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:
To find the probability that a randomly selected stock will close up \$0.75 or more, we need to calculate the probability that a value from a normal distribution with:
- Mean (μ) = \$0.35
- Standard deviation (σ) = \$0.33
is greater than or equal to \$0.75.
Step 1: Use the Z-score formula
The Z-score formula is:
$$
Z = \frac{X – \mu}{\sigma}
$$
Where:
- $X$ is the value we are interested in (\$0.75)
- $\mu$ is the mean (\$0.35)
- $\sigma$ is the standard deviation (\$0.33)
$$
Z = \frac{0.75 – 0.35}{0.33} = \frac{0.40}{0.33} \approx 1.21
$$
Step 2: Use the Z-score to find the probability
Now, we use the standard normal distribution table (Z-table) or a calculator to find the probability that Z is greater than 1.21.
From the Z-table:
$$
P(Z < 1.21) \approx 0.8869
$$
So,
$$
P(Z > 1.21) = 1 – 0.8869 = 0.1131
$$
Final Answer:
$$
\boxed{0.1131} \text{ or } \boxed{11.31\%}
$$
Explanation (300+ words):
This problem is a classic example of using the normal distribution to find probabilities related to real-world data. Here, we are told that the change in daily closing stock prices is normally distributed — this implies that most changes will cluster around the mean (\$0.35), with fewer observations occurring as we move further away from this average.
In statistics, when data follow a normal distribution, we can standardize any value using the Z-score, which tells us how many standard deviations a particular value is from the mean. In this case, we are asked to find the probability that a stock will close up by \$0.75 or more. First, we calculate how far \$0.75 is from the mean (\$0.35), measured in standard deviations. The result, a Z-score of approximately 1.21, means that \$0.75 is 1.21 standard deviations above the mean.
To find the probability of this occurring, we look up the Z-score in the standard normal distribution table, which provides the probability that a value is less than a given Z-score. Because we want the probability of being greater than \$0.75, we subtract that value from 1.
The final result, 0.1131 or 11.31%, tells us that there’s about an 11.31% chance that a randomly selected stock will have a price increase of \$0.75 or more on a given day. This relatively low probability reflects the fact that \$0.75 is significantly above the average change of \$0.35, making it a less common outcome.