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Math Help - 2 parts to this question..

  1. #1
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    Angry 2 parts to this question..

    I hate these question that involve part A and B part A I get but part B not so sure?

    Data was collected outside a popular new restaurant to determine the mean waiting time to be seated at a table. Assume the data was normally distributed with a mean of 45 min and standard deviation of 12 min
    determine the probability that a randomly selected person has to wait less than 20 min
    So for this questionI would use:
    normalcdf(0,20,45,12)
    = 0.0185219157

    on to part B

    determine the probability that a randomly selected person has to wait more than 1 h?
    they have to wait more than an hour
    so I cant use the normalCDF thing

    confused?
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  2. #2
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    Am I understanding correctly that you know how to figure out the probability of somebody waiting less than a certain time, but not how to figure out the probability of them waiting MORE than a certain time?

    First find the probability of waiting less than an hour. Everybody waits either more than an hour, or less than an hour. Important concept: the sum of the probability of all outcomes is always going to be 1 (in technical language, the distribution is normalized). So just subtract your probability of waiting less than an hour from one, and you have the answer.

    An alternative approach is to notice that the normal distribution is symmetrical around the mean, so the probability of waiting more than 1 hour is the same as the probability of waiting less than 30 minutes (both are 15 minutes from the mean).
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  3. #3
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    Quote Originally Posted by justin View Post
    I hate these question that involve part A and B part A I get but part B not so sure?

    Data was collected outside a popular new restaurant to determine the mean waiting time to be seated at a table. Assume the data was normally distributed with a mean of 45 min and standard deviation of 12 min
    determine the probability that a randomly selected person has to wait less than 20 min
    So for this questionI would use:
    normalcdf(0,20,45,12)
    = 0.0185219157

    on to part B

    determine the probability that a randomly selected person has to wait more than 1 h?
    they have to wait more than an hour
    so I cant use the normalCDF thing

    confused?
    A good idea for doing "less than" is to use -E99 instead of 0 (well, in the case of time it is usually more logical to use 0), as -E99 is the calculator's estimate of infinity (just the lowest number it can handle). Type "E" by hitting 2nd, then the comma.
    The same applies when doing a "more than" question...just use E99 to represent infinity.
    With that being said, all you have to do is
    normalCDF(60,E99,45,12)

    (Start,end,mean,sd)

    Hope this helps
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