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Math Help - [SOLVED] CLT- need help with a problem...

  1. #1
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    [SOLVED] CLT- need help with a problem...

    Hello to any kind and wise soul(s) who might be able to help me with this problem!

    My question involves this the Central Limit Theorem- and what do I do with the age 20 in this problem? I'm sure I need to enter this into my equation, but I'm just not sure how or where.

    Here's the problem...

    According to the NCHS (National Center for Health Statistics), the average height of men over the age of 20 is 69.1 inches with a standard deviation of 5.3 inches. Assume the population distribution is normal.
    a. What is the probability that a single randomly selected individual man will be shorter than 69 inches?

    b. What is the probability that the mean height of 19 randomly selected men will be shorter than 69 inches?

    Here's how I came about my wrong answers:
    a. n= 1 x= 69 sigma= 5.3 mu= 69.1

    Z= x - mu / (sigma/ quare root of n) So...

    Z = 69 - 69.1 / (5.3 / 1) = -.0189 = .4286 or 43%

    And I did the same process for b., but n = 19.

    What did I do wrong? Thanks so much to whoever can help!
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  2. #2
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    Part a:

    z=\frac{x-{\mu}}{\sigma}=\frac{69-69.1}{5.3}\approx{-0.019}

    Look this up in the table and we see the corresponding value is 0.4925

    or 49.25% probability.

    part b:

    This is a t-distribution, since n < 30.

    t=\frac{(x-{\mu})\sqrt{n}}{\sigma}

    \frac{(69-69.1)\sqrt{19}}{5.3} = -0.0822

    This corresponds to 0.4677 or 46.77%
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  3. #3
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    Hi Galactus- Thanks so much- I looked it up on the table wrong! I'm so impressed with your help- can't thank you enough!

    Can I pick your brain on another one? This is also CLT and what I lovingly call a "many, many" problem. We did these recently, but they required a slightly different equation. I've emailed my question on this to my prof, but haven't heard back. Maybe you'll know?

    Here's an example:

    Percent of students who were in the Top Quarter of High School Class 41%.

    If we take many,many samples of 125 students 90% of the time the sample percentage of those who were in the top quarter of high school class will be between____ and_____ .

    The way I would've done this is:

    X= mu +- Z(sigma/n)

    But I'm not sure what the values are. I know Z= 1.96 and n=125. I'm stuck as to where to go from there. I feel like this involves p, but I've managed to really confuse myself!

    Thanks a million!
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  4. #4
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    I've figured it out! (Well, guess I can't take all the credit- my prof emailed and told me where to find the equation I need to use! :0)

    p^ = p +- Z * square root of [p(1-p) / n]

    :0)
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