First, we need to find the z-score usign the next formula
[tex]z=(x-\mu)/\sigma[/tex]x is the score
μ is the mean
σ is the standard deviation
In our case
The mean (μ) = 4
The standard division (σ )= 1
The calls lasted less than (x)=3
we substitute the values
[tex]z=\frac{3-4}{1}=-1[/tex]the with tables we can find the probability for the value of z given
[tex]P(x<3)=0.15866\times100\text{\%=15.87\%}=16\text{\%}[/tex]percentage of the calls that lasted less than 3 min is 15.87% approximately 16%
