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Math Help - Normal distribution with population knee heights

  1. #1
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    Normal distribution with population knee heights

    Given: Men have sitting knee heights that are normally distributed with mean of 22.0 in. and std dev of 1.1 in. Women have mean of 20.3 in and std dev of 1.0 in.

    Question: Find min and max sitting knee heights that include at least 95% of all men and at at least 95% of all women.

    What I did: I originally took the avg of the means given and the avg of std devs given and using invNorm found the min and max using .025 from either side of the curve. Min: 19.10 in and Max: 23.21 in. After testing these limits with the men's average I found that I was cutting out 20.9% of the men, so obviously I did this wrong. NOTE: I also found 95% max and min for each gender, but then didn't know what to do with this information.

    What should I have done to find the answers?

    Thanks.
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    Quote Originally Posted by stats08 View Post
    Given: Men have sitting knee heights that are normally distributed with mean of 22.0 in. and std dev of 1.1 in. Women have mean of 20.3 in and std dev of 1.0 in.

    Question: Find min and max sitting knee heights that include at least 95% of all men and at at least 95% of all women.

    What I did: I originally took the avg of the means given and the avg of std devs given and using invNorm found the min and max using .025 from either side of the curve. Min: 19.10 in and Max: 23.21 in. After testing these limits with the men's average I found that I was cutting out 20.9% of the men, so obviously I did this wrong. NOTE: I also found 95% max and min for each gender, but then didn't know what to do with this information.

    What should I have done to find the answers?

    Thanks.
    To get the minimum for men you require the value of x_{\min} such that \Pr(X < x_{\min}) = 0.025.

    Note that z^* = \frac{x_{\min} - 22.0}{1.1} where \Pr(Z < z^*) = 0.025 \Rightarrow z^* = -1.96.

    Then -1.96 = \frac{x_{\min} - 22.0}{1.1} \Rightarrow x_{\min} = 19.84.

    In a similar way I get x_{\max} = 24.16.

    Do the same thing for the women.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    To get the minimum for men you require the value of x_{\min} such that \Pr(X < x_{\min}) = 0.025.

    Note that z^* = \frac{x_{\min} - 22.0}{1.1} where \Pr(Z < z^*) = 0.025 \Rightarrow z^* = -1.96.

    Then -1.96 = \frac{x_{\min} - 22.0}{1.1} \Rightarrow x_{\min} = 19.84.

    In a similar way I get x_{\max} = 24.16.

    Do the same thing for the women.
    Is there some confusion here?

    Are we looking for two knee height intervals one containing at least 95% of men and the other 95% of women, or one interval which contains at least 95% of men and 95% of women simultaneously?

    Working this numerically I find that the smallest interval symmetric about the mean knee height of men and women combined is approximately [18.48, 23.82] inches. This is the larger of the two symmetric intervals that contain 95% of womens knees centred at the grand mean and that contain 95% of mens knees centred at the grand mean.

    The interval given contains about 95% of mens knees and 96.5% of womens knees.

    (of course if we had chossen to centre the interval at another point we would have found a different result, now the question is what is the smallest interval that meets our requirements? I'm pretty sure that the interval found here is not the smallest that will do the job)

    The shortest interval meeting our requirements is approximatly [18.65, 23.82] which contains about 95% of womens knees and 95% of mens knees. This was found by brute force by computing the probability of a range of intervals for men and women plotting the 95% contours and keeping the point at which the contours cross!

    CB
    Last edited by CaptainBlack; November 19th 2008 at 04:47 AM. Reason: to add the latest numerical results
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    Population Knee Heights

    You are correct, Captain, I need to find common max and min that will contain 95% of all men and all women combined (as if I'm designing a carseat).

    When you say you found a smaller interval using "brute force by computing the probability of a range of intervals", can you give an example on how to even pick an interval to try and how you know when contours cross?

    I was aware that some guessing is involved, but I'm not sure if it is lack of sleep or lost brain cells, but I'm not sure where to begin guessing outside of givens.

    Thanks.
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    Quote Originally Posted by stats08 View Post
    You are correct, Captain, I need to find common max and min that will contain 95% of all men and all women combined (as if I'm designing a carseat).

    When you say you found a smaller interval using "brute force by computing the probability of a range of intervals", can you give an example on how to even pick an interval to try and how you know when contours cross?

    I was aware that some guessing is involved, but I'm not sure if it is lack of sleep or lost brain cells, but I'm not sure where to begin guessing outside of givens.

    Thanks.
    Depend on what software you have available. This can be done approximatly in Excel (not that I used Excel).

    Take a rectangular region with the interval half width down one edge, and the centre along the bottom edge. In each cell compute the proportion of mens knee heights that fall in the interval defined by the figures in the two edges with the interval parameters.

    Repeat in another rectangular region but this time for women's knee heights.

    In a third rectangular region record if the corresponding cells in the first two regions are greater than 0.95, select the cell with the with both greater than 0.95.

    CB
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    Nice! Thank you. I'll try it out.
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    Quote Originally Posted by CaptainBlack View Post
    Is there some confusion here?

    Are we looking for two knee height intervals one containing at least 95% of men and the other 95% of women, or one interval which contains at least 95% of men and 95% of women simultaneously?

    Working this numerically I find that the smallest interval symmetric about the mean knee height of men and women combined is approximately [18.48, 23.82] inches. This is the larger of the two symmetric intervals that contain 95% of womens knees centred at the grand mean and that contain 95% of mens knees centred at the grand mean.

    The interval given contains about 95% of mens knees and 96.5% of womens knees.

    (of course if we had chossen to centre the interval at another point we would have found a different result, now the question is what is the smallest interval that meets our requirements? I'm pretty sure that the interval found here is not the smallest that will do the job)

    The shortest interval meeting our requirements is approximatly [18.65, 23.82] which contains about 95% of womens knees and 95% of mens knees. This was found by brute force by computing the probability of a range of intervals for men and women plotting the 95% contours and keeping the point at which the contours cross!

    CB
    My mistake. Yes there was confusion.

    However, I'll note that if you calculate seperate 95% intervals for men and for women you get [19.84, 24.16] and [18.34, 22.16] respectively.

    Then an interval (not the shortest) that contains at least 95% of both men and women is clearly [18.34, 24.16] which is not too much longer than [18.65, 23.82]. It would only add a small number of extra dollars to the total production cost of the car seat.

    And at the current rate of obesity, the interval [18.65, 23.82] will probably be obselete (or should I say obese-lete) before the first car seats come off the production line.
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  8. #8
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    Quote Originally Posted by mr fantastic View Post
    My mistake. Yes there was confusion.

    However, I'll note that if you calculate seperate 95% intervals for men and for women you get [19.84, 24.16] and [18.34, 22.16] respectively.

    Then an interval (not the shortest) that contains at least 95% of both men and women is clearly [18.34, 24.16] which is not too much longer than [18.65, 23.82]. It would only add a small number of extra dollars to the total production cost of the car seat.

    And at the current rate of obesity, the interval [18.65, 23.82] will probably be obselete (or should I say obese-lete) before the first car seats come off the production line.
    It would depend on the marginal costs of providing a greater range (including consideration of the profit margin on the product). It is often supprising how large the impact of a 12.5% increase in some parameter in a product design can sometimes be (often because of their impact on other aspects of design).

    I know of at least one product with some design parameters that could not be increased by even 5% without an additional development cost in the order of 10's if not 100's of millions of pounds!

    CB
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  9. #9
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    Quote Originally Posted by mr fantastic View Post
    And at the current rate of obesity, the interval [18.65, 23.82] will probably be obselete (or should I say obese-lete) before the first car seats come off the production line.
    I always knew things were different in Oz. In the Northern Hemisphere the obese become wider not taller!

    CB
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