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I am working on a self-study question where

A study indicates that the typical American woman spends USD 340 per year for personal care products. The distribution of the amount follows a right-skewed distribution with a standard deviation of USD 80 per year. If a random sample of 100 women is selected, what is the probability that the sample mean of this sample will be between USD320 and USD350?

I made two attempts to answer the question: one using normal distribution and another using CLT. However, neither approach has helped me achieve the goal answer of 0.8882.

My Normal Distribution Approach

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My CLT Approach

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Appreciate some guidance and advice please

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  • $\begingroup$ Please correct the typos in your question. $\endgroup$
    – Glen_b
    Commented Apr 28, 2014 at 8:09
  • $\begingroup$ As per your drawing, "right skewed" means "positively skewed", i.e. that the right tail is longer and probability mass concentrates to the left of the density graph. $\endgroup$
    – Alecos Papadopoulos
    Commented Apr 28, 2014 at 10:16
  • $\begingroup$ @Glen_b I've corrected the question. Do let me know if there's any more corrections needed. $\endgroup$
    – user1275515
    Commented Apr 28, 2014 at 11:46

2 Answers 2

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The normal distribution approach given is failing to take into account how the standard deviation of the sample mean decreases with the sample size.

The CLT approach is looking good until you evaluate H(1.25). Draw a picture of the area you are trying to calculate and try to figure out what is wrong. Would it make sense to say that the area to the left of 1.25 was .1056?

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  • $\begingroup$ thanks for the pointer. I get the hint. Since the question states that the distribution is right-skewed, therefore, we can't assume that the left and right area are the same. $\endgroup$
    – user1275515
    Commented Apr 28, 2014 at 11:51
  • $\begingroup$ That's not the only problem with that approach! You have simply got to stop throwing random calculations at these problems until you reproduce an answer. That's not how to understand what you're doing. You need to consider how to do the calculations that relate to the question. So if it's asking about the distribution sample averages, and your calculation doesn't consider the distribution sample averages, your answer can't be right. Since your first attempt didn't deal with the distribution of averages, it's a complete waste of time to even consider it. $\endgroup$
    – Glen_b
    Commented Apr 28, 2014 at 11:53
  • $\begingroup$ jsk's first paragraph above is pointing out that in your first attempt you're not dealing with the distribution of averages. You need to ponder the error in ignoring the difference between a distribution for an individual and the distribution of some calculation based on more than one individual carefully, because you've made that error more than once, even after that type of error has already been pointed out before. Since the two approaches must give different answers, rather than do both and see if one was right, figure out which one cannot be right before you start. $\endgroup$
    – Glen_b
    Commented Apr 28, 2014 at 12:01
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    $\begingroup$ Thanks for the advice Glen. Will make more conscious attempts to improve my problem solving methods. I am confident I will be able to continually improve and contribute to this awesome community which has really helped me a lot in my quest of self-learning stats. $\endgroup$
    – user1275515
    Commented Apr 28, 2014 at 12:02
  • $\begingroup$ @Glen_b Im complete clueless as how to determine the Z-value for measuring skewness. I was wondering if sites.google.com/site/fundamentalstatistics/appendix-b is something along the lines that I can look into? $\endgroup$
    – user1275515
    Commented Apr 28, 2014 at 12:05
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Thank you for the support and advice Glen & JSK.

I'm managed to work out the answer which I thought I share with everyone (and also get some pointers if I had any steps wrong despite getting to the right answer).

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