find the normal distribution

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 54 and a standard deviation of 9. Using the 68-95-99.7 rule, what is the approximate percentage of lightbulb replacement requests numbering between 54 and 72?

Answer 1

X∼N(54,81) where μ = 54 and σ = 9.

Use the z-score formula to find the probability for the lower and upper values.

z₁ = (54 - 54) / 9 = 0/9 = 0 ⇒ P(z₁ < 0) = 0.50

z₂ = (72 - 54) / 9 = 18/9 = 2 ⇒ P(z₂ < 2) = 0.9772

P(0 < z < 2) = 0.9772 - 0.50 = 0.4772 or 47.72%

The approximate percentage of lightbulb replacement requests numbering between 54 and 72 is about 47.72%.