Assuming that the change in daily closing prices for stocks on the New York Stock Exchange is a random variable that is normally distributed with a mean of $.35 and a standard deviation of $.33. Based on this information, what is the probability that a randomly selected stock will close up $.75 or more?
The Correct Answer and Explanation is:
Answer: Approximately 11.3%
Explanation:
We begin with the assumption that the daily change in a stock’s closing price is normally distributed with a mean (μ) of $0.35 and a standard deviation (σ) of $0.33. Our goal is to calculate the probability that a stock closes up by $0.75 or more, which can be written as P(X ≥ 0.75).
To solve this, we first convert the problem to the standard normal distribution by using the Z-score formula:
Z = (X – μ) / σ
Here, X is the value of interest ($0.75). Plugging in the values, we get:
Z = (0.75 – 0.35) / 0.33 = 0.40 / 0.33 ≈ 1.21
This Z-score of approximately 1.21 tells us that $0.75 is about 1.21 standard deviations above the mean. The next step is to find the cumulative probability corresponding to Z = 1.21. Standard normal distribution tables or calculators show that P(Z < 1.21) is roughly 0.887. This cumulative probability represents the chance that a stock’s daily change is less than $0.75.
Since we need the probability for stocks closing at $0.75 or more, we take the complement of this value:
P(X ≥ 0.75) = 1 – P(Z < 1.21) ≈ 1 – 0.887 ≈ 0.113
Thus, there is about an 11.3% chance that a randomly selected stock will close up by $0.75 or more. This approach of converting to a standard normal variable is a fundamental technique in statistics, as it simplifies calculations by allowing the use of standardized tables or software. In summary, using the Z-score transformation and standard normal distribution properties, we determined that the probability is approximately 11.3%.
NYSE Stock Price Change Problem
Change in daily closing prices is normally distributed.
Mean (μ) = $0.35, Standard Deviation (σ) = $0.33.
Question: What is the probability that a stock closes up $0.75 or more?
Steps:
1. Compute Z-score:
Z = (0.75 – 0.35) / 0.33 ≈ 1.21
2. Find cumulative probability P(Z < 1.21) ≈ 0.887.
3. Therefore, P(X ≥ 0.75) = 1 – 0.887 ≈ 0.113 or 11.3%.
Final Answer: ~11.3%