Samples of size n = 5 are taken from a manufacturing process every hour. A quali

Question

Samples of size n = 5 are taken from a manufacturing process every hour. A quality characteristic is measured, and X and R are computed for each sample. After 25 samples have been analyzed, we have sigma_i = 1^25 x_i = 662.50 and sigma_i = 1^25 R-I = 9.90 The quality characteristic is normally distributed Find the control limits for the X and R charts. Assume that both charts exhibit control. If the specifications are 26.40 plusminus 0.50, estimate the fraction nonconforming. If the mean of the process were 26.40, what fraction nonconforming would result?

Explanation / Answer

Solution

Back-up Theory

Let xij represent the jth measurement in the ith sub-group, j = 1, 2,   , ni and j = 1, 2,   , k where ni = size of the ith sub-group and k = number of sub-groups.

Average of the ith sub-group, say xi(bar) = (sum over j of xij)/ni

Grand average, say x(double bar) = {sum over i of xi(bar)}/k

Sub-group Range, say Ri = Max measurement – min measurement of ith sub-group.

Average Range = (sum over i of Ri)/k

Control Limits for x(bar) = x(double bar) ± A2.R(bar) [values of A2 are available in Standard Tables of Control Charts Constants]

Control Limits for R = lower: D3.R(bar); upper: D4.R(bar) [values of D3 and D4 are available in Standard Tables of Control Charts Constants]

Standard Deviation, = R(bar)/d2 [values of d2 are available in Standard Tables of Control Charts Constants]

Under the assumption that measurements follow Normal Distribution, given the specification for measurement as

s ± t, proportion of measurement not conforming to specification is given by

P[Z < {(s - t) - x(double bar)}/] + P[Z > {(s + t) - x(double bar)}/], where Z is the Standard Normal Variate whose probabilities can be read off from the Standard Normal Tables]

Now, to work out the solutions,

We have ni = 5 for i, k = 25, A2(5) = 0.577, d2(5) = 2.326, D3(5) = 0, D4(5) = 2.114

Part (a)

x(double bar) = 662.50/25 = 26.5; R(bar) = 9.00/25 = 0.36

Control Limits for x(bar) = 26.5 ± 0.577x0.36 = 26.5 ± 0.2077 = (26.2923, 26.7077) ANSWER

Control Limits for R = lower: 0; upper: 2.114x0.36 = 0.7610.

So, the limits are: (0, 0.761) ANSWER

Part (b)

= 0.36/2.326 = 0.1548

Given specification is: (25.90, 26.90), proportion of measurement not conforming to specification

= P[Z < {(- 0.6/0.1548)/ + P[Z >{(0.4/0.1548)}]

= P(Z < - 3.876) + P(X > 2.584) = 0.0001 + 0.0049 = 0.005 = 0.5% ANSWER

Part (c)

Given setting is at 26.40, replace x(double bar) by 26.40 in the above (b) part calculations.

Proportion of measurement not conforming to specification = P[Z < {(- 0.5/0.1548)/ + P[Z >{(0.3/0.1548)}]

= P(Z < - 3.230) + P(X > 1.937) = 0.0006 + 0.0244 = 0.025 = 2.5% ANSWER

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.