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I have following investment data: mean-8 standard deviation 15 Assume that if it is normal distribution, what is the probability that returns will exceed 23%. How to calculate this?

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2 Answers 2

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This is just a simple normal distribution with z-scores. We can already calculate the z-score easily. The z-score is modeled as:

$$z=\frac{\bar{x}-\mu}{\sigma}$$

In this case, $\bar{x}=23\%$ because this is the statistic you want to learn about. $\mu$ is the mean which in this problem is $8\%$, and $\sigma$ is the standard deviation which is also given to be $15\%$. For this case, I'll just treat them all as numbers instead of percents (easier and doesn't matter for this case). Plug it all in:

$$z=\frac{23-8}{15}=1$$

The z-score is an indication of how many standard deviation away you are from the mean. In this case, it turned out to be perfectly $1$ standard deviation. From the Empirical Rule (AKA "68-95-99.7"), you know that between $z=-1$ and $z=1$ you can find 68% of the data. The data to the right of that interval will be $\frac{1-0.68}{2}=16\%$ of the population. This is how I got it:

enter image description here

The total area under a curve is 1 (that's just definition). From the Empirical Rule, 68% of the distribution falls between 1 standard deviation to the left and 1 standard deviation to the right of the mean. The leftover 32% is split into two areas - one to the right of $z=1$, and another equal one to the left of $z=-1$. So you divide the total area, $32\%$, by 2 to get what the area when $z>1$ (when the returns exceed $23\%$, in this problem).

When the z-scores aren't so pretty, you will need to use your calculator's normalcdf() or maybe a z-score table. normalcdf() is very convenient since you don't even need to calculate the z-scores in most cases (there's an overloaded method in which you can put the actual values).

Using the calculator, we see that normalcdf(1, E99)=normalcdf(23, E99, 8, 15)=15.8655% which is about $16\%$ (Empirical Rule is an estimation). The answer is therefore $\approx 15.8655\%$ but I'm sure your teacher expects you to use the Empirical Rule and get $16\%$.

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  • $\begingroup$ Please, can you provide illustration! This looks very good explained! :) $\endgroup$
    – truе
    Apr 3, 2014 at 22:47
  • $\begingroup$ @EvaSpring Included. Ignore the 95%/99.7%, they're irrelevant to this problem. $\endgroup$
    – Shahar
    Apr 3, 2014 at 22:54
  • $\begingroup$ OK, it is more clear to me now. But what for example if they wanted to calculate probability between 23% and lets say 54%? How that can be done? $\endgroup$
    – truе
    Apr 3, 2014 at 23:13
  • $\begingroup$ @EvaSpring Then you use normalcdf() or z-score table. $\endgroup$
    – Shahar
    Apr 3, 2014 at 23:46
  • $\begingroup$ OK, but we use cumulative for both or not? Because cumulative means only values less than or equal to x. $\endgroup$
    – truе
    Apr 3, 2014 at 23:53
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Hint: Presumably your mean and standard deviations are measured in percent to match the question. You are asking for how much of a normal distribution is more than one standard deviation above the mean. Do you know how to use a Z score table? It has the data you need.

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  • $\begingroup$ Hmm.... Yep! But Z-score is a measure for only particular numbers? But your approach is waking up my gray cells now... $\endgroup$
    – truе
    Apr 3, 2014 at 22:03
  • $\begingroup$ Can you elaborate on this? OK, I have a normal distribution here, how to find what is required? :) $\endgroup$
    – truе
    Apr 3, 2014 at 22:14
  • $\begingroup$ If you were asked the chance that returns exceed $8\%$, the answer would be $50\%$ because half the normal distribution is above the mean and half below. Do you know how much of a normal distribution is within one standard deviation of the mean? $\endgroup$ Apr 3, 2014 at 22:19

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