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According to Salary Wizard, the average base salary for a brand manager in Houston, Texas, is \(\$ 88,592\) and the average base salary for a brand manager in Los Angeles, California, is \(\$ 97,417\) (Salary Wizard website, February 27,2008 ). Assume that salaries are normally distributed, the standard deviation for brand managers in Houston is \(\$ 19,900,\) and the standard deviation for brand managers in Los Angeles is \(\$ 21,800\) a. What is the probability that a brand manager in Houston has a base salary in excess of \(\$ 100,000 ?\) b. What is the probability that a brand manager in Los Angeles has a base salary in excess of \(\$ 100,000 ?\) c. What is the probability that a brand manager in Los Angeles has a base salary of less than \(\$ 75,000 ?\) d. How much would a brand manager in Los Angeles have to make in order to have a higher salary than \(99 \%\) of the brand managers in Houston?

Short Answer

Expert verified
a) 0.2839 b) 0.4522 c) 0.1526 d) $134,084

Step by step solution

01

Calculate Z-score for Houston base salary > $100,000

To find the probability of a brand manager in Houston earning more than $100,000, we first calculate the z-score. The z-score formula is: \[ z = \frac{X - \mu}{\sigma} \] For Houston: \( \mu = 88,592 \) and \( \sigma = 19,900 \). Substitute \( X = 100,000 \):\[ z = \frac{100,000 - 88,592}{19,900} \approx 0.572 \]
02

Find Probability for Z-score (Houston)

Using the standard normal distribution table, find the probability corresponding to the z-score of 0.572. This gives the probability of earning less than \(100,000. For \( z = 0.572 \), the cumulative probability \( P(Z < 0.572) \approx 0.7161 \).The probability of earning more than \)100,000 is:\[ P(Z > 0.572) = 1 - 0.7161 = 0.2839 \]
03

Calculate Z-score for Los Angeles base salary > $100,000

For Los Angeles, with \( \mu = 97,417 \) and \( \sigma = 21,800 \), substitute \( X = 100,000 \):\[ z = \frac{100,000 - 97,417}{21,800} \approx 0.118 \]
04

Find Probability for Z-score (Los Angeles, $100,000)

Find the cumulative probability for \( z = 0.118 \) which is \( P(Z < 0.118) \approx 0.5478 \).The probability of earning more than $100,000 is:\[ P(Z > 0.118) = 1 - 0.5478 = 0.4522 \]
05

Calculate Z-score for Los Angeles base salary < $75,000

Now, find the z-score for a salary less than $75,000. Substitute \( X = 75,000 \) in the z-score formula.\[ z = \frac{75,000 - 97,417}{21,800} \approx -1.029 \]
06

Find Probability for Z-score (Los Angeles, $75,000)

Look up the cumulative probability for \( z = -1.029 \), which is \( P(Z < -1.029) \approx 0.1526 \). This is the probability a brand manager in Los Angeles earns less than $75,000.
07

Find Salary Above 99% in Houston

To find the salary such that 99% of Houston brand managers earn less, use the z-score for the 99th percentile, approximately 2.33.Solve for \( X \) in the z-score formula:\[ X = \mu + z \times \sigma = 88,592 + 2.33 \times 19,900 \approx 134,084 \]A brand manager in Los Angeles would need to earn \( \approx 134,084 \) to be higher than 99% of those in Houston.

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Key Concepts

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

Normal Distribution
In the world of statistics, the normal distribution is a cornerstone concept. Imagine it as a bell curve. It is symmetrical and depicts how data points are distributed in a set. For most of the real-world processes, data tends to cluster around a central value with no bias to the left or right. That's what a normal distribution represents.

For example, when considering salaries in a certain profession, we might find that most people earn around the average salary, with fewer individuals earning extremely high or low salaries. This clustering around the mean is the essence of normal distribution.
  • The **mean** (\( \mu \) ) is the center of the curve where the peak is positioned.
  • The **standard deviation** (\( \sigma \) ) measures the spread or width of the curve, influencing how spread out the values are.
This concept helps in assessing probabilities and calculating outcomes in various professional analyses like salary assessments.
Z-score Calculation
Calculating a z-score is a valuable statistical method used to understand the relationship of a particular data point within a set. A z-score is essentially a transformation that tells us how many standard deviations a given data point is from the mean.

Consider the formula for calculating a z-score: \[ z = \frac{X - \mu}{\sigma} \]
  • \( X \) represents the specific data point of interest.
  • \( \mu \) is the mean of the dataset.
  • \( \sigma \) is the standard deviation.
The result, the z-score, provides insight into the position of the data point.
  • A positive z-score means the data point is above the mean.
  • A negative z-score suggests it is below the mean.
  • A z-score of zero indicates it falls exactly at the mean.
Using z-scores, one can quickly gauge where a salary stands in relation to the mean salary within a city or profession.
Probability Analysis
Probability analysis involves finding the likelihood of different outcomes within a normal distribution. It's used for determining how likely it is for a data point to occur given the normal distribution. By using z-scores, we can navigate the standard normal distribution table to find cumulative probabilities.

Here’s how one performs a probability analysis using z-scores:
  • After calculating a z-score, use a Z-table (standard normal distribution table) to find the cumulative probability.
  • Cumulative probability gives the likelihood of a data point being less than the z-score value.
  • To find the probability of being greater than a particular value, subtract the cumulative probability from 1.
This approach helps in predicting probabilities, such as the chance that a brand manager in Houston earns more than $100,000.
Salary Analysis
Salary analysis using statistical methods provides insights into the distribution of salaries within a profession or region. With tools like normal distribution, z-score, and probability analysis, one can make educated guesses about salaries without having complete data.

Suppose we wish to understand how salaries compare in two different cities – for instance, Houston and Los Angeles for brand managers. By examining:
  • The mean salary, which gives an idea of what the typical salary is.
  • The spread (standard deviation) which tells us how salaries vary.
Employing z-scores and probability analysis can help determine the proportion of data points falling above or below a certain salary. For instance, estimating what it takes for a Los Angeles brand manager to earn more than 99% of Houston managers allows for informed decision-making in salary negotiations and expectations.

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Most popular questions from this chapter

The website for the Bed and Breakfast Inns of North America gets approximately seven visitors per minute. Suppose the number of website visitors per minute follows a Poisson probability distribution. a. What is the mean time between visits to the website? b. Show the exponential probability density function for the time between website visits. c. What is the probability that no one will access the website in a 1-minute period? d. What is the probability that no one will access the website in a 12 -second period?

Given that \(z\) is a standard normal random variable, compute the following probabilities. a. \(\quad P(-1.98 \leq z \leq .49)\) b. \(\quad P(.52 \leq z \leq 1.22)\) c. \(\quad P(-1.75 \leq z \leq-1.04)\)

Consider a multiple-choice examination with 50 questions. Each question has four possible answers. Assume that a student who has done the homework and attended lectures has a \(75 \%\) probability of answering any question correctly. a. \(\quad\) A student must answer 43 or more questions correctly to obtain a grade of \(\mathrm{A}\). What percentage of the students who have done their homework and attended lectures will obtain a grade of A on this multiple- choice examination? b. A student who answers 35 to 39 questions correctly will receive a grade of \(C .\) What percentage of students who have done their homework and attended lectures will obtain a grade of \(\mathrm{C}\) on this multiple-choice examination? c. A student must answer 30 or more questions correctly to pass the examination. What percentage of the students who have done their homework and attended lectures will pass the examination? d. Assume that a student has not attended class and has not done the homework for the course. Furthermore, assume that the student will simply guess at the answer to each question. What is the probability that this student will answer 30 or more questions correctly and pass the examination?

Consider the following exponential probability density function. \\[ f(x)=\frac{1}{3} e^{-x / 3} \quad \text { for } x \geq 0 \\] a. Write the formula for \(P\left(x \leq x_{0}\right)\) b. Find \(P(x \leq 2)\) c. \(\quad\) Find \(P(x \geq 3)\) d. Find \(P(x \leq 5)\) e. Find \(P(2 \leq x \leq 5)\)

A random variable is normally distributed with a mean of \(\mu=50\) and a standard deviation of \(\sigma=5\) a. Sketch a normal curve for the probability density function. Label the horizontal axis with values of \(35,40,45,50,55,60,\) and \(65 .\) Figure 6.4 shows that the normal curve almost touches the horizontal axis at three standard deviations below and at three standard deviations above the mean (in this case at 35 and 65 ). b. What is the probability that the random variable will assume a value between 45 and \(55 ?\) c. What is the probability that the random variable will assume a value between 40 and \(60 ?\)

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