- CFA Exams
- CFA Level I Exam
- Topic 1. Quantitative Methods
- Learning Module 3. Statistical Measures of Asset Returns
- Subject 2. Measures of Dispersion
CFA Practice Question
Variable X is distributed normally and has a mean of 10. If the probability that an observation of X will be negative is 0.16, what is the coefficient of variation of X?
A. 0.1
B. 1.0
C. 0.32
Explanation: The probability that X lies a distance of 10 below the mean is given as 0.16. A normal distribution is symmetric with regard to mean; this implies that the probability that X will be 10 greater than the mean is also 0.16. Thus, the probability that X lies between 0 and 20 is 1-0.16-0.16 = 0.68. For a normal distribution, 68% of the observations lie within one standard deviation of the mean. Hence, the standard deviation of X equals 10. The coefficient of variation is then equal to standard deviation/mean = 10/10 = 1.
User Contributed Comments 11
| User | Comment |
|---|---|
| chuong | Apply Z =(X-m)s => 1= (0-10)/s => s= -10 => s=10 => s/m=1 |
| whoi | The approach from mean/variance set to coefficient of variation: a. given information: P(X<0) = 0.16 and E(X) = 10 (Mean) b. calculation. Difference from Mean to 0 = 10 Difference from Mean to 20 = 10 P(X in the Range from 0 to 20) = 1 - P(X<0) - P(X>20) = 0.68 Because P(X in the Range from 0 to 20) = 68% = Mean +/- 1 STD the STD must be 10. c. Coefficient variation: Mean/STD = 0.10/0.10 =1 |
| bobert | Why use 20? is it just some benchmark number? |
| steved333 | because 20 is 10 above the mean of 10. The reason the range is 0-20 is because the probability referred to a negative value. A negative value is <0. Since the mean is 10, and 0 is 10 below the mean, you have to look at the same distance above the mean, as well to see what the confidence interval is. Since it's .68, you know based on the rules of a Bernoulli trial that 10 is the standard deviation since .68 represents one SD in either direction. |
| bdaguy | We are solving for: Pr(X<0) = 16% Adjust this to a standardized normal variable so that we can read off of a z-table: Pr(X<0) = Pr((X-10)/sigma < (0-10)/sigma)) = Pr(Z < -10/sigma) = 16% Therefore, from a z-table, we need to find the value in the left part of the distribution (known because Z < 0 given that sigma is a positive number) that equates to a probability of 16%. Since z-tables only provide the right-half of the distribution, look for Pr(Z>z) = 16%. This occurs where z = 1.0. To bring this back over to the left-side of the distribution, Pr(Z<-z) = Pr(Z<-1) = 16% Therefore, -10/sigma = -1 ==> sigma = 10 Co-efficient of variation = sigma/mu = 10/10 = 1 |
| aakash1108 | ...this is an awsome question. |
| Joel1980 | Or awesome. |
| HTale | I think people are overcomplicating this question. The mean is 10, therefore, if the probability of getting negative is 0.16, it must mean it is one standard deviation away from the mean. Therefore, the standard deviation is 10. This means that CV = SD/Mean = 10/10 = 1. That is how you would think about it under an exam situation, I wouldn't even do all of what was done above...and I'm a Math major :P |
| poomie83 | so if the mean was 20 would we be checking to see if X lies somewhere between 0 and 40? |
| bsm9 | This sucks. I hate this. I would rather be out with friends. I deserve better. This exam is a time waster; better to network/market. |
| edrei7 | For normal distributions, just take into heart the 68-95-99.7 rule. For any other distribution, the Chebyshev's theorem will hold true. |