Question

Assume a normal distribution and that the average phone call in a certain town lasted 4 min, with a standard deviation of 1 min. What percentage of the calls lasted lessthan 3 min?eBooks/eResourcesPeopleWhat percentage of calls last less than 3 min?

Answer 1

First, we need to find the z-score usign the next formula

`z=(x-\mu)/\sigma`

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

`z=(3-4)/(1)=-1`

the with tables we can find the probability for the value of z given

`P(x<3)=0.15866*100\text{\%=15.87\%}=16\text{\%}`

percentage of the calls that lasted less than 3 min is 15.87% approximately 16%