Suppose a call center receives 4000 calls on average during their Friday evening shift. Next Friday several people have put in for time off, so the center will be staffed by 178 call center employees. When working continuously, the average call center employee can handle 22 calls per evening shift (with a standard deviation of 5 calls). Assuming the center receives its expected 4000 calls next Friday, what is the probability that the center will have enough employees to handle this call volume

The probability that the center will have enough employees to handle this call volume is 0.2764

Step-by-step explanation:

Probability is defined as the ratio of favorable outcomes to the total number of outcomes.

Here the total mean expected number of outcomes=u = 4000

The favorable  expected outcomes are = x= 22* 178= 3916

The  standard deviation= σ= 5

The random variable X =178 has normal distribution with mean 4000/22 and standard deviation 5

So z= x-u/σ

Z= (3916-4000)/22/5

z= -0.764

From the table of areas under normal curve

P (  x= 3916)=  P( -0.764 ≤ Z≤ 0)

P (  x= 3916)=  0.2764

The probability that the center will have enough employees to handle this call volume is 0.2764

y=3x+4

step-by-step explanation:

given the points a(0,4) and b(-4, -8)

eqn of line ab: (y-4)/(x-0)=(-8-4)/(-4-0)

(y-4)/x=-12/-4=3

y=3x+4