chapter 1

CHAPTER 16Section 16.11. All ten values of the quality statistic are between the two control limits, so no out-of-controlsignal is generated.2. All ten values are between the two control limits. However, it is readily verified that all butone plotted point fall below the center line (at height .04975). Thus even though no singlepoint generates an out-of-control signal, taken together, the observed values do suggest thatthere may be a decrease in the average value of the quality statistic. Such a “small” change ismore easily detected by a CUSUM procedure (see section 16.5) than by an ordinary chart.3. P(10 successive points inside the limits) = P(1 st inside) x P(2 nd inside) x…x P(10 th inside) =(.998) 10 = .9802. P(25 successive points inside the limits) = (.998) 25 = .9512. (.998) 52 =.9011, but (.998) 53 = .8993, so for 53 successive points the probability that at least one willfall outside the control limits when the process is in control is 1 - .8993 = .1007 > .10.Section 16.24. For Z, a standard normal random variable, ( − c ≤ Z ≤ c) = . 995Φ.005( c ) = P( Z ≤ c) = . 995 + = . 9975appropriate control limits are therefore2P implies thatµ 2. 81σ. Table A.3 then gives c = 2.81. The± .5.a. P(point falls outside the limits when µ = µ +. 05σ)⎛ 3σ3σ⎞= 1−P ⎜µ 0− < X < µ0+ whenµ= µ0+ . 5σ⎟⎝ nn⎠( − 3 − .5 n < Z < 3 . n )= 1−P − 5( − 4.12 < < 1.882) = 1−.9699. 0301= 1 − Z=P .⎛ 3σ3σ⎞b. 1−P ⎜ µ0− < X < µ0+ whenµ= µ0− σ ⎟⎝ nn⎠= − P − 3 + n < Z < 3 + n = 1 − P −.76< Z < 5.24 = .( )1 ( ) 2236c. 1 − P ( − 3 − 2 n < Z < 3 − 2 n ) = 1 − P( − 7.47 < Z < −1.47) = . 6808469