Question

Due in 6 hours, 2 minutes. Due Sun 09/08/2019 11:59 pm 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 49 and a standard deviation of 10. Using the 68-95-99.7 rule, what is the approximate percentage of lightbulb replacement requests numbering between 29 and 49? Do not enter the percent symbol. ans = License Points possible: 1 Unlimited attempts.

Answer 1

$\mathbf{Solution}$

$\mathbf{given:}\;\mu=49,\;\sigma=10$

$\mathbf{Use\; 68-95-99.7 \;rule}$

$\mu-3\sigma=49-(3*10)=\mathbf{19}$

$\mu-2\sigma=49-(2*10)=\mathbf{29}$

$\mu-1\sigma=49-(1*10)=\mathbf{39}$

$\mu=\mathbf{49}$

$\mu+1\sigma=49+(1*10)=\mathbf{59}$

$\mu+2\sigma=49+(2*10)=\mathbf{69}$

$\mu+3\sigma=49+(3*10)=\mathbf{79}$

0.15% 0.15% 2.35% 13.5% 34% 34% 13.5% 2.35% ود و3 وو و و6 و7 68% 95% 9979

Approximate percentage of lightbulb replacement requests numbering between 29 and 49 is $13.5\%+34\%=\mathbf{{\color{Red} 47.5}}\%$

$\therefore \mathbf{ans}=\mathbf{{\color{Red} 47.5}}\%$