# Normal Distribution Question

An investment strategy has an expected return of 12% and a standard deviation of 10%. If investment returns are normally distrubuted, the prob. of getting a return less than 2% is closest to: A) 10% B) 16% C) 32% D) 34%

B Probability of observation lying more than 1 std. deviation away from mean is 32% (1 - .68). 32%/2 = 16% Divide by 2 because you want to know probability of lying beneath 1 std deviation away from mean… not above and beneath. Make sense?

Edit: I’m a moron.

How do you know automatically that its 1 std. deviation from the mean?

The given mean is 12 and the given std. deviation is 10. They are inquiring about 2. 12 - 10 = 2 2 is exactly 1 std. deviation (10) away from the mean (12). Does that help? Not sure I understood your question clearly.

but how do you know 2 is exactly 1 std. deviation (10) away from the mean (12)? thats where i am getting confused?

I’m still not following your questions. Perhaps you are getting confused about the mean? The expected value (12) is the mean. The mean is given… it is 12. The std. dev is given… it is 10. Mean - 1*Std Dev = 2 12 - 1*(10) = 2 There is a rule of thumb for normal distributions. 68% fall within 1 std. deviation 95% fall within 2 std. deviations 99% fall within 3 std. deviations P(x > than 1 std. deviation away from mean) = 1 - .68 = .32 P(x is > than 1 std. BELOW mean) = .32 / 2 = .16 Can you try to be more detailed in your question?

I think you need to find the z-score: z= observation - pop mean / std deviation z= 2-12/10 = -10/10 = -1 z-value = -1 we need to find the prob that z < -1, since we’re looking for the probability of a return of less than 2% z score of -1 = .1587, which is the area under the curve to the left of this so 16% probability that you get a return less than 2%

rbford - I believe you and I did the same thing. My method is approximating your precision.

i finally get it. Thank you everryone !