Answer: 49.85%
Step-by-step explanation:
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.
and 
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
.
i.e. i.e. The approximate percentage of lightbulb replacement requests numbering between
and
. (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
and
= 49.85%
⇒ The approximate percentage of lightbulb replacement requests numbering between 34 and 61.= 49.85%
There's no question there. You didn't attach the pic
104 bottles 2.288 divided by .022 equals 104
Amount of sales of newspapers for the month of January = $8341.50
Percentage of profit for which the newspaper is sold = 0.5%
Then
Amount of profit made in the month of January = 0.5% * 8341.50 dollars
= (0.5/100) * 8341.50 dollars
= 4170.75/100 dollars
= 41.707 dollars
= 41.71 dollars
So the shop makes a profit of $41.71 in the month of January by selling newspapers worth $8341.50. I hope the procedure is perfectly clear for you to understand.