Answer:
P₂ [ 32 ≤ X ≤ 46 ] = 47,71 %
Step-by-step explanation:
The rule:
68-95-99.7
establishes
P₁ [ μ₀ - σ ≤ X ≤ μ₀ + σ] ≈ 0,683
P₂ [ μ₀ - 2σ ≤ X ≤ μ₀ + 2σ] ≈ 0,954
P₃ [ μ₀ - 3σ ≤ X ≤ μ₀ + 3σ] ≈ 0,997
For our paticular case we have
μ₀ = 46
σ = 7
Then we get:
μ₀ - σ = 39 μ₀ + σ = 53
μ₀ - 2σ = 32 μ₀ + 2σ = 60
μ₀ - 3σ = 25 μ₀ + 3σ = 67
We were asked by % of replacement request numbering between 32 and 46.
Normal curve is symmetric therefore μ₀ - 2σ = 32 is just half of the interval P₂ [ μ₀ - 2σ ≤ X ≤ μ₀ + 2σ] ≈ 0,954, then between 32 and 46 the porcentage of lightbulb is 0,954/2 , is 0,4771 or 47,71 %
P₂ [ μ₀ - 2σ ≤ X ≤ μ₀ ] = P₂ [ 32 ≤ X ≤ 46 ] = 47,71 %
