Question

11. Based on very old but actual figures, the average price of stocks in the S & P 500 index is 30 $ with a standard deviation of $ 8.20. Suppose that the prices of these stocks are a normally distributed random variable.

a) What is the probability that a stock price is at least $ 40?
b) What is the probability that a stock price is less than $ 20?
c) What is the minimum price of a stock that places it in the most valuable 10 % of stocks?

Answer 1

Let’s assume that “X” be the “stock price” follows normal distribution with parameter “μ=30” and “σ=8.2”.

So, the probability that the stock price is at least “$40”, is given below.

=> P(X ≥ 40) = 1 – P(X ≤ 40) = 1 – P[(X-μ)/σ ≤ (40-30)/8.2] = 1 – P[t ≤ 1.22], where “t = (X-μ)/σ “, follows standard normal distribution.

=> P(X ≥ 40) = 1 – Φ(1.22), where Φ(1.22) = P[t ≤ 1.22], form the “standard normal table we will find out the value of “Φ(1.22)”.

=> P(X ≥ 40) = 1 – Φ(1.22) = 1 – 0.8887676 = 0.1112324 = 0.1112, => P(X ≥ 40) = 0.1112, be the required probability.

b).

Now, the probability that a stock price is less than “$20” is given below.

=> P(X < 20) = P[(X-μ)/σ ≤ (20-30)/8.2] = P[t ≤ (-1.22)], where “t = (X-μ)/σ “, follows standard normal distribution.

=> P(X < 20) = P[t < (-1.22)] = P[t > 1.22] = 1 – P[t ≤ 1.22] = 1 – Φ(1.22) = 1 – 0.8887676 = 0.1112324 = 0.1112, => P(X < 20) = 0.1112, be the required probability.

c).

Now, we need to find out the minimum price of a stock that places it in most valuable “10%” of the stock. So, given the statement the following condition should hold.

=> P(X ≥ X0) = 10% = 0.1, where “X0” be the minimum price of most valuable stock.

=> P(X ≤ X0) = 90% = 0.9, => P[(X-μ)/σ ≤ (X0 – 30)/8.2] = 0.9, => Φ[(X0 – 30)/8.2] = 0.9 = Φ(1.29).

=> (X0 – 30)/8.2 = 1.29, => X0 – 30 = 1.29*8.2 = 10.578, => X0 = 30 + 10.578 = 40.578 = 40.58.

=> “X0 = 40.58”, be the minimum price of the most valuable 10% stock.