Question

The physical plant at the main campus of a large state university recieves daily requests to replace florecent lightbulbs. The distribution of the number of daily requests is bell-shaped and has a mean of 55 and a standard deviation of 4. Using the 68-95-99.7 rule, what is the approximate percentage of lightbulb replacement requests numbering between 43 and 55?

Answer 1

50%

Explanation:

68-95-99.7 rule

68% of all values lie within the 1 standard deviation from mean
`(\mu-\sigma,\mu+\sigma)`

95% of all values lie within the 1 standard deviation from mean
`(\mu-1\sigma,\mu+1\sigma)`

99.7% of all values lie within the 1 standard deviation from mean
`(\mu-3\sigma,\mu+3\sigma)`

The distribution of the number of daily requests is bell-shaped and has a mean of 55 and a standard deviation of 4.

`\mu = 55\\\sigma = 4`

68% of all values lie within the 1 standard deviation from mean
`(\mu-\sigma,\mu+\sigma)` =
`(55-4,55+4)`=
`(51,59)`

95% of all values lie within the 2 standard deviation from mean
`(\mu-1\sigma,\mu+1\sigma)`=
`(55-2(4),55+2(4))`=
`(47,63)`

99.7% of all values lie within the 3 standard deviation from mean
`(\mu-3\sigma,\mu+3\sigma)`=
`(55-3(4),55+3(4))`=
`(43,67)`

Refer the attached figure

P(43<x<55)=2.5%+13.5%+34%=50%

Hence The approximate percentage of light bulb replacement requests numbering between 43 and 55 is 50%