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Math Help - Distrubition help needed!

  1. #1
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    Distrubition help needed!

    I'm sorry for asking a maths question again. XD

    Q8. The smallest 84% of stainless steel nails are less than 49.7mm long, and the largest 97.5% are longer than 41.6mm long.

    a. Caculate the likely mean length of the nails and an appropriate value for the standard deviation.

    What do I do? >.<

    Thanks in advance! =]
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  2. #2
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Hikaru View Post
    I'm sorry for asking a maths question again. XD

    Q8. The smallest 84% of stainless steel nails are less than 49.7mm long, and the largest 97.5% are longer than 41.6mm long.

    a. Caculate the likely mean length of the nails and an appropriate value for the standard deviation.

    What do I do? >.<

    Thanks in advance! =]
    This is a normal distribution.

    We can say that the chances that a nail is shorter than 41.6mm is 2.5%
    And the chances that a nail is longer than 49.7mm is 16%

    P(X < 41.6) = 0.025

    P \left( \frac{X - \mu}{\sigma} < \frac{41.6 - \mu}{\sigma} \right) = 0.025

    P \left( Z < \frac{41.6 - \mu}{\sigma} \right) = 0.025

    \Phi \left( \frac{41.6 - \mu}{\sigma} \right) = 0.025

    \frac{41.6 - \mu}{\sigma} = -1.96

    =====

    P(X > 49.7) = 0.16

    1 - P(X \leq 49.7) = 0.16

    P \left( \frac{X - \mu}{\sigma} \leq \frac{49.7 - \mu}{\sigma} \right) =

    P \left( Z \leq \frac{49.7 - \mu}{\sigma} \right) = 0.84

    \Phi \left( \frac{49.7 - \mu}{\sigma} \right) = 0.84

    \frac{49.7 - \mu}{\sigma} = 0.996

    ====


    From here on it's simply a matter of setting those two equations equal to each other and solving.
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  3. #3
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    Quote Originally Posted by Hikaru View Post
    I'm sorry for asking a maths question again. XD

    Q8. The smallest 84% of stainless steel nails are less than 49.7mm long, and the largest 97.5% are longer than 41.6mm long.

    a. Caculate the likely mean length of the nails and an appropriate value for the standard deviation.

    What do I do? >.<

    Thanks in advance! =]
    Let X be the random variable length of nail and assume X ~ Normal( \mu, ~ \sigma).

    \Pr(X < 49.7) = 0.84 \Rightarrow \Pr\left( Z < \frac{49.7-\mu}{\sigma}\right) = 0.84

    But \Pr(Z < a) = 0.84 \Rightarrow a \approx 0.99446.

    Therefore:

    \frac{49.7-\mu}{\sigma} = 0.99446 .... (1)


    \Pr(X > 41.6) = 0.975 \Rightarrow \Pr\left( Z > \frac{41.6-\mu}{\sigma}\right) = 0.975

    But \Pr(Z > a) = 0.975 \Rightarrow a \approx -1.95996.

    Therefore:

    \frac{41.6 -\mu}{\sigma} = -1.95996 .... (2)


    Now solve equations (1) and (2) simultaneously.


    For checking purposes: I get \mu = 46.97, correct to two decimal places.
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  4. #4
    Bar0n janvdl's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Let X be the random variable length of nail and assume X ~ Normal( \mu, ~ \sigma).

    \Pr(X < 49.7) = 0.84 \Rightarrow \Pr\left( Z < \frac{49.7-\mu}{\sigma}\right) = 0.84

    But \Pr(Z < a) = 0.84 \Rightarrow a \approx 0.99446.

    Therefore:

    \frac{49.7-\mu}{\sigma} = 0.99446 .... (1)


    \Pr(X > 41.6) = 0.975 \Rightarrow \Pr\left( Z > \frac{41.6-\mu}{\sigma}\right) = 0.975

    But \Pr(Z > a) = 0.975 \Rightarrow a \approx -1.95996.

    Therefore:

    \frac{41.6 -\mu}{\sigma} = -1.95996 .... (2)


    Now solve equations (1) and (2) simultaneously.
    Mr F, your distribution table seems to be more accurate than mine...
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  5. #5
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    Quote Originally Posted by janvdl View Post
    This is a normal distribution.

    We can say that the chances that a nail is shorter than 41.6mm is 2.5%
    And the chances that a nail is longer than 49.7mm is 16%

    P(X < 41.6) = 0.025

    P \left( \frac{X - \mu}{\sigma} < \frac{41.6 - \mu}{\sigma} \right) = 0.025

    P \left( Z < \frac{41.6 - \mu}{\sigma} \right) = 0.025

    \Phi \left( \frac{41.6 - \mu}{\sigma} \right) = 0.025

    \frac{41.6 - \mu}{\sigma} = -1.96

    =====

    P(X > 49.7) = 0.16

    1 - P(X \leq 49.7) = 0.16

    P \left( \frac{X - \mu}{\sigma} \leq \frac{49.7 - \mu}{\sigma} \right) =

    P \left( Z \leq \frac{49.7 - \mu}{\sigma} \right) = 0.84

    \Phi \left( \frac{49.7 - \mu}{\sigma} \right) = 0.84

    \frac{49.7 - \mu}{\sigma} = 0.996

    ====


    From here on it's simply a matter of setting those two equations equal to each other and solving.
    That's a first. Getting beaten to the punch on a statistics question
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  6. #6
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    Quote Originally Posted by janvdl View Post
    Mr F, your distribution table seems to be more accurate than mine...
    And my TI-89 is even more accurate .....
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  7. #7
    Bar0n janvdl's Avatar
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    Quote Originally Posted by mr fantastic View Post
    And my TI-89 is even more accurate .....
    Ah okay I get it

    My calculator isn't capable of doing that...
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  8. #8
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    Quote Originally Posted by janvdl View Post
    Ah okay I get it

    My calculator isn't capable of doing that...
    There are many on-line calculators that can do it for you. For example:

    z score Calculator
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