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:

* 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%

hit or miss i guess they never miss huh

step-by-step explanation:

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