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mariaaking1413
17.07.2019 •
Mathematics
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?
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Ответ:
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%
Ответ:
5g
Step-by-step explanation:
the first half life would be 20 the next, 10. and lastly 5