Assume a normal distribution and that the average phone call in a certain town lasted 9 min, with a standard deviation of 1 min. What percentage of the calls lasted less than 8 min?

The percentage of the calls lasted less than 8 min is 16%.

Step-by-step explanation:

We are dealing with a normal distribution with an average phone call of 9 min and a standard deviation of 1 min. Below we can observe the empirical rule applied with a mean of 9 and a standard deviation of 1. The number 8 represents one standard deviation below the mean, so, the percentage of observations below 8 is 16%. Therefore the percentage of the calls lasted less than 8 min is 16%.

The percentage of the calls lasted less than 8 min is 16%

Step-by-step explanation:

* Lets explain how to solve the problem

- To find the percentage of the calls lasted less than 8 min, find the

z-score for the calls lasted

∵ The rule of z-score is z = (x - μ) / σ, where

# x is the score

# μ is the mean

# σ is the standard deviation

* Lets solve the problem

- The average phone call in a certain town lasted is 9 min

∴ The mean (μ) = 9

- The standard deviation is 1 min

∴ σ = 1

- The calls lasted less than 8 min

∴ x = 8

∵ z = (x - μ) / σ

∴ z = (8 - 9) / 1 = - 1/1 = - 1

∴ P (z < 8) = - 1

- Use z-table to find the percentage of x < 8

∴ P (x < 8) = 0.15866 * 100% = 15.87% ≅ 16%

* The percentage of the calls lasted less than 8 min is 16%