Question

Bhola Bhikhu is thinking about adding a new stock to her portfolio. Based on advice from a friend who claimed to make a lot of money, she decides to use the price to earnings ratio (P/E) as the only measurement of the performance of a stock. Bhola selects several possible stocks based on this measurement and will select one stock from the bundle. She assigns each of the selected stocks an equal chance of being selected. Stock one has a price to earnings ratio of 20, Stock two has an P/E of 22, Stock three has a P/E of 20, Stock four has a P/E of 18, and Stock five has a P/E of 16, and Stock six has a P/E of 20. Let ? denote the random variable representing the price to earnings ratio for the selected stock.

a) Create a PMF table for the random variable ?.

b) What is the probability that Bhola selects a stock with a P/E greater than 17?

c) Given that Bhola selects a stock with a P/E of at least 19, what is the probability that the P/E is over 20?

d) Given that Bhola selects a stock with a P/E greater than 17, what is the probability that the P/E is at most 22?

e) Find the expected value of ? and standard deviation of �

Answer 1

a)

There are total 6 stocks with P/E values, 20, 22, 20 , 18, 16, 20

Since, each of the selected stocks an equal chance of being selected, the PMF table for the random variable ? is,

P(X = 16) = 1/6

P(X = 18) = 1/6

P(X = 20) = 3/6 = 1/2

P(X = 22) = 1/6

b)

Probability that Bhola selects a stock with a P/E greater than 17 = P(X > 17) = P(X = 18) + P(X = 20) + P(X = 22)

= (1/6) + (1/2) + (1/6)

= 5/6

c)

Given that Bhola selects a stock with a P/E of at least 19, the probability that the P/E is over 20 = P(X > 20 | X 19)

= P(X > 20 and X 19) / P(X 19)

= P(X > 20) / P(X 19)

= {P(X = 22)} / {P(X = 20) + P(X = 22)}

= (1/6) / (1/2 + 1/6)

= 1/4

d)

Given that Bhola selects a stock with a P/E greater 17, the probability that the P/E is at most 22 = P(X 22 | X > 17)

= P(X 22 and X > 17) / P(X > 17)

= {P(X = 18) + P(X = 20) + P(X = 22)} / {P(X = 18) + P(X = 20) + P(X = 22}

= 1

e)

Expected value of X is,

E[X] = (1/6) * 16 + (1/6) * 18 + (1/2) * 20 + (1/6) * 22 = 19.33333

Now,

E[X²] = (1/6) * 16² + (1/6) * 18² + (1/2) * 20² + (1/6) * 22² = 377.33

Var(X) = E[X²] - E[X]²  = 377.33 - 19.33²   = 3.6811

Standard deviation of X = = 1.9186