Answer:
I have no clue
Step-by-step explanation:
i dont lnow
Answer:
swag
Step-by-step explanation:
SOLUTION
The mean is 4min
standard deviation is 1min
the z score is

where

then we have

The probability the call lasted less than 3 min will be
Therefore, the probability that (z < -1 ) is
[tex]\begin{gathered} Pr(z<-1)=Pr(0
Hence, the percentage of the calls that lasted less than 3 min is 16%
Thats 3.85714285714 because if you add the numbers and divide by the numbers they are it’ll be 3.85714285714