На испытание поставлено N0=1000 образцов изделий число отказов фиксировались через каждые Δt=100 часов

Определим вероятность безотказной работы:
Р(t)= N0-n(t)N0 ;
Р(100)= 1000-601000 = 0,94; Р(200)= 1000-1101000 = 0,89;
Р(300)= 1000-1521000 = 0,848; Р(400)= 1000-1871000 = 0,813;
Р(500)= 1000-2171000 = 0,783; Р(600)= 1000-2401000 = 0,76;
Р(700)= 1000-2601000 = 0,74; Р(800)= 1000-2781000 = 0,722;
Р(900)= 1000-2941000 = 0,706; Р(1000)= 1000-3091000 = 0,691;
Р(1100)= 1000-3231000 = 0,677; Р(1200)= 1000-3361000 = 0,664;
Р(1300)= 1000-3491000 = 0,651; Р(1400)= 1000-3611000 = 0,639;
Р(1500) = 1000-3761000 = 0,624; Р(1600) = 1000-3901000 = 0,61;
Р(1700) = 1000-4041000 = 0,596; Р(1800) = 1000-4181000 = 0,582;
Р(1900) = 1000-4331000 =0,567; Р( 2000) = 1000-4461000 = 0,554;
Р( 2100) = 1000-4581000 = 0,542; Р(2200 ) = 1000-4711000 = 0,529;
Р(2300) = 1000-4831000 = 0,517; Р( 2400) = 1000-4971000 = 0,503;
Р( 2500) = 1000-5121000 =0,488; Р( 2600) = 1000-5301000 = 0,47;
Р( 2700) = 1000-5531000 =0,447; Р( 2800) = 1000-5801000 = 0,42;
Р( 2900) = 1000-6201000 =0,38; Р( 3000) = 1000-6801000 = 0,32.
Определим вероятность отказа:
q(t)=1- Р(t);
q(100)=1-0,94=0,06;
q(200)=1-0,89=0,11;
q(300)=1-0,848=0,152;
q(400)=1-0,813=0,187;
q(500)=1-0,783=0,217;
q(600)=1-0,76=0,24;
q(700)=1-0,74=0,26;
q(800)=1-0,722=0,278;
q(900)=1-0,706=0,294;
q(1000)=1-0,691=0,309;
q(1100)=1-0,677=0,323;
q(1200)=1-0,664=0,336;
q(1300)=1-0,651=0,349;
q(1400)=1-0,639=0,361;
q(1500)=1-0,624=0,376;
q(1600)=1-0,61=0,39;
q(1700)=1-0,596=0,404;
q(1800)=1-0,582=0,418;
q(1900)=1-0,567=0,433;
q(2000)=1-0,554=0,446;
q(2100)=1-0,542=0,458;
q(2200)=1-0,529=0,471;
q(2300)=1-0,517=0,483;
q(2400)=1-0,503=0,497;
q(2500)=1-0,488=0,512;
q(2600)=1-0,47=0,53;
q(2700)=1-0,447=0,553;
q(2800)=1-0,42=0,58;
q(2900)=1-0,38=0,62;
q(3000)=1-0,32=0,68.
Найдем частоту отказов по формуле:
f(t)= n(Δt)N0·Δt, 1/час;
f(100)= 60100000 = 0,0006; f(200)= 50100000 = 0,0005;
f(300)= 42100000 = 0,00042; f(400)= 35100000 = 0,00035;
f(500)= 30100000 = 0,0003; f(600)= 23100000 = 0,00023;
f(700)= 20100000 = 0,0002; f(800)= 18100000 = 0,00018;
f(900)= 16100000 = 0,00016; f(1000)= 15100000 = 0,00015;
f(1100)= 14100000 = 0,00014; f(1200)= 13100000 = 0,00013;
f(1300)= 13100000 = 0,00013; f(1400)= 12100000 = 0,00012;
f(1500)= 15100000 = 0,00015; f(1600)= 14100000 = 0,00014;
f(1700)= 14100000 = 0,00014; f(1800)= 14100000 = 0,00014;
f(1900)= 15100000 = 0,00015; f(2000)= 13100000 = 0,00013;
f(2100)= 12100000 = 0,00012; f(2200)= 13100000 = 0,00013;
f(2300)= 12100000 = 0,00012; f(2400)= 14100000 = 0,00014;
f(2500)= 15100000 = 0,00015; f(2600)= 18100000 = 0,00018;
f(2700)= 23100000 = 0,00023; f(2800)= 27100000 = 0,00027;
f(2900)= 40100000 = 0,0004; f(3000)= 60100000 = 0,0006.
Найдем интенсивность отказов по формуле:
λ(t) = n(Δt)Δt∙Nср, 1/час;
λ (100)= 60100(1000+9402) = 0,0006185; λ (200)= 50100(1000+8902) = 0,0005291;
λ (300)= 42100(1000+8482) = 0,0004545; λ (400)= 35100(1000+8132) = 0,0003861;
λ (500)= 30100(1000+7832) = 0,0003365; λ (600)= 23100(1000+7602) = 0,0002613;
λ (700)= 20100(1000+7402) = 0,0002298; λ (800)= 18100(1000+7222) = 0,000209;
λ (900)= 16100(1000+7062) = 0,0001875; λ (1000)= 15100(1000+6912) = 0,0001774;
λ (1100)= 14100(1000+6772) = 0,0001669; λ (1200)= 13100(1000+6642) = 0,0001562;
λ (1300)= 13100(1000+6512) = 0,0001574; λ (1400)= 12100(1000+6392) = 0,0001464;
λ (1500)= 15100(1000+6242) = 0,0001847; λ (1600)= 14100(1000+6102) = 0,0001739;
λ (1700)= 14100(1000+5962) = 0,0001754; λ (1800)= 14100(1000+5822) = 0,0001792;
λ (1900)= 15100(1000+5672) = 0,0001914; λ (2000)= 13100(1000+5542) = 0,0001673;
λ (2100)= 12100(1000+5422) = 0,0001556; λ (2200)= 13100(1000+5292) = 0,00017;
λ (2300)= 12100(1000+5172) = 0,0001582; λ (2400)= 14100(1000+5032) = 0,0001862;
λ (2500)= 15100(1000+4882) = 0,0002016; λ (2600)= 18100(1000+4702) = 0,0002448;
λ (2700)= 23100(1000+4472) = 0,0003178; λ (2800)= 27100(1000+4202) = 0,0003802;
λ (2900)= 40100(1000+3802) = 0,0005797; λ (3000)= 60100(1000+3202) = 0,000909.