GATE ME 2023 Set 1 — Question 53

Let D1 and D2 be the two dimensions which are independent and follow a normal distribution.
Given:
Mean of D1, $\mu_{D1} = 20$ mm
Standard deviation of D1, $\sigma_{D1} = 0.3$ mm
Mean of D2, $\mu_{D2} = 35$ mm
Standard deviation of D2, $\sigma_{D2} = 0.4$ mmThe dimension C is given by the difference between D2 and D1:
$C = D2 - D1$Since D1 and D2 are independent normal random variables, their difference C also follows a normal distribution.The mean of C is the difference of the means of D2 and D1:
$\mu_C = \mu_{D2} - \mu_{D1} = 35 - 20 = 15$ mmThe variance of C is the sum of the variances of D1 and D2 (since they are independent):
$\sigma_C^2 = \sigma_{D2}^2 + \sigma_{D1}^2$
$\sigma_C^2 = (0.4)^2 + (0.3)^2 = 0.16 + 0.09 = 0.25$ mm$^2$The standard deviation of C is the square root of its variance:
$\sigma_C = \sqrt{0.25} = 0.5$ mmThe question asks for the upper limit of dimension C to be considered under six-sigma process control. In statistical process control, the control limits are typically set at $\pm 3\sigma$ from the mean. The term "six-sigma" refers to the overall quality level where the process variation ($6\sigma$) is contained within the specification limits, but the control limits for monitoring the process are set at $\mu \pm 3\sigma$.The upper control limit (UCL) for dimension C is given by:
$UCL = \mu_C + 3\sigma_C$Substituting the calculated values:
$UCL = 15 + 3(0.5) = 15 + 1.5 = 16.5$ mmRounding to one decimal place, the upper limit is 16.5 mm.