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Thread: CLT question

  1. #1
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    CLT question

    A soil scientist wants estimate the average pH for a large field by randomly selecting 40 core samples. Although the population standard deviation of pH is not know, past experiences indicates that most soils have a pH value between 5 and 8. Find the approximate probability that the sample mean will be within .2 units of the true average pH.

    $\displaystyle P\left( -\frac{0.2}{\sigma/\sqrt{40}} \leq \frac{\overline{Y}-\mu}{\sigma/\sqrt{n}} \leq \frac{0.2}{\sigma/\sqrt{40}}\right)$

    $\displaystyle P\left( -\frac{0.2}{\sigma/\sqrt{40}} \leq Z \leq \frac{0.2}{\sigma/\sqrt{40}}\right)$

    I'm just have trouble deciding on a $\displaystyle \sigma$, since it's between 5 and 8, I'm not sure if I should just take an average which is 6.5, or if there's some other way of getting it.
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  2. #2
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    Quote Originally Posted by lllll View Post
    A soil scientist wants estimate the average pH for a large field by randomly selecting 40 core samples. Although the population standard deviation of pH is not know, past experiences indicates that most soils have a pH value between 5 and 8. Find the approximate probability that the sample mean will be within .2 units of the true average pH.

    $\displaystyle P\left( -\frac{0.2}{\sigma/\sqrt{40}} \leq \frac{\overline{Y}-\mu}{\sigma/\sqrt{n}} \leq \frac{0.2}{\sigma/\sqrt{40}}\right)$

    $\displaystyle P\left( -\frac{0.2}{\sigma/\sqrt{40}} \leq Z \leq \frac{0.2}{\sigma/\sqrt{40}}\right)$

    I'm just have trouble deciding on a $\displaystyle \sigma$, since it's between 5 and 8, I'm not sure if I should just take an average which is 6.5, or if there's some other way of getting it.
    "past experiences indicates that most soils have a pH value between 5 and 8." does not mean $\displaystyle 5 < \sigma < 8$. It means that 5 < Y < 8, more or less.

    Perhaps you could use the fact that for many distributions $\displaystyle \Pr(\mu - 3 \sigma < Y < \mu + 3 \sigma) \approx 1$. Then an estimate for $\displaystyle \sigma$ could be found by solving $\displaystyle \mu - 3 \sigma = 5$ and $\displaystyle \mu + 3 \sigma = 8$ for $\displaystyle \sigma$.

    A competent second opinion might confirm or deny this suggestion .....
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