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anonymous
 one year ago
The time required to finish a test in normally distributed with a mean of 40 minutes and a standard deviation of 8 minutes. What is the probability that a student chosen at random will finish the test in more than 56 minutes?
84%
2%
34%
16%
anonymous
 one year ago
The time required to finish a test in normally distributed with a mean of 40 minutes and a standard deviation of 8 minutes. What is the probability that a student chosen at random will finish the test in more than 56 minutes? 84% 2% 34% 16%

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amistre64
 one year ago
Best ResponseYou've already chosen the best response.1what do you have to find a solution with?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1do you know the empirical rule?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1you have some studying to do then .... look up the empirical rule and let me know what you get

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1what does it say ...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0In statistics, the socalled 68–95–99.7 rule is a shorthand used to remember the percentage of values that lie within a band around the mean in a normal distribution with a width of one, two and three standard deviations, respectively; more accurately, 68.27%, 95.45% and 99.73% of the values lie within one, two and three standard deviations of the mean, respectively. In mathematical notation, these facts can be expressed as follows, where x is an observation from a normally distributed random variable, μ is the mean of the distribution, and σ is its standard deviation:

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1copy paste eh, at least you found it :) now, we need to know how many standard deviation are between the mean, and the desired value. we start by finding the difference between them. what is the difference?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got it form here, I just need to know how to start it out. Thank you so much for your help!

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1ok, if you need more help on it, just let me know.
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