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

Answer:

50%

Step-by-step 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%