The maintenance department at the main campus of a large state university receives daily requests to replace fluorecent lightbulbs. the distribution of the number of daily requests is bell-shaped and has a mean of 40 and a standard deviation of 4. using the 68-95-99.7 rule, what is the approximate percentage of lightbulb replacement requests numbering between 40 and 48?

47.5%

Step-by-step explanation:

Given : The distribution of the number of daily requests is bell-shaped and has a mean of and a standard deviation of 40.

To find : The approximate percentage of lightbulb replacement requests numbering between 40 and 48.

We can see that i.e. 48 is two standard deviations ( to the right) from the mean.               (1)

And 40 is the mean value () itself . It means the lower limit lies at the center of the normal curve (at z=0).

It means we need only right the region which is 2 standard deviations from the mean (at z=2).

According to the 68-95-99.7 rule, 95% of the population falls within two deviation from the mean in both left and right side of the normal curve.

Then, the required area will be :

Hence,  the approximate percentage of lightbulb replacement requests numbering between 40 and 48 = 47.5%

muyb ajo

step-by-step explanation:

jegosu

answer:   (s+1)(s+4)

step-by-step explanation:

factor s2+5s+4

s2+5s+4

=(s+1)(s+4)