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:
The probability that a randomly selected stock will close up by $0.75 or more is 0.1127 (or 11.27%).
Explanation:
The problem states that the daily changes in stock closing prices on the New York Stock Exchange follow a normal distribution with:
- Mean (μ\mu) = $0.35
- Standard deviation (σ\sigma) = $0.33
- Target value (XX) = $0.75
To find the probability P(X≥0.75)P(X \geq 0.75), we calculate the Z-score: Z=X−μσ=0.75−0.350.33=0.400.33≈1.21Z = \frac{X – \mu}{\sigma} = \frac{0.75 – 0.35}{0.33} = \frac{0.40}{0.33} \approx 1.21
Using the standard normal table, the cumulative probability for Z=1.21Z = 1.21 is 0.8873. Since we need P(X≥0.75)P(X \geq 0.75), we take the complement: P(X≥0.75)=1−P(Z≤1.21)=1−0.8873=0.1127P(X \geq 0.75) = 1 – P(Z \leq 1.21) = 1 – 0.8873 = 0.1127
This means there is an 11.27% chance that a randomly selected stock will have a closing price increase of at least $0.75.
Interpretation:
- Most stock price changes will be close to the mean ($0.35).
- The probability of a price increase of $0.75 or more is relatively low.
- The probability is visualized in the normal distribution graph, where the shaded red area represents the portion of stocks closing above $0.75.
This result can help investors understand the likelihood of significant stock price movements based on historical volatility.