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Chapter 7: Problem 50

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 percent of the days was the price over \(\$ 45.00 ?\) How many days would you estimate? b. What percent of the days was the price between \(\$ 38.00\) and \(\$ 40.00 ?\) c. What was the stock's price on the highest 15 percent of days?

Short Answer

Expert verified
a) 9.18%, about 22 days. b) 14.92%. c) Approximately \$44.34.

Step by step solution

01

Understanding the Normal Distribution

We are given a normal distribution for the price of shares with a mean \( \mu = 42.00 \) and a standard deviation \( \sigma = 2.25 \). We will use the properties of the normal distribution to solve parts a, b, and c.
02

Calculating Z-score for Part a

For part a, we want to find the percent of days the price was over \( \$45.00 \). The Z-score is calculated using the formula: \[ Z = \frac{X - \mu}{\sigma} \].Substitute \( X = 45.00 \), \( \mu = 42.00 \), \( \sigma = 2.25 \):\[ Z = \frac{45.00 - 42.00}{2.25} = \frac{3.00}{2.25} = 1.33 \].
03

Finding Probability for Z > 1.33

Using the standard normal distribution table, find the probability of Z being less than 1.33. This probability is approximately 0.9082. Thus, the probability of Z being greater than 1.33 is \(1 - 0.9082 = 0.0918\). So, 9.18% of days have the price over \$45.00. For part a (second question), multiply 9.18% by 240 to estimate the number of days: \(0.0918 \times 240 \approx 22 \text{ days.}\)
04

Calculating Z-scores for Part b

For part b, we find the percent of days with prices between \\(38.00 and \\)40.00. Calculate the Z-score for both values:\[ Z_{38} = \frac{38.00 - 42.00}{2.25} = \frac{-4.00}{2.25} \approx -1.78 \]\[ Z_{40} = \frac{40.00 - 42.00}{2.25} = \frac{-2.00}{2.25} \approx -0.89 \].
05

Calculating Probability for Z Between -1.78 and -0.89

Using the standard normal distribution table, find the probabilities:- Probability of Z < -1.78 is approximately 0.0375.- Probability of Z < -0.89 is approximately 0.1867.The probability of Z between -1.78 and -0.89 is \(0.1867 - 0.0375 = 0.1492\). Hence, 14.92% of the days have the price between \\(38.00 and \\)40.00.
06

Calculating Z-score for Part c

For part c, we want the stock's price on the highest 15% of days. We need to find the Z-score corresponding to the top 15%, which is the 85th percentile. From the standard normal distribution table, the Z-score for the 85th percentile is approximately 1.04.
07

Calculating the Corresponding Price for Z = 1.04

Use the Z-score formula to find the corresponding stock price:\[ X = \mu + Z\sigma = 42.00 + 1.04 \times 2.25 = 42.00 + 2.34 = 44.34 \].So, the stock's price on the highest 15 percent of days is approximately \$44.34.

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Key Concepts

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

Z-score
A Z-score gives you insight into how far a particular value deviates from the mean of a set of data, in terms of the number of standard deviations. It is a useful way to determine the position of a value within a normal distribution. The formula for calculating Z-score is \( Z = \frac{X - \mu}{\sigma} \), where:
  • \( X \) is the value for which you want to calculate the Z-score.
  • \( \mu \) is the mean of the data set.
  • \( \sigma \) is the standard deviation.
A positive Z-score indicates the value is above the mean, while a negative Z-score shows it is below the mean. A Z-score of 1 denotes a value that is one standard deviation above the mean, for instance. In the exercise, we calculate Z-scores to determine how frequently stock prices exceed or fall between certain values.
Probability
Probability helps us determine the likelihood of a particular event occurring within a set distribution. In a standard normal distribution, probabilities are represented as areas under the curve. Calculating the probability requires using Z-scores and referring to standard normal distribution tables.
For instance, once we know the Z-score of a particular share price, we can use this information to look up how likely that price can occur compared to the average. Using the table, we determine the probability of prices above or between particular values. This reflects actual percentage chances in the solution: the stock price being more than \( \\(45.00 \) or between \( \\)38.00 \) and \( \$40.00 \). These probabilities can be converted into estimates of actual days when certain prices are seen.
Standard Deviation
Standard deviation measures how spread out the numbers in a data set are. It tells us about the variability of your data: whether numbers are closely clustered around the mean or significantly spread out.
  • If the standard deviation is low, data points tend to be close to the mean.
  • Conversely, a high standard deviation indicates data points are more spread out.
In the exercise example, the standard deviation of \( \\(2.25 \) indicates how much the share prices deviate from the mean share price, \( \\)42.00 \). Knowing the standard deviation is crucial in calculating Z-scores as it shows the typical deviation of values either side of the mean. This in turn, assists us in evaluating probabilities for specific price ranges, giving insights into share price behavior over the given period.

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Most popular questions from this chapter

A normal population has a mean of 12.2 and a standard deviation of \(2.5 .\) a. Compute the \(z\) value associated with 14.3 . b. What proportion of the population is between 12.2 and \(14.3 ?\) C. What proportion of the population is less than \(10.0 ?\)

Many retail stores offer their own credit cards. At the time of the credit application the customer is given a 10 percent discount on the purchase. The time required for the credit application process follows a uniform distribution with the times ranging from 4 minutes to 10 minutes. a. What is the mean time for the application process? b. What is the standard deviation of the process time? c. What is the likelihood a particular application will take less than 6 minutes? d. What is the likelihood an application will take more than 5 minutes?

According to a government study among adults in the 25 - to 34 -year age group, the mean amount spent per year on reading and entertainment is \(\$ 1,994\) (www. infoplease.com/ipa/A0908759.html. Assume that the distribution of the amounts spent follows the normal distribution with a standard deviation of \(\$ 450 .\) a. What percent of the adults spend more than \(\$ 2,500\) per year on reading and entertainment? b. What percent spend between \(\$ 2,500\) and \(\$ 3,000\) per year on reading and entertainment? c. What percent spend less than \(\$ 1,000\) per year on reading and entertainment?

A recent study of the hourly wages of maintenance crew members for major airlines showed that the mean hourly salary was \(\$ 20.50,\) with a standard deviation of \(\$ 3.50 .\) If we select a crew member at random, what is the probability the crew member earns: a. Between \(\$ 20.50\) and \(\$ 24.00\) per hour? b. More than \(\$ 24.00\) per hour? c. Less than \(\$ 19.00\) per hour?

A recent article in the Cincinnati Enquirer reported that the mean labor cost to repair a heat pump is \(\$ 90\) with a standard deviation of \(\$ 22 .\) Monte's Plumbing and Heating Service completed repairs on two heat pumps this morning. The labor cost for the first was \(\$ 75\) and it was \(\$ 100\) for the second. Compute \(z\) values for each and comment on your findings.
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