Question

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.

Answer 1

Answer:

the result in terms of absolute error and percentage error

Answer 2

Answer:

• Mean = D1 + D2+ D3 + D4 + D5 + D6/6

0.161/6 = 0.0268 = 0.027 cm

• Mean absolute error

ΔD1 = | D1 - D bar |

ΔD2 = |D2 - D bar |

ΔD3 = |D3 - D bar|

and write next three numbers as same it

Mean absolute error :

ΔD1 + ΔD2 + ΔD3 + ΔD4 + ΔD5 + ΔD6/6