HELP! Any statisticians out there?

hailzp
hailzp Posts: 903 Member
I am doing my stats assignment and I find the theory chapter very difficult.

The question is: The owner of a book store has analysed the sales data from customers over the last year and found that the mean amount of money spent by a customer is $21.45 with a standard deviation of $17.87.

a) Using only the statistical information above, briefly explain why X is not Normally distributed.

b) In fact, the distribution of X is strongly skewed. Would X be positively skewed or negatively skewed? Justify your answer.


PLEASE help me if you can. I am tearing my hair out :(
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Replies

  • Carl01
    Carl01 Posts: 9,307 Member
    It is a flashback to high school!! smiley-sad025.gif
  • hailzp
    hailzp Posts: 903 Member
    The **** news is, is this a uni course!!! I have to do it for psychology. :( My brain finds it very difficult to work this way.
  • xcrushx28
    xcrushx28 Posts: 182 Member
    I am doing my stats assignment and I find the theory chapter very difficult.

    The question is: The owner of a book store has analysed the sales data from customers over the last year and found that the mean amount of money spent by a customer is $21.45 with a standard deviation of $17.87.

    a) Using only the statistical information above, briefly explain why X is not Normally distributed.

    b) In fact, the distribution of X is strongly skewed. Would X be positively skewed or negatively skewed? Justify your answer.


    PLEASE help me if you can. I am tearing my hair out :(

    Well I only took one statistic class so I'm no expert, but that's a very huge standard deviation..... That's why its not normally distributed. Think if you added two or subtracted two standard deviations from your mean.

    Probably positively skewed and I don't have any reasoning for that.....

    Not sure if I'm helping at this point :) Google?
  • bigdawg025
    bigdawg025 Posts: 774 Member
    I really want to help... but I'd have to dig in my business stats book from last semester to give you a good answer. I know the standard deviation is extremely high in comparison to the mean meaning there are several extreme data in the sample used. This would be why it's not normally distributed.

    I don't have a clue about part b... because I would think the standard deviation could apply to either end... probably positively skewed, though.
  • hailzp
    hailzp Posts: 903 Member
    OK guys. Good so far, gives me a base to work from. I tried googling stuff but it is a bit difficult if your not really sure what you are looking for :frown:
  • xcrushx28
    xcrushx28 Posts: 182 Member
    So you could say that in relation to the mean the standard deviation is very high and thus cannot be normalized.

    Its positively skewed because the actual deviation should be much lower.
  • I think it has to do with the fact that the distribution is truncated at 0. In other words, nobody in the store's sales data will be recorded as spending negative dollars...thus the distribution will be heavily skewed to the left...the size of the standard deviation is not relevant for this question.
  • j4nash
    j4nash Posts: 1,719 Member
    Standard deviation is extremely high relative to the mean.. this would imply skewness.

    I don't think the negative or postiive skew can be derived without a sample as a positive skew and a negative skew can both have the same mean and standard deviation.

    but it's been 10 yrs.
  • zipnguyen
    zipnguyen Posts: 990 Member
    I am doing my stats assignment and I find the theory chapter very difficult.

    The question is: The owner of a book store has analysed the sales data from customers over the last year and found that the mean amount of money spent by a customer is $21.45 with a standard deviation of $17.87.

    a) Using only the statistical information above, briefly explain why X is not Normally distributed.

    b) In fact, the distribution of X is strongly skewed. Would X be positively skewed or negatively skewed? Justify your answer.


    PLEASE help me if you can. I am tearing my hair out :(


    Wow, really on MFP? Well Xbar is 21.45 with 1 std dev of 17.87. So if 99% of all oberservations fall within 3 std dev, then 17.87*3=upper and lower range of said 99%. Note how when you multiply 3*17.87 you get something close to $48. So how can sales of books be $21.45-$48= some negative number when supposedly all 99% of oberservations is supposed to occur within 3 std dev and assumed in business to be a positive number.

    Oh, btw the way, I have a Ph.D. In finance and BA and MA in applied economics and statistics. :bigsmile:

    Can't give you the answer but the hint above pretty much solved the problem for you... Of skewness, etc.
  • zipnguyen
    zipnguyen Posts: 990 Member
    You know how hard it is to type with thumbs...
  • nikolaim5
    nikolaim5 Posts: 233
    DOh!
  • hailzp
    hailzp Posts: 903 Member
    You know how hard it is to type with thumbs...

    (Yus, really, on MFP, I have no where else to go ): )

    I don't doubt your academic achievements! Thank you very much, helped a lot. :bigsmile:
  • zipnguyen
    zipnguyen Posts: 990 Member
    You know how hard it is to type with thumbs...

    (Yus, really, on MFP, I have no where else to go ): )

    I don't doubt your academic achievements! Thank you very much, helped a lot. :bigsmile:

    Good luck.
  • luv_lea
    luv_lea Posts: 1,094 Member
    The answer is C, always choose C. Or, at least that's what I always did in high school when I didn't know the answer.
  • significance
    significance Posts: 436 Member
    What percent of customers would have to have a negative spend (less than zero $) if it were normally skewed, with that standard deviation?
  • nnylee
    nnylee Posts: 811 Member
    I'm not a statistician or anything near to that, but I think it's so cool that you actually got answers here.
  • caveats
    caveats Posts: 493 Member
    Wow, really on MFP? Well Xbar is 21.45 with 1 std dev of 17.87. So if 99% of all oberservations fall within 3 std dev, then 17.87*3=upper and lower range of said 99%. Note how when you multiply 3*17.87 you get something close to $48. So how can sales of books be $21.45-$48= some negative number when supposedly all 99% of oberservations is supposed to occur within 3 std dev and assumed in business to be a positive number.

    Oh, btw the way, I have a Ph.D. In finance and BA and MA in applied economics and statistics. :bigsmile:

    Can't give you the answer but the hint above pretty much solved the problem for you... Of skewness, etc.

    Damn, I could've used your "help" last week at work when we were compiling reports for a client in Excel since we're waiting on a new SPSS license from corporate. >_<
  • karagav
    karagav Posts: 172 Member
    this scares me...i'm just starting the stats portion of my research and statistics course...i ended the research methods term with a 93 and nottttt so sure this semester is going to be such a happy ending!! haha
  • becoming_a_new_me
    becoming_a_new_me Posts: 1,860 Member
    I am doing my stats assignment and I find the theory chapter very difficult.

    The question is: The owner of a book store has analysed the sales data from customers over the last year and found that the mean amount of money spent by a customer is $21.45 with a standard deviation of $17.87.

    a) Using only the statistical information above, briefly explain why X is not Normally distributed.

    b) In fact, the distribution of X is strongly skewed. Would X be positively skewed or negatively skewed? Justify your answer.


    PLEASE help me if you can. I am tearing my hair out :(


    Wow, really on MFP? Well Xbar is 21.45 with 1 std dev of 17.87. So if 99% of all oberservations fall within 3 std dev, then 17.87*3=upper and lower range of said 99%. Note how when you multiply 3*17.87 you get something close to $48. So how can sales of books be $21.45-$48= some negative number when supposedly all 99% of oberservations is supposed to occur within 3 std dev and assumed in business to be a positive number.

    Oh, btw the way, I have a Ph.D. In finance and BA and MA in applied economics and statistics. :bigsmile:

    Can't give you the answer but the hint above pretty much solved the problem for you... Of skewness, etc.

    You have a PhD aaaaaand you look like that? Well, you can solve my problems anytime. :blushing: :laugh:
  • hailzp
    hailzp Posts: 903 Member
    Well guys, you have all been outstanding. Assignment completed. I may have to come back to the forums next week :/