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

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 %

b

step-by-step explanation:

answer:

step-by-step explanation: