Explanation Given info: In the double sampling plan the value of the parameters are n 1 = 50 , n 2 = 100 , c 1 = 1 and r 1 = 4 . It is also given that r 1 = c 2 + 1 , that is the lot is rejected if at either stage the number of defectives us at least r 1 . Calculation: Double sampling plan: In a double sampling plan the number of defective item in initial sample is x 1 with the sample of size is n 1 . The three possible courses of action in this type of sampling plan are, immediately accept the lot A , immediately reject the lot A or take a second sample of size if n 2 items and reject or accept the lot A depending on the total number x 1 + x 2 of defective items in the two samples. A specific plan with c 1 , r 1 and c 2 is designed as, Reject the lot is x 1 ≥ r 1 . Accept the lot is x ≤ c 1 . If c 1 < x 1 < r 1 , take a second sample and the accept the lot if x 1 + x 2 ≤ c 2 or reject it otherwise. It is given that c 1 = 1 . Thus, the lot is accepted after the first sample if the number of defectives is at most 1. It is given that r 1 = 4 . Thus, the lot is rejected if the number of defective items is at least 4. It is given that r 1 = c 2 + 1 . Thus, 4 = c 2 + 1 c 2 = 4 − 1 = 3 Thus, the lot is rejected also if the total number of defectives is 6 or more after the sixth sample. Hence, the probability of accepting the lot is defined as, P ( A ) = P ( X 1 = 0 or 1 ) + P ( X 1 = 2 , X 2 = 0 or 1 ) + P ( X 1 = 3 , X 2 = 0 ) The first sample size is 50 and the second sample size is 100. Thus, n 1 = 50 and n 2 = 100 . The lot size is large enough for the binomial approximation to apply. Thus, P ( A ) = P ( X 1 = 0 or 1 ) + P ( X 1 = 2 , X 2 = 0 or 1 ) + P ( X 1 = 3 , X 2 = 0 ) = P ( X 1 = 0 or 1 ) + P ( X 1 = 2 ) P ( X 2 = 0 or 1 ) + P ( X 1 = 3 ) P ( X 2 = 0 ) = ∑ x = 0 1 ( 50 x 1 ) p x 1 ( 1 − p ) 50 − x 1 + ( 50 2 ) p 2 ( 1 − p ) 50 − 2 ∑ x 2 = 0 1 ( 100 x 2 ) p x 2 ( 1 − p ) 50 − x 2 + [ ( 50 3 ) p 3 ( 1 − p ) 50 − 2 ] [ ( 100 x 2 ) p x 2 ( 1 − p ) 100 − x 2 ] For p = 0.02 : Probability value: Software procedure: Step by step procedure to obtain cumulative probability using MINITAB software is given as, Choose Calc > Probability Distributions > Binomial Distribution . Choose Cumulative Probability . Enter Number of trials as 50 and Event probability as 0.02 . In Input constant , enter 1. Click OK . The output using MINITAB software is given as, Thus, P ( X 1 = 0 or 1 ) = 0.735771 Probability value: Software procedure: Step by step procedure to obtain probability using MINITAB software is given as, Choose Calc > Probability Distributions > Binomial Distribution . Choose Probability . Enter Number of trials as 50 and Event probability as 0.02 . In Input constant , enter 2. Click OK . The output using MINITAB software is given as, Thus, P ( X 1 = 2 ) = 0.185801 Probability value: Software procedure: Step by step procedure to obtain cumulative probability using MINITAB software is given as, Choose Calc > Probability Distributions > Binomial Distribution . Choose Cumulative Probability . Enter Number of trials as 100 and Event probability as 0.02 . In Input constant , enter 1. Click OK . The output using MINITAB software is given as, Thus, P ( X 2 = 0 or 1 ) = 0.403272 Software procedure: Step by step procedure to obtain probability using MINITAB software is given as, Choose Calc > Probability Distributions > Binomial Distribution . Choose Probability . Enter Number of trials as 50 and Event probability as 0.02 . In Input constant , enter 3. Click OK . The output using MINITAB software is given as, Thus, P ( X 1 = 3 ) = 0.0606697 Probability value: Software procedure: Step by step procedure to obtain probability using MINITAB software is given as, Choose Calc > Probability Distributions > Binomial Distribution . Choose Cumulative Probability . Enter Number of trials as 100 and Event probability as 0.02 . In Input constant , enter 0. Click OK . The output using MINITAB software is given as, Thus, P ( X 2 = 0 ) = 0.132620 Thus, the probability of accepting the lot when p = 0.01 is, P ( A ) = P ( X 1 = 0 or 1 ) + P ( X 1 = 2 , X 2 = 0 or 1 ) + P ( X 1 = 3 , X 2 = 0 ) = P ( X 1 = 0 or 1 ) + P ( X 1 = 2 ) P ( X 2 = 0 or 1 ) + P ( X 1 = 3 ) P ( X 2 = 0 ) = 0

WEBASSIGN ACCESS FOR PROBABILITY & STATS

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ISBN: 9780357893111

Author: DEVORE

Publisher: CENGAGE L

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Chapter 16.6, Problem 38E

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Find the probability of accepting the lot when p=0.02,0.05 and 0.10.

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Chapter 16.6, Problem 37E

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Chapter 16 Solutions

WEBASSIGN ACCESS FOR PROBABILITY & STATS

Ch. 16.1 - A control chart for thickness of rolled-steel...Ch. 16.1 - Refer to Exercise 1 and suppose the ten most...Ch. 16.1 - Suppose a control chart is constructed so that the...Ch. 16.1 - A cork intended for use in a wine bottle is...Ch. 16.1 - If a process variable is normally distributed, in...Ch. 16.2 - In the case of known and , what control limits...Ch. 16.2 - Consider a 3-sigma control chart with a center...Ch. 16.2 - The table below gives data on moisture content for...Ch. 16.2 - Refer to the data given in Exercise 8, and...Ch. 16.2 - When installing a bath faucet, it is important to...

Ch. 16.2 - The accompanying table gives sample means and...Ch. 16.2 - The accompanying table gives sample means and...Ch. 16.2 - Prob. 13E Ch. 16.2 - Prob. 14E Ch. 16.2 - Prob. 15E Ch. 16.3 - A manufacturer of dustless chalk instituted a...Ch. 16.3 - Subgroups of power supply units are selected once...Ch. 16.3 - Prob. 18E Ch. 16.3 - The accompanying table gives sample means and...Ch. 16.3 - Prob. 20E Ch. 16.4 - On each of the previous 25 days, 100 electronic...Ch. 16.4 - A sample of 200 ROM computer chips was selected on...Ch. 16.4 - Prob. 23E Ch. 16.4 - Prob. 24E Ch. 16.4 - Prob. 25E Ch. 16.4 - Prob. 26E Ch. 16.4 - Prob. 27E Ch. 16.4 - Prob. 28E Ch. 16.5 - Prob. 29E Ch. 16.5 - Prob. 30E Ch. 16.5 - The standard deviation of a certain dimension on...Ch. 16.5 - Prob. 32E Ch. 16.6 - Consider the single-sample plan with c = 2 and n =...Ch. 16.6 - A sample of 50 items is to be selected from a...Ch. 16.6 - Prob. 35E Ch. 16.6 - Develop a single-sample plan for which AQL = .02...Ch. 16.6 - Consider the double-sampling plan for which both...Ch. 16.6 - Prob. 38E Ch. 16.6 - Refer to Example 16.11, in which a single-sample...Ch. 16.6 - Consider the single-sample plan that utilizes n =...Ch. 16 - Observations on shear strength for 26 subgroups of...Ch. 16 - The number of scratches on the surface of each of...Ch. 16 - The following numbers are observations on tensile...Ch. 16 - Prob. 44SE Ch. 16 - Prob. 46SE

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Find the probability of accepting the lot when p = 0.02 , 0.05 and 0.10 .

Solution Summary: The author calculates the probability of accepting the lot when p=0.01,0.02-and0.10 is 0.8188, 0.2904 and 0.0038.

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Chapter 16 Solutions