A probability question from notes SS.2 reading 8

P=0.4, RB=0.5 P=0.6, RB=0 RB stands for return of stock B P is the probability associated with it. 1. What is the standard deviation of stock B? this SD seems different from the SD in reading 7 STAT. 2. How to use BAIIplus STAT function to do the calculation?

Variance can be expressed as : Var = E(RB^2) - E(RB)^2 where E(RB) is the expectation value of the return of stock B (ie, the mean), and E(RB^2) is the expectation value of the return of stock be squared. Ie, the mean of the square (NOT the square of the mean). Expectation value of any variable X is the sum over all values X can take of that particular value of X times the probability of finding X in that particular value. In our case there are only two cases, one with probability 0.4 and one with probability 0.6, so we can find the two expectation values as follows: E(RB) = 0.4 * 0.5 + 0.6 * 0 = 0.2 E(RB^2) = 0.4 * (0.5^2) + 0.6 * (0^2) = 0.1 Therefore the Variance is Var = 0.1 - 0.2^2 = 0.06 Standard Deviation is square root of variance, so one gets StdDev = 0.245 I’ve never used the BAII plus, so I wouldn’t know how to use the statistics functions on that, but this is a simple enough problem to just multiply and add and take the square root manually.

Another way: Var = .4 * (.5-.2)^2 + .6 (0-.2)^2 = .4*.09 + .6*.04 = .036+.024 = .06 sd = sqrt(…06) = .245 I am not sure if this can be solved with the BA II Plus calc. CP

cpk’s answer looked a bit easier for me. but i guess mathematically you were talking about the same equation. 3x to both of you.

Here is another method via the BA II Plus… Based on: P=0.4, RB=0.5 P=0.6, RB=0 think of a return of .5 occuring 4 times and a return of 0.0 occuring 6 times. In the [DATA] function enter .5 4 times and 0.0 6 times. The 1 variable [STAT] will display the std. dev.

I feel that it is a moral imperative for me to point out that the above solution, while correct, makes me gag.

Joey, out of curiousity, which of the several solutions above is the one that makes you gag?

Jack’s solution because it is not very general. In particular, if p = 0.162875439, it’s going to take jack a long time to do this problem on his BA II (but Jack seems to know the other way too).