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Use the following information to answer the next five exercises: A company wants to evaluate its attrition rate, in other words, how long new hires stay with the company. Over the years, they have established the following probability distribution. Let \(X=\) the number of years a new hire will stay with the company. Let \(P(x)=\) the probability that a new hire will stay with the company \(x\) years. Complete Table 4.20 using the data provided. $$\begin{array}{|c|c|}\hline x & {P(x)} \\ \hline 0 & {0.12} \\ \hline 1 & {0.18} \\ \hline 2 & {0.30} \\ \hline 3 & {0.15} \\ \hline 4 & {} \\\ \hline 5 & {0.10} \\ \hline 6 & {0.05} \\ \hline\end{array}$$

Short Answer

Expert verified
The missing probability is \( P(4) = 0.10 \).

Step by step solution

01

Understanding the problem

The table represents the probability distribution of a random variable \( X \), where \( X \) is the number of years a new employee stays with the company. We need to find the missing probability \( P(4) \) in the distribution so that the probabilities sum to 1, which is essential for any probability distribution.
02

Write the equation for total probability

Since the sum of all probabilities in a probability distribution must equal 1, we can set up the equation: \[ P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1 \]. We know the values of all probabilities except \( P(4) \): \[ 0.12 + 0.18 + 0.30 + 0.15 + P(4) + 0.10 + 0.05 = 1 \].
03

Calculate the sum of known probabilities

Calculate the sum of the given probabilities: \[ 0.12 + 0.18 + 0.30 + 0.15 + 0.10 + 0.05 = 0.90 \].
04

Solve for the missing probability

Subtract the sum of known probabilities from 1 to find \( P(4) \): \[ P(4) = 1 - 0.90 = 0.10 \].
05

Complete the probability distribution table

The completed table is: \[\begin{array}{|c|c|}\hline x & {P(x)} \hline 0 & {0.12} \hline 1 & {0.18} \hline 2 & {0.30} \hline 3 & {0.15} \hline 4 & {0.10} \hline 5 & {0.10} \hline 6 & {0.05} \hline\end{array}\]

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Key Concepts

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

Random Variable
A random variable is a fundamental concept in probability and statistics. It is a variable whose possible outcomes are numerical values of an experiment or random event. Although it might sound complex, imagining it as a way to summarize or simplify all possible outcomes of an event can be helpful.
In our exercise, the random variable, \( X \), represents the number of years a new hire stays with the company. Each possible value of \( X \) is mapped to a probability, which shows the likelihood of a new hire staying for that number of years. Here are some key points about random variables:
  • They can be either discrete or continuous.
  • Discrete random variables have a set of specific possible values (like our years of service, 0 to 6).
  • Continuous random variables can take any value within a range.
Understanding random variables is crucial because they allow us to create models and make predictions based on data. For the company, modeling employee tenure as a random variable helps to estimate attrition rates and make informed decisions.
Attrition Rate
The term attrition rate, often referred to as employee turnover, describes the pace at which employees leave the company over a period. In our scenario, calculating how long new hires remain with the company helps the organization understand this rate.
Attrition rates can provide valuable insight into company dynamics and help identify underlying issues or successes in employee retention strategies. It's essential to understand the difference between voluntary and involuntary attrition:
  • Voluntary Attrition: Employees leave by choice, often for new career opportunities, better pay, or a more suitable work-life balance.
  • Involuntary Attrition: The company decides to let employees go due to layoffs, terminations, or reorganizations.
By analyzing the probability distribution of how long employees stay, the company gets a clearer picture of its attrition rates. This helps in tailoring retention efforts and improving workforce stability.
Probability Calculation
Probability calculation involves determining the likelihood of different outcomes for a random variable. In probability distributions, all the probabilities must add up to 1, which mathematically implies certainty that one of the outcomes must occur.
For the given probability distribution of employee tenure, calculating the missing probability was essential. Here's a brief overview of the process used in our exercise:
  • First, sum up all known probabilities to see how much of the 'probability space' is covered.
  • If any probabilities are missing, like \( P(4) \) in our exercise, the sum of known probabilities is subtracted from 1.
  • The result is the missing probability, ensuring the entire set of probabilities equals 1.
This technique can be handy in various fields, whether predicting business trends, assessing risk, or planning strategies. The main takeaway is how probabilities tie together to form complete distributions, aiding in better understanding and predicting outcomes.

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Most popular questions from this chapter

Use the following information to answer the next four exercises. Recently, a nurse commented that when a patient calls the medical advice line claiming to have the flu, the chance that he or she truly has the flu (and not just a nasty cold) is only about 4%. Of the next 25 patients calling in claiming to have the flu, we are interested in how many actually have the flu. Define the random variable and list its possible values.

Use the following information to answer the next six exercises: The Higher Education Research Institute at UCLA collected data from 203,967 incoming first-time, full-time freshmen from 270 four-year colleges and universities in the U.S. 71.3% of those students replied that, yes, they believe that same-sex couples should have the right to legal marital status. Suppose that you randomly select freshman from the study until you find one who replies 鈥測es.鈥 You are interested in the number of freshmen you must ask. Construct the probability distribution function (PDF). Stop at \(x = 6\). $$\begin{array}{|c|c|}\hline x & {P(x)} \\ \hline 1 & {} \\ \hline 2 & {} \\\ \hline 3 & {} \\ \hline 4 & {} \\ \hline 5 & {} \\ \hline x & {P(x)} \\\ \hline 6 & {} \\ \hline\end{array}$$

Use the following information to answer the next eight exercises: The Higher Education Research Institute at UCLA collected data from 203,967 incoming first-time, full-time freshmen from 270 four year colleges and universities in the U.S. 71.3% of those students replied that, yes, they believe that same-sex couples should have the right to legal marital status. Suppose that you randomly pick eight first-time, full-time freshmen from the survey. You are interested in the number that believes that same sex-couples should have the right to legal marital status. What is the probability that at most five of the freshmen reply 鈥測es鈥?

A venture capitalist, willing to invest \(\$ 1,000,000\), has three investments to choose from. The first investment, a software company, has a 10% chance of returning \(\$ 5,000,000\) profit, a 30% chance of returning \(\$ 1,000,000\) profit, and a 60% chance of losing the million dollars. The second company, a hardware company, has a 20% chance of returning \(\$ 3,000,000\) profit, a 40% chance of returning \(\$ 1,000,000\) profit, and a 40% chance of losing the million dollars. The third company, a biotech firm, has a 10% chance of returning \(\$ 6,000,000\) profit, a 70% of no profit or loss, and a 20% chance of losing the million dollars. a. Construct a PDF for each investment. b. Find the expected value for each investment. c. Which is the safest investment? Why do you think so? d. Which is the riskiest investment? Why do you think so? e. Which investment has the highest expected return, on average?

Suppose that you are performing the probability experiment of rolling one fair six-sided die. Let F be the event of rolling a four or a five. You are interested in how many times you need to roll the die in order to obtain the first four or five as the outcome. 鈥 \(p\) = probability of success (event F occurs) 鈥 \(q \)= probability of failure (event F does not occur) a. Write the description of the random variable X. b. What are the values that \(X\) can take on? c. Find the values of \(p\) and \(q\). d. Find the probability that the first occurrence of event F (rolling a four or five) is on the second trial.

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