37
Ejercicio 1 DISTRIBUCION POISSON p ƛ= np L n 60 N 400 0.01 0.6 0.87809862 c 1 0.02 1.2 0.66262727 0.03 1.8 0.46283689 0.04 2.4 0.30844104 0.05 3 0.19914827 0.06 3.6 0.12568912 0.07 4.2 0.077977 0.08 4.8 0.04773253 0.09 5.4 0.02890612 0.1 6 0.01735127 DISTRIBUCION BINOMIAL DISTRIBUCION HIPERGEOMETRICO p q L p D L 0.01 0.99 0.87876673 0.01 4 0.89138365 0.02 0.98 0.66190393 0.02 8 0.65699005 0.03 0.97 0.4592108 0.03 12 0.44024147 0.04 0.96 0.3022331 0.04 16 0.27801163 0.05 0.95 0.19155337 0.05 20 0.1683672 0.06 0.94 0.11792319 0.06 24 0.0987454 0.07 0.93 0.0708943 0.07 28 0.05642082 0.08 0.92 0.04177129 0.08 32 0.03152959 0.09 0.91 0.02418085 0.09 36 0.01727851 0.1 0.9 0.01377708 0.1 40 0.00930279 DETERMINE U N PLAN DE MUESTREO SIMPLE PARA AQL = 0.01 α = 0.0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 poison BINOMI HIPERG Axis Title

Practica 1

Embed Size (px)

DESCRIPTION

control de calidad

Citation preview

Hoja1Ejercicio 1 DISTRIBUCION POISSONp= npLn60N4000.010.60.8780986178c10.021.20.66262726620.031.80.4628368870.042.40.30844104120.0530.19914827350.063.60.12568912330.074.20.07797699950.084.80.04773253290.095.40.0289061180.160.0173512652DISTRIBUCION BINOMIALDISTRIBUCION HIPERGEOMETRICOpqLpDL

0.010.990.87876672870.0140.89138364780.020.980.66190392970.0280.6569900520.030.970.45921079710.03120.44024147440.040.960.30223310070.04160.27801162960.050.950.19155337470.05200.16836719560.060.940.1179231890.06240.09874540210.070.930.07089430180.07280.05642082340.080.920.04177128560.08320.03152958570.090.910.02418084680.09360.01727850980.10.90.0137770790.1400.0093027887DETERMINE U N PLAN DE MUESTREO SIMPLE PARA AQL = 0.01 = 0.05 LTPD = 0.1 y = 0.1

LTPD 0.1NOS VALDREMOS DE LAS TABLAS DE GRUBBS PARA REALIZAR EL EJERCICIO AQL 0.010.05LTPD/AQL 100.1

AQLLTPDc=5%=10%LTPD/AQL00.0512.30345.156862745110.3553.8910.957746478920.8185.3226.506112469431.3666.6814.890922401241.977.9944.057868020352.6139.2753.549559892863.28510.5323.206088280173.98111.7712.956794775284.69512.9952.767838125795.42514.2062.6186175115106.16915.4072.4974874372116.92416.5982.3971692663n35.536n38.939PLAN A PLAN BLTPD 0.1LTPD 0.1AQL 0.01AQL 0.010.050.1ltpd3.6aql0.390.12568912331-0.94110905560.0588909444PLAN A 0.050.1256891233PLAN B0.05889094440.1El plan B se acepta con n=39 c=1

Hoja34) con la caracteristica de los dos planes siguientes, calcular: Plan A n=150 y c=4 Plan B n=180 y c= 5

a) el plan mas reguiroso

n150n180Plan Aplan BP= npLp= npL0.011.50.98142406380.011.80.98962196310.0230.81526324450.023.60.84411853380.034.50.53210357640.035.40.54613210440.0460.28505650030.047.20.27589745270.057.50.13206185630.0590.11569052080.0690.05496364150.0610.80.04225517360.0710.50.02109356560.0712.60.01390305770.08120.00760039070.0814.40.00421839430.0913.50.00260434030.0916.20.00120006060.1150.00085664120.1180.0003239935a) el plan mas riguroso es el plan A trazando la linea de 45 desde el origen hasta el primero en ser tocado por la linea (plan A)

b) El lote con menor costo de inspeccion si los lotes tienen una calidad promedio de 4% y el costo de inspeccion es de 0.8 Bs/unidad

Despejando n de la ecuacion ya que el valor de p = 0.04p0.04PLAN APLAN Bn100n125Costo de muestreo80Costo de muestreo100

costo por unidad (Bs/un)0.8

b) el plan menos costoso seria el plan A con 80 Bs por 100 unidades inspeccionadas

5) Deduzca un plan de muestreo mediante el metodo de Grubb, con AQL =8 y LTPD = 14%

Asumiremos = 5% y =10%AQL0.08Nos valdremos de las tablas de grubbsLTPD0.14LTPD/AQL1.75Buscaremos este valor 0.050.1 = 5% = 10%}cAQLLTPDLTPD/AQL2014.07227.0451.92190164872114.89428.1841.89230562642215.71929.321.86525860422316.54930.4531.84017161162417.38231.5841.81705212292518.21932.7111.79543333882619.05833.8361.77542239482719.90134.9591.75664539472820.74636.081.7391304348

PLAN APLAN AC27n249AQL0.08c27LTPD0.145%n248.762524910.2872392432LTPD34.860.1028723924

PLAN BPLAN BC27n250AQL0.08c27LTPD0.14(%)5.2480713228n249.7071428571250(%)10AQL201-0.94751928680.0524807132

PLAN CPLAN C C28n260AQL0.08c28LTPD0.14(%)5n259.325260(%)9.1298129033LTPD36.40.091298129

PLAN D PLAN C C28n258AQL0.08c28LTPD0.14(%)4.7503778136n257.7142857143258(%)10AQL20.641-0.95249622190.0475037781

= 5% = 10%} PLAN A510.2872392432PLAN B5.248071322810 PLAN C59.1298129033 PLAN D4.7503778136106) Deduzca y grafique un plan de muestreo secuencial elemento a elemento, con AQL =2% y LTPD=8% a) un inspector prueba los 40 primeras unidades sin encontrar defectuosos, tendria que aceptar el lote antes de llegar a la unidad 40 o tendria que tomar una decision

AQL0.02LTPD 0.080.050.1h11.5531792524h21.9940842182S0.0435874902

nXLXRnXLXR0-1.55317925241.994084218221-0.63784195842.90942151221-1.50959176222.037671708422-0.59425446822.95300900242-1.4660042722.081259198623-0.5506669782.99659649263-1.42241678182.124846688724-0.50707948783.04018398284-1.37882929162.168434178925-0.46349199763.08377147295-1.33524180152.212021669126-0.41990450743.12735896316-1.29165431132.255609159327-0.37631701733.17094645337-1.24806682112.299196649528-0.33272952713.21453394358-1.20447933092.342784139729-0.28914203693.25812143379-1.16089184072.386371629930-0.24555454673.301708923910-1.11730435052.429959120131-0.20196705653.345296414111-1.07371686032.473546610332-0.15837956633.388883904312-1.03012937012.517134100533-0.11479207613.432471394513-0.98654187992.560721590734-0.07120458593.4760588847l14-0.94295438972.604309080835-0.02761709573.519646374815-0.89936689952.647896571360.01597039453.56323386516-0.85577940942.6914840612370.05955788473.606821355217-0.81219191922.7350715514380.10314537483.650408845418-0.7686044292.7786590416390.1467328653.693996335619-0.72501693882.8222465318400.19032035523.737583825820-0.68142944862.865834022a) se deberan inspeccionar 35 elementos antes de llegar a los 40

b) se rechazan el elemento 10, 18 y 23 pero no son rechazados de la grafica, pero la cantidad de unidades defectuosos llega a 3 por lo que se rechaza el lote

se deberan inspeccionar almenos 22 elementos y como maximo 23 para poder encontrar las tres unidades defectuosas

7) Deduzca un plan de muestreo mediante el metodo de Grubbs en AQL 1.999% y LTPD 9.788% y compare con un plan de muestreo secuencial Cual seria el plan mas conveniente si se tiene lotes de 5000 unidades con una calidad promedio de 1%?5%?

AQL0.01999LTPD0.097880.050.1AQL/LTPD4.896

=5%=10%CaqlltpdLTPD/AQL00.0512.30345.15710.3553.8910.95820.8185.3226.50631.3666.6814.89141.977.9944.05852.6139.2753.550C3nn68.33416708356968.257049448369PLAN POR GRUBBSn69c3Plan Secuencial h11.3470099303n27.1845152875h21.7293890835n34.9014531513S0.0495506326nXlXr1-1.29745929771.77893971612-1.24790866511.82849034873-1.19835803241.87804098144-1.14880739981.9275916145-1.09925676721.97714224666-1.04970613452.02669287937-1.00015550192.07624351198-0.95060486932.12579414459-0.90105423662.175344777210-0.8515036042.224895409811-0.80195297142.274446042412-0.75240233872.323996675113-0.70285170612.373547307714-0.65330107352.423097940315-0.60375044082.47264857316-0.55419980822.522199205617-0.50464917562.571749838218-0.45509854292.621300470919-0.40554791032.670851103520-0.35599727772.720401736121-0.3064466452.769952368822-0.25689601242.819503001423-0.20734537982.86905363424-0.15779474712.918604266725-0.10824411452.968154899326-0.05869348193.017705531927-0.00914284923.0672561646280.04040778343.1168067972290.0899584163.1663574298300.13950904873.2159080625310.18905968133.2654586951320.23861031393.3150093277330.28816094663.3645599604340.33771157923.414110593

Hoja2A) CUAL ES LA PROBABILIDAD DE QUE LA RESISTENCIA DE LA MUESTRA SEA MENOR QUE 6250

MEDIA U6000DESVIACION ESTANDAR 100P(X