Given : 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 ( normal distribution ) and has a mean of 61 and a standard deviation of 9.

i.e. $\mu=61$ and $\sigma=9$

To find : The approximate percentage of lightbulb replacement requests numbering between 34 and 61.

i.e. The approximate percentage of lightbulb replacement requests numbering between 34 and $34+3(9)$ .

i.e. i.e. The approximate percentage of lightbulb replacement requests numbering between $\mu$ and $\mu+3(\sigma)$ . (1)

According to the 68-95-99.7 rule, about 99.7% of the population lies within 3 standard deviations from the mean.

i.e. about 49.85% of the population lies below 3 standard deviations from mean and 49.85% of the population lies above 3 standard deviations from mean.

i.e.,The approximate percentage of lightbulb replacement requests numbering between $\mu$ and $\mu+3(\sigma)$ = 49.85%

⇒ The approximate percentage of lightbulb replacement requests numbering between 34 and 61.= 49.85%

4 0

2 years ago

Solve it by substitution method:- 5x - 6y=-9 and 3x + 4y= 25

user100 [1]

$\left\{\begin{array}{ccc}-5x-6y=-9&/\cdot2\\3x+4y=25&/\cdot3\end{array}\right\\+\left\{\begin{array}{ccc}-10x-12y=-18\\9x+12y=75\end{array}\right\\------------\\.\ \ \ \ \ \ \ -x=57\\.\ \ \ \ \ \ \ \ \ x=-57\\\\3\cdot(-57)+4y=25\\-171+4y=25\\4y=25+171\\4y=196\ \ \ \ /:4\\y=49\\\\Solution:x=-57;\ y=49.$

8 0

3 years ago