Question

The maintenance department at the main campus of a large state university receives daily requests to replace fluorecent lightbulbs. The distribution of the number of daily requests is bell-shaped and has a mean of 46 and a standard deviation of 7. Using the 68-95-99.7 rule, what is the approximate percentage of lightbulb replacement requests numbering between 32 and 46

Answer 1

P₂ [ 32 ≤ X ≤ 46 ] = 47,71 %

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 %