24 |
| N = `sumf_i = 400` | `sumf_imu_i = -138` |
Here,
N = 400, A = 749·5, h = 100 and `sumf_imu_i = -138`
∴ `bar"X" = "A" + "h"/"N" sumf_imu_i`
⇒ `bar"X" = 749·5 + 100 xx ((-138)/400)`
= `749·5 - (138)/(4)`
= 749·5 - 34·5
= 715.
Hence, the average life time of a tube is 715 hours.
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