By rounding to one significant number, estimate the answer to these questions:
1 answer:
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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(0Hence, the percentage of the calls that lasted less than 3 min is 16%