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?

Answer: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%