A drill drills holes with standard deviation of depth 0.05 cms and mea

1.	A drill drills holes with standard deviation of depth 0.05 cms and mean depth of holes is 5.70 cms. Find 3 σ control limits for mean of samples of size 4. Given A = 1.500.
Answer» Given: X̄¹ =5.70 cms : σ₁ =0.05 cms, n=4 The 3σ – control limits for X̄ – chart (standards are given) are C.L = X̄¹ = 5.70 L.C.L = X̄¹ – Aσ¹ = 5.70 – 1,5 × 0.05 = 5.625 U.C.L = X̄¹+ Aσ¹ = 5-70 + 1.5 × 0.05 = 5.775.

A drill drills holes with standard deviation of depth 0.05 cms and mean depth of holes is 5.70 cms. Find 3 σ control limits for mean of samples of size 4. Given A = 1.500.

Answer»

Given: X̄¹ =5.70 cms :

σ₁ =0.05 cms, n=4

The 3σ – control limits for X̄ – chart (standards are given) are

C.L = X̄¹ = 5.70

L.C.L = X̄¹ – Aσ¹

= 5.70 – 1,5 × 0.05 = 5.625

U.C.L = X̄¹+ Aσ¹

= 5-70 + 1.5 × 0.05

= 5.775.