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Math Help - So confused - Normal Distribution

  1. #1
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    So confused - Normal Distribution

    Hello all,

    This is a straight forward (or so I thought) textbook question. I'll quote it word for word so I don't leave anything out.

    You will need to look at this distribution of Reaction Times for the question:

    http://img.photobucket.com/albums/v6...tablegraph.jpg

    The question is then:

    "By simple counting, you can calculate exactly what percentage of the sample lies above or below +/- 1, 1.5, 2, 2.5, and 3 Standard Deviations using the tables/graphs. You can also calculate, from the normal distribution table, what percentage of scores would lie above or below those cutoffs if the distribution were perfectly normal. Calculate these values and consider the implications of using a normal distribution to approximate a distribution that is not normal. Where are you off by a quite a bit, and where are you reasonably acurrate? Consider this from the point of view of both one-tailed and two-tailed areas. How would your answers change if the sample had been very much larger or smaller?"

    The Mean of this distribution is: 60.26
    The Standard Deviation of this distribution is: 13.01
    N=300
    Distribution is obviously not normal.

    I have no idea what this question wants me to do exactly. If anyone can decipher the steps exactly I need to do to answer this question so I can receive full credit, I will forever be in your debt. I'm pretty sure I can manage the actual math behind it, but I just need clear instruction. You know its bad when your professor dislikes the text because of its poor wording. so again, a general outline of what I need to do to answer this fully will be much appreciated. If possible, if you can do an example of how this question needs to be answered using +/- 1SD, I'm sure I can follow the rest.

    Thank you so much in advance.

    James
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  2. #2
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    Quote Originally Posted by jamesmartinn View Post
    Hello all,

    This is a straight forward (or so I thought) textbook question. I'll quote it word for word so I don't leave anything out.

    You will need to look at this distribution of Reaction Times for the question:

    http://img.photobucket.com/albums/v6...tablegraph.jpg

    The question is then:

    "By simple counting, you can calculate exactly what percentage of the sample lies above or below +/- 1, 1.5, 2, 2.5, and 3 Standard Deviations using the tables/graphs. You can also calculate, from the normal distribution table, what percentage of scores would lie above or below those cutoffs if the distribution were perfectly normal. Calculate these values and consider the implications of using a normal distribution to approximate a distribution that is not normal. Where are you off by a quite a bit, and where are you reasonably acurrate? Consider this from the point of view of both one-tailed and two-tailed areas. How would your answers change if the sample had been very much larger or smaller?"

    The Mean of this distribution is: 60.26
    The Standard Deviation of this distribution is: 13.01
    N=300
    Distribution is obviously not normal.

    I have no idea what this question wants me to do exactly. If anyone can decipher the steps exactly I need to do to answer this question so I can receive full credit, I will forever be in your debt. I'm pretty sure I can manage the actual math behind it, but I just need clear instruction. You know its bad when your professor dislikes the text because of its poor wording. so again, a general outline of what I need to do to answer this fully will be much appreciated. If possible, if you can do an example of how this question needs to be answered using +/- 1SD, I'm sure I can follow the rest.

    Thank you so much in advance.

    James
    The +1 sd value is: \bar{x}+1 \times s=60.26+13.01=73.27, now count the number of items in the data less (or above) than this value. Now convert this into a proportion by dividing by 300. Compare this with the corresponding value for a standard normal distribution 0.8413.

    RonL
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  3. #3
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    Awesome thanks. Do you think its necessary I should compute all these values?

    Area above and below (1 SD)
    Area below and below (-1 SD)


    Is this question asking for both or just one?
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  4. #4
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    Quote Originally Posted by jamesmartinn View Post
    Awesome thanks. Do you think its necessary I should compute all these values?

    Area above and below (1 SD)
    Area below and below (-1 SD)


    Is this question asking for both or just one?
    For the normal distribution rhe areas are related (area below (-1 SD) is 1-area below (1 SD)). But you will need to count the cases below -1 SD as well as below +1 SD as the distribution is skew.

    RonL
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    K i think i understand, can i post my solutions later and you have a look at it?
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  6. #6
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    Quote Originally Posted by jamesmartinn View Post
    K i think i understand, can i post my solutions later and you have a look at it?
    Yes

    RonL
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    This will seem overly redundant, but apparently, thats how my prof likes it. Sorta sad that I'm in an advanced statistics class and still doing problems like this, without statistical packages no less... but anyway.

    The empirical distribution in the chart had these characteristics:
    M=60.26
    SD = 13.01
    N=300

    a) +1 SD = 60.26 + 13.01 (1) = 73.27

    Now looking at the distribution table, there are a total of 44 scores at and above 73.27. Therefore, 44/300 = 14.67% of the scores in this distribution lie above +1 SD

    -1 SD = 60.26 - 13.01 (1) = 47.25

    There are exactly 256 scores at and above 47.25. Therefore, 256/300 = 85.33% of the scores lie above - 1 SD.

    I did this for +/- 1.5, 2, 2.5, 3.0 SD's. Yeah, very tedious, but i suppose necessary.


    So the question is asking me to compare the percentage of the scores in this distribution which is skewed, to a standard normal distribution. So consulting my normal table, I find the following percentages.

    P (z > 1) = 15.87%
    P (z > - 1) = 84.13%


    I only feel like tying out that example, but does everything look okay so far? Is my reasoning okay? lol did I even do the question properly?!

    i'll summarize the rest of the stats i found:

    From my skewed distribution:

    8% of scores lie above +1.5 SD
    97.67% of scores lie above -1.5 SD

    4% of scores lie above +2 SD
    100% of scores lie above -2 SD

    1.33% of scores lie above +2.5 SD
    100% of scores lie above -2.5 SD

    1% of scores lie above +3 SD
    100% of scores lie above -3 SD.

    From my standard normal distribution table:

    15.87% of scores lie above +1 SD
    84.13% of scores lie above -1 SD

    6.68% of scores lie above +1.5 SD
    93.32% of scores lie above -1.5 SD

    2.28% of scores lie above +2 SD
    97.72% of scores lie above -2 SD

    0.62% of scores lie above +2.5 SD
    99.38% of scores lie above -2.5 SD

    0.13% of scores lie above +3 SD
    99.87% of scores lie above -3 SD

    So to answer this question further, it appears that the area that lies above negative standard deviations are similar in both distributions. I'm really lost when trying to finish the rest of this question off, i'd be greatful for some assistance/insight. Thanks.
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  8. #8
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    Quote Originally Posted by jamesmartinn View Post
    This will seem overly redundant, but apparently, thats how my prof likes it. Sorta sad that I'm in an advanced statistics class and still doing problems like this, without statistical packages no less... but anyway.

    The empirical distribution in the chart had these characteristics:
    M=60.26
    SD = 13.01
    N=300

    a) +1 SD = 60.26 + 13.01 (1) = 73.27

    Now looking at the distribution table, there are a total of 44 scores at and above 73.27. Therefore, 44/300 = 14.67% of the scores in this distribution lie above +1 SD

    -1 SD = 60.26 - 13.01 (1) = 47.25

    There are exactly 256 scores at and above 47.25. Therefore, 256/300 = 85.33% of the scores lie above - 1 SD.

    I did this for +/- 1.5, 2, 2.5, 3.0 SD's. Yeah, very tedious, but i suppose necessary.


    So the question is asking me to compare the percentage of the scores in this distribution which is skewed, to a standard normal distribution. So consulting my normal table, I find the following percentages.

    P (z > 1) = 15.87%
    P (z > - 1) = 84.13%


    I only feel like tying out that example, but does everything look okay so far? Is my reasoning okay? lol did I even do the question properly?!

    i'll summarize the rest of the stats i found:

    From my skewed distribution:

    8% of scores lie above +1.5 SD
    97.67% of scores lie above -1.5 SD

    4% of scores lie above +2 SD
    100% of scores lie above -2 SD

    1.33% of scores lie above +2.5 SD
    100% of scores lie above -2.5 SD

    1% of scores lie above +3 SD
    100% of scores lie above -3 SD.

    From my standard normal distribution table:

    15.87% of scores lie above +1 SD
    84.13% of scores lie above -1 SD

    6.68% of scores lie above +1.5 SD
    93.32% of scores lie above -1.5 SD

    2.28% of scores lie above +2 SD
    97.72% of scores lie above -2 SD

    0.62% of scores lie above +2.5 SD
    99.38% of scores lie above -2.5 SD

    0.13% of scores lie above +3 SD
    99.87% of scores lie above -3 SD

    So to answer this question further, it appears that the area that lies above negative standard deviations are similar in both distributions. I'm really lost when trying to finish the rest of this question off, i'd be greatful for some assistance/insight. Thanks.
    That looks good, but the discusion of the differences is not quite right.

    The proportion that is below -1.5, -2, ... SD's is much lower in the empirical
    distribution than in the normal, while that above 1.5, 2, .. SD's is greater in
    the empirical distribution compared with the normal.

    RonL
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  9. #9
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    I sorta arrived at a similar conclusion when i organized all the data in a chart so i could do a direct comparison with the empirical vs hypothetical distributions, so i'm glad im on track!

    i can't figure out how sample size factors into this all, can you comment on that?
    again, thank you for your help, it is very much appreciated
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