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%

for this pair of functions g(x)=x-2 and f(x)=|x| is (g o f)(a)= |a| -2

step-by-step explanation:

g(x)=x-2

f(x)=|x|

(g o f)(a)=?

(g o f)(x)=g(f(x))

(g o f)(x)=g(|x|)

(g o f)(x)=|x|-2

(g o f)(a)=|a|-2

a

step-by-step explanation: