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    A group of students believes that the time taken to travel to college, T minutes can be assumed to be normally distributed. Within the college 5% of students take at least 55 minutes to travel to college and 0.1% takes less than 10 minutes. Find the mean and standard deviation of T..
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    Sorry you've not had any responses about this. Are you sure you've posted in the right place? Here's a link to our subject forum which should help get you more responses if you post there.


    Just quoting in Fox Corner so she can move the thread if needed :wizard:
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    (Original post by sunshinewothaya)
    Within the college 5% of students take at least 55 minutes to travel to college
    In other words, P(T >= 55) = 0.05. Which can be rephrased as P(T < 55) = 0.95.

    0.1% takes less than 10 minutes.
    and P(T < 10) = 0.001.

    Recall that z = \dfrac{x-\mu}{\sigma}. You can find z with the probabilities above, your x's are 55 and 10, and the formula I gave will let you use this data to construct a pair of simultaneous equations which will yield the mean and standard deviation.
 
 
 
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