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The price of shares of Bank of Florida at the end of trading each day for the last year followed the normal distribution. Assume there were 240 trading days in the year. The mean price was \(\$ 42.00\) per share and the standard deviation was \(\$ 2.25\) per share. a. What is the probability that the end-of-day trading price is over \(\$ 45.00 ?\) Estimate the number of days in a year when the trading price finished above \(\$ 45.00 .\) b. What percent of the days was the price between \(\$ 38.00\) and \(\$ 40.00 ?\) c. What is the minimum share price for the top \(15 \%\) of end-of-day trading prices?

Short Answer

Expert verified
a) 22 days; b) 14.92%; c) $44.34.

Step by step solution

01

Calculate Z-score for $45.00

The Z-score formula is \( Z = \frac{(X - \mu)}{\sigma} \), where \( X \) is the value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. For \( X = 45.00 \), \( \mu = 42.00 \), and \( \sigma = 2.25 \), calculate the Z-score:\[ Z = \frac{(45.00 - 42.00)}{2.25} = \frac{3.00}{2.25} \approx 1.33 \].
02

Find Probability of Z > 1.33

Using the standard normal distribution table, find the probability that \( Z > 1.33 \). The table value for \( Z = 1.33 \) is 0.9082, which is the cumulative probability \( P(Z < 1.33) \). Thus, \( P(Z > 1.33) = 1 - 0.9082 = 0.0918 \).
03

Estimate Number of Days Over $45.00

To estimate the number of days with prices above \$45.00, multiply the probability by the total number of trading days: \( 0.0918 \times 240 \approx 22 \) days.
04

Calculate Z-scores for $38.00 and $40.00

Use the Z-score formula for both values:\( Z_{38} = \frac{(38.00 - 42.00)}{2.25} = -1.78 \) and for \( 40.00 \), \( Z_{40} = \frac{(40.00 - 42.00)}{2.25} = -0.89 \).
05

Find Probability for $38.00 < Price < $40.00

Using the standard normal distribution table, find probabilities for \( Z < -1.78 \) and \( Z < -0.89 \). The table values are approximately 0.0375 for \( Z = -1.78 \) and 0.1867 for \( Z = -0.89 \). Thus, \( P(-1.78 < Z < -0.89) = 0.1867 - 0.0375 = 0.1492 \), which equates to 14.92% of the days.
06

Find Minimum Share Price for Top 15%

For the top 15%, we need the 85th percentile of the normal distribution (since 100% - 15% = 85%). From the standard normal distribution table or using a calculator, the Z-value for 85% is approximately 1.04. Solve for the price using \( X = \mu + Z\sigma \):\( X = 42.00 + 1.04\times2.25 = 44.34 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Z-score
The Z-score is a statistical measure that tells you how many standard deviations a data point is from the mean of the dataset.
  • It is calculated using the formula: \[ Z = \frac{(X - \mu)}{\sigma} \]where \(X\) represents the data value you are analyzing, \(\mu\) is the mean of the dataset, and \(\sigma\) is the standard deviation.
  • A positive Z-score indicates the data point is above the mean, while a negative Z-score tells us the data point is below the mean.
In the context of stock pricing, like in the exercise, finding the Z-score can help determine how unusual or typical a specific stock price is compared to historical prices. For example, a Z-score of 1.33, calculated for a stock price of \(45.00, means this price is 1.33 standard deviations above the mean trading price of \)42.00. This value can then be used for further analysis, such as calculating probabilities or determining percentiles.
probability calculation
Probability calculation in the context of the normal distribution involves determining the likelihood of a certain range of outcomes.
  • This is often done using the Z-score in conjunction with a standard normal distribution table, which provides the probability that a Z-score is less than a certain value.
  • Once you know the Z-score for a certain data point, you can find the cumulative probability of observing that value, or any value less, using the table.
For example, in the exercise, once the Z-score for a stock price of $45.00 was found to be 1.33, the standard normal table gave the cumulative probability as 0.9082. So, the probability of the price being more than $45.00 is 1 minus this value, leading to a result of 0.0918. This means there is about a 9.18% chance that the day's closing price will exceed $45.00. Similarly, understanding the probability distribution helps estimate the number of trading days for specific price ranges or outcomes.
percentiles
Percentiles are used in statistics to understand and interpret data distribution. They indicate the value below which a given percentage of observations fall.
  • For example, the 85th percentile is the value below which 85% of the data points lie.
  • In the context of normal distribution, percentiles can be found using the Z-score corresponding to that percentile.
In the exercise, to find the stock price that marks the top 15% of trading days, it was imperative to find the 85th percentile, as 100% minus 15% equals 85%.The Z-score for the 85th percentile is approximately 1.04. Using the formula \( X = \mu + Z\sigma \), the value was calculated as approximately \(44.34. This means that on 85% of the trading days, the price was lower than \)44.34, while on the remaining 15% of days, the price was higher.
trading days analysis
Trading days analysis involves examining stock prices over a certain period to make informed trading decisions or understand market trends.
  • The goal is to derive meaningful insights from the past trading data, which follows a normal distribution in this exercise.
  • By calculating Z-scores and probabilities for specific prices, analysts can identify patterns and set strategies for buying or selling shares.
Analyzing the probability of a price exceeding $45.00 or determining the percentage of days prices stayed within $38.00 to $40.00 is an example of trading day analysis. For investors and traders, this analysis helps determine how historical price trends could influence future price movements. It informs decisions like setting price targets, which could be crucial for maximizing profits or minimizing risks. Understanding the analysis of 240 trading days, as provided in the exercise, allows investors to predict occurrences and gain a competitive edge in the market.

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Most popular questions from this chapter

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