The diameter of a wire as measured by a screw gauge was found to be 0.026 cm, 0.028 cm, 0.029 cm, 0.027cm, 0.024cm and 0.027 cm. Calculate
(i) mean value of diameter
(ii) mean absoulte error
(iii) relative error (iv) percentage error. Also express the result in terms of absolute error and percentage error.

Explanation

To solve the problem step by step, we will calculate the mean value of the diameter, mean absolute error, relative error, and percentage error based on the measurements provided.

### Given Data:
Measurements of the diameter of the wire:
- \( L_1 = 0.026 \, \text{cm} \)
- \( L_2 = 0.028 \, \text{cm} \)
- \( L_3 = 0.029 \, \text{cm} \)
- \( L_4 = 0.027 \, \text{cm} \)
...