Let X= the variable representing the life of tube lights: X∼N(μ,σ2).
Given μ=4500h,σ=1500h.
P(4000<X<6000)=P(X<6000)−P(X≤4000)
=P(Z<15006000−4500)−P(Z≤15004000−4500)
=P(Z<1)−P(Z≤−31)
≈0.8413447−0.3694413
≈0.4719
1000(0.4719)=472The number of tubes lasting between 4000 hrs and 6000 hrs is 472.
Comments