Assume the hold time of callers to a cable company is normally distributed with a mean of 3.53.5 minutes and a standard deviation of 0.20.2 minute. determine the percent of callers who are on hold between 3.43.4 minutesminutes and 3.83.8 minutes.

62.47%

Step-by-step explanation:

Mean =

Standard deviation =

We are supposed to find the percent of callers who are on hold between 3.4 minutes and 3.8 minutes.

Formula :

At x = 3.4

At x =3.8

We are supposed to find P(3.4<z<3.8)

Using z table

So, P(3.4<z<3.8)=P(-0.5<z<1.5)=P(z<1.5)-P(z<-0.5) = 9332-0.3085=0.6247= 62.47%

So, the percent of callers who are on hold between 3.4 minutes and 3.8 minutes is 62.47%

g = (2, 2)

step-by-step explanation:

given the ratio is 1 : 1 then f is the midpoint of eg

let the coordinates of g = (x, y)

using the midpoint formula

(0 + x) = 1 ( multiply both sides by 2 )

0 + x = 2 ⇒ x = 2

similarly

(4 + y) = 3 ( multiply both sides by 2 )

4 + y = 6 ( subtract 4 from both sides )

y = 2

coordinates of g = (2, 2)

boxplot e is the answer