only one bulb is needed to achive sufficient lighting
with a circuit that will turn on the next bulb in line when one fails
What is theprobability it will last for 50k hrs
the expected mean lifetime of the lighting will beμtotal=μ1+μ2+μ3=15000+15000+15000=45000standard deviation:σtotal=σ12+σ22+σ32=2600hrsthis doesnt give significant soα=0.05,ie 95 conifidencethis is a upper tail hypothesis testH0:μ=50000hrs nullH1:μ<50000hrs alternatez1=σx−μ=260045000−50000=−1.92P(z1)=21[1+erf(−21.92)]=0.0190.019<<α so must reject the H0 hypothesis, thus only 1.9