A bell-shaped, symmetrical frequency distribution has a mean of 10. If 16% of the observations in the distribution are negative, what is the coefficient of variation of X? A: 1.0 B: 0.32 C: 10.0 D: 0.10
A. I can put down my work if you want.
please do mwvt9
yep answer is A. 1 sigma limit = 10 bell shaped. 16% negative means on a normal curve, -1 sigma point is 0 and below. 50% is on 1 side. (of the normal curve), and if 16% is below 0 – means 1 sigma = 10 therefore sigma = 10 cv = sigma / mu = 10/10 = 1.
Right. Since you know that 68% of all returns are within one standard deviation or sigma then you know that 32% of returns are still left. So you have 16% returns in each tail. Since you know 16% of returns in the left tail are negative, that means that one SD from the mean has to be a SD of 10. Then just calc CV as cpk123 did above.
These are the types of questions you will see on the exam. The calculations themselves are not hard at all. Many are done without a calculator, but you must understand concepts. Here they are testing two concepts: The basics of SD Formula for CV
i just read this and got scared… these are typical exam q’s for quant?
Answer is A… Took me like 5 minutes to do it though…but once I figured it out…it felt good
Does that problem look hard? BTW - It’s a typical Schweser question - everyone’s answer implies “bell-shaped symmetrical frequency distribution” is a normal probability density function. Of course it’s not because: a) a frequency distribution is a set of observations from a sample not a population distribution b) there are lots of distributions that can be described as bell-shaped and symmetrical that are not normal so the answer could be anything.
Suppose the question had stated that 23% of observations in the distribution are negative (instead of 16%). How would you solve the problem then?
P(X < 0) =0.23 P((X - mu)/sigma < -10/sigma) =0.23 P(Z < -10/sigma) = 0.23 From z-tables P(Z < -0.4090) = 0.23 => -10/sigma = -0.4090 => sigma =10/0.4090 = 24 (or something) sigma/mu = 24/10 = 2.4
Hey Joey, Iam mighty confused, Iam hoping you can help me out here. So here’s how I approached it, 23% of the values are -ve, so 77% are between 0 and +ve infinity. Looking in the body of the Z tables I located 0.7704, which relates to a z value of 0.74. Thus “my” Standard Deviation is 10/0.74 = 13.5, and CV = 13.5/10 = 1.35 What am I doing wrong?
Credit goes to Joey: For a value of x=0 (because we are looking at values below zero), we have: Z = x-u/sigma ==> Z= 0-u/sigma = -u/sigma So, P(Z <= -u/sigma) = 0.23 That corresponds to a z of 0.74, thus: -u/sigma = 0.74 = - 10/sigma = 0.74 ====> sigma = 10/0.74 = 13.51 CV = 13.51/10 = 1.35 Dreary
can someone please explain the answer to the second problem. I understood the first one. thanks,
JoeyDVivre Wrote: ------------------------------------------------------- > P(X < 0) =0.23 > P((X - mu)/sigma < -10/sigma) =0.23 > > P(Z < -10/sigma) = 0.23 > > From z-tables P(Z < -0.4090) = 0.23 > > => -10/sigma = -0.4090 > > => sigma =10/0.4090 = 24 (or something) > > sigma/mu = 24/10 = 2.4 oops - new z-tables > From z-tables P(Z < -0.74) = 0.23 > > => -10/sigma = -0.74 > > => sigma =10/0.74 = 13 (or something) > > sigma/mu = 13/10 = 1.3