4.1: Discrete Random Variables, the Discrete Uniform Distribution and the Binomial Distribution

A bond analyst is predicting the number of bond defaults that will occur in a mutual fund's portfolio of low-rated municipal bonds. The portfolio is comprised of 100 bonds that are rated between CCC and C. Historically the default rate on similarly-rated municipal bonds has been 17.5 percent. A two standard deviation interval around the historical default rate is closest to: A) 3.1 to 31.9 B) 8.7 to 35 C) 9.9 to 25.1

· Bond default is a binomial variable with mean (μ) equal to np, where n = 100, p(default) = 17.5%, and variance equal to np(1 − p) = 14.4375. So the standard deviation (σ) = 3.7997. Use the formula μ ± 2(σ) for the two standard deviation interval, the low end of the 95 confidence interval is 17.5 − 2(3.7997) = 9.9, and the high end is 17.5 + 2(3.7997) = 25.1.

Describe skewness for binomial distributions with probability of success =.5 ; >.5 ; <.5:

· Equal to .5 à Perfectly symmetric · Greater than .5 à Skewed left · Less than .5 à Skewed right

Given graph on back... What is the probability the stock will be $110?

· sequence containing two up moves and one down move is calculated as p2(1 − p) · .42(.6) = .096 · 3*.096=.288