Q. An investment strategy has an expected return of 12 percent and a standard deviatio of returns of 10 percent. If investment returns are normally distributed, the probability of earning a return less than 2 percent is closest to:
A. 10%
B. 16%
C. 32%
The answer is B and the explanation as follows: Approximately 68 percent of the returns fall within one standard deviation (plus or minus) of the mean:
(100%-68%) / 2 = 16%
I don’t get it. How do you know whether 68% of the returns fall within one standard deviation from the information given above and…
i just don’t get both question and explanation would anyone else please help me or just tell me which page I should read either in the curriculum or Schweser notes.
For a standard normal distribution 68% of values lie within +- 1 standard deviation of the mean - i think this is a fact you should know. (95% lie within approx 2 sd of the mean) Therefore 32% of values lie outside of +- 1 standard deviation of the mean ie 16% in each tail because a normal distribution is symmetrical. In this case they want to know what is the probability of a return of less than 2 which is the probability of being in the lower tail which is 16%
the answer above is a bit confusing… according to me the answer is as we know z value = x - arithmetic mean / standard deviation so we will get 12-2/ 10 = 1 so the % age is 68 as the question said the profitability of getting less than 2 is 16 ( 68% is the confidence interval and the values lie outside is 100 -68 =32% and since it is 2 tailed is 16 … i hope this helps…
Here is a little picture that I drew to help to illustrate this problem better. Please take a look: http://www.valtinho.com/cummulative_standard_normal.gif Here is something which might be useful for you to remember: The z-score is “the number of standard deviations a given observation is from the population mean.” So, z-score = (2 - 12) / 10 z-score = -1std So what you are looking for is the probability of the shaded area of the picture. You need to use common sense to arive at the answer, because you are given very few details in the problem. In this case, you need the left tail of the distribution (probability UNDER 2%). So, you need to memorize the z-score formula and a few of the most commonly used z-scores (you are not likely to be given a z-score table in the exam). The commonly used z-scores are (std is my abbreviation for standard deviation): 1std = ±68% 1.65std = ±90% 1.96std = ±95% and 2.58std = ±99%
