SOLUTION: The average time a person spends in a library is 30 minutes.The standard deviation is 6 minutes.if a visitor is selected at random.Assume the variable to be normally distributed.

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Question 1180150: The average time a person spends in a library is 30 minutes.The standard deviation is 6 minutes.if a visitor is selected at random.Assume the variable to be normally distributed.
a. What is the probability that he/she spends between 25 and 35 minutes?
b. What is the probability that he/she spends at most 40 minutes?
c. What is the probability that he/she spends at least 35 minutes?
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!

Hi
Normal Distribution:  µ = 30 and σ = 6 
Using TI or similarly an inexpensive calculator like a Casio fx-115 ES plus
Continuous curve
Chose 'a person'
P(25> x > 35) = normalcdf(25, 35, 30, 6) = .5953
P(x < 40) = P(x ≤ 40) = normalcdf(-9999, 40, 30, 6) = .9522
P(x > 35 = P(x ≥ 35) = normalcdf( 35, 9999, 30, 6) = .2023

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