Question

The physical plant at the main campus of a large state university receives daily requests to replace flourecent light bulbs. The distribution of the number of daily requests is bell-shaped and has a mean of 37 and a standard deviation of 4. Using the 68-95-99.7 rule, what is the approximate percentage of light bulb replacement requests numbering between 37 and 49?

ans = ______________%

Answer 1

soultion

Given that,

$\mu$ = 37

$\sigma$ = 4

So,

Using empirical rule

P($\mu$ - 3$\sigma$< X < $\mu$ + 3$\sigma$) = 99.7%

P(37 - 12< X < 37 + 12) = 99.7%

(25 < X < 49 ) = 99.7%

P ( x ≤ 25) =0.0015

P ( x ≤ 49 ) = 1 - 0.0015 =0.9985

P ( x ≤ 25 ) =0.0015

P ( x ≤ 29 ) =0.025

P ( x ≤ 33) =0.16

P ( x ≤ 41 ) =0.84

P ( x ≤ 45 ) =0.975

P ( x ≤ 49 ) =0.9985