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Math Help - [SOLVED] Normal Distrubtion Question

  1. #1
    Member looi76's Avatar
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    [SOLVED] Normal Distrubtion Question

    Question
    The time taken by a garage to replace worn-out brake pads follows a normal distribution with mean 90 minutes and standard deviation 5.8 minutes. The garage claims to complete the replacement in 'a to b minutes'. If this claim is to be correct for 90% of the repairs, find a and b correct to 2 significant figures, based on a symmetrical intervals centred on the mean.

    Attempt:
    I only know that \mu = 90 and \sigma = 5.8. I can't understand what the question exactly means. How am I suppose to find the values of a and b ?
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  2. #2
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    Quote Originally Posted by looi76 View Post
    Question
    The time taken by a garage to replace worn-out brake pads follows a normal distribution with mean 90 minutes and standard deviation 5.8 minutes. The garage claims to complete the replacement in 'a to b minutes'. If this claim is to be correct for 90% of the repairs, find a and b correct to 2 significant figures, based on a symmetrical intervals centred on the mean.

    Attempt:
    I only know that \mu = 90 and \sigma = 5.8. I can't understand what the question exactly means. How am I suppose to find the values of a and b ?
    Let a = \mu - c and b = \mu + c.

    Then \Pr(\mu - c < X < \mu + c) = 0.9 \Rightarrow \Pr(X > \mu + c) = 0.05.

    Alternatively, \Pr(\mu - c < X < \mu + c) = 0.9 \Rightarrow \Pr(X < \mu - c) = 0.05.

    Therefore \mu - c = 80.4598.

    I leave the remaining details for you to finish.
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  3. #3
    Member looi76's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Let a = \mu - c and b = \mu + c.

    Then \Pr(\mu - c < X < \mu + c) = 0.9 \Rightarrow \Pr(X > \mu + c) = 0.05.

    Alternatively, \Pr(\mu - c < X < \mu + c) = 0.9 \Rightarrow \Pr(X < \mu - c) = 0.05.

    Therefore \mu - c = 80.4598.

    I leave the remaining details for you to finish.
    I should solve the problem using the normal distribution table. I have attached the table to this post.
    For 90% the value is 1.282, why did you choose 1.645?
    Attached Thumbnails Attached Thumbnails [SOLVED] Normal Distrubtion Question-normal_distrubtion_table.gif  
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  4. #4
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    Quote Originally Posted by looi76 View Post
    I should solve the problem using the normal distribution table. I have attached the table to this post.
    For 90% the value is 1.282, why did you choose 1.645?
    The 10% (0.1) must be evenly divided between the two tails ..... 5% (0.05) each end .... So \Pr(X > \mu + c) = 0.05 \, \Rightarrow \Pr (X < \mu + c) = 0.95 ..... Therefore the corresponding z-value of X = \mu + c is 1.645 .....

    So z = \frac{X - \mu}{\sigma} \Rightarrow 1.645 = \frac{c}{\sigma} \Rightarrow c = .....
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