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Thread: Normal distribution question

  1. #1
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    Normal distribution question

    My question is as follows: Speed driven by motorists are normally distributed with mean 63 and standard deviation unknown.

    1.) If 2.5% of all motorists are traveling at speed 70 or more, what is the value for standard deviation?

    2.) Mean and standard deviation are unknown. It is observed that 10% of all motorists travel at less than speed 36 and 10% of all motorists travel at more than speed 70. What must mean and standard deviation be?
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  2. #2
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    Re: Normal distribution question

    Quote Originally Posted by romsek View Post
    My question is as follows: Speed driven by motorists are normally distributed with mean 63 and standard deviation unknown.

    1.) If 2.5% of all motorists are traveling at speed 70 or more, what is the value for standard deviation?

    2.) Mean and standard deviation are unknown. It is observed that 10% of all motorists travel at less than speed 36 and 10% of all motorists travel at more than speed 70. What must mean and standard deviation be?
    $\Phi()$ below refers to the CDF of the standard normal distribution.

    1) $P[s>70] = 0.025$

    $1-P[s<70] = 0.025$

    $P[s<70] = 0.975$

    so we know that

    $\Phi\left(\dfrac{70-63}{\sigma} \right) = 0.975$

    Using a table, or software, or what have you, you then get

    $\dfrac{7}{\sigma} \approx 1.96$

    $\sigma \approx \dfrac{7}{1.96} = 3.57$

    2) $P[s<36]=0.1,~P[s>70]=0.1$

    well you have 2 equations now in $\mu$ and $\sigma$

    $\Phi\left(\dfrac{36-\mu}{\sigma}\right) = 0.1$

    $1-\Phi\left(\dfrac{70-\mu}{\sigma}\right) = 0.1$

    apply the same table lookup done in (1) to get rid of $\Phi()$ and solve the resulting algebraic system.

    I leave that to you.
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    Re: Normal distribution question

    Essentially, this is a matter of being able to read a table of the "standard normal distribution". A good one is here: https://www.stat.tamu.edu/~lzhou/sta...ormaltable.pdf.

    "2.5% of them are larger than this number" corresponds to probability 0.025 larger than the number so 1- 0.025= 0.975 corresponds to z= 1.96. That is, z is 1.96 standard deviations above the mean. Here, we are told that the mean is 63 so, taking the standard deviation to be " \sigma", 70= 53+ 1.96\sigma. Solve that equation for \sigma.
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    Re: Normal distribution question

    Does anyone think that the OP was meant to use the 68-95-99.7 Approximation Rule?
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    Re: Normal distribution question

    Quote Originally Posted by Prove It View Post
    Does anyone think that the OP was meant to use the 68-95-99.7 Approximation Rule?
    If the question asked for an approximation or rough figure, yes. Otherwise, it would drive me NUTS to make that assumption when 95 is closer to 95.5.
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    Re: Normal distribution question

    Can someone please help me better understand #2 of the original question / provide the solution?
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    Re: Normal distribution question

    Again, look at a table. The lowest 10% in a standard normal distribution is z= -2.3. That is, 2.3 and more standard deviations below the mean. \frac{36- \mu}{\sigma}= -2.3. The highest 10% is at z= 1.29 above the mean. \frac{70- \mu}{\sigma}= 1.29. Solve those two equations for \mu and \sigma.
    Last edited by HallsofIvy; Oct 2nd 2017 at 04:20 AM.
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    Re: Normal distribution question

    For example, would I do 36/-2.3 to get the std and then find the mean accordingly from that? How do I combine the two (highest & lowest)?
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