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Chapter 3: Problem 25

The distributor of a machine for cytogenics has developed a new model. The company estimates that when it is introduced into the market, it will be very successful with a probability \(0.6,\) moderately successful with a probability \(0.3,\) and not successful with probability \(0.1 .\) The estimated yearly profit associated with the model being very successful is \(\$ 15\) million and with it being moderately successful is \(\$ 5\) million; not successful would result in a loss of \(\$ 500,000\). Let \(X\) be the yearly profit of the new model. Determine the probability mass function of \(X\).

Short Answer

Expert verified
The PMF is: \(P(X = 15) = 0.6\), \(P(X = 5) = 0.3\), \(P(X = -0.5) = 0.1\).

Step by step solution

01

Introduction to Probability Mass Function

A probability mass function (PMF) for a discrete random variable provides the probabilities associated with each of its possible outcomes. In this problem, we need to determine the PMF for the estimated profits based on the success level of the machine.
02

Defining Random Variable Outcomes

The outcomes for the random variable \(X\), representing yearly profit, based on the company's estimates, are: 1. Very successful: \(X = 15\) million dollars 2. Moderately successful: \(X = 5\) million dollars 3. Not successful: \(X = -0.5\) million dollars.
03

Assigning Probabilities to Outcomes

Assign the probabilities to each outcome of \(X\) as follows, using the company's estimates:- Probability the model is very successful \(P(X = 15) = 0.6\).- Probability the model is moderately successful \(P(X = 5) = 0.3\).- Probability the model is not successful \(P(X = -0.5) = 0.1\).
04

Writing the Probability Mass Function

The Probability Mass Function of \(X\) can be represented as:\[P(X = x) = \begin{cases} 0.6, & \text{if } x = 15 \ 0.3, & \text{if } x = 5 \ 0.1, & \text{if } x = -0.5 \end{cases}\]This PMF summarizes the probabilities for the different profit outcomes based on the success of the new model.

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Key Concepts

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

Discrete Random Variables
A discrete random variable is a type of variable that takes on a countable number of distinct values. These values are usually outcomes of a random process, and each has a specific probability attached to it. In the context of business analysis, a discrete random variable can represent profit outcomes among other possibilities.
For the machine distributor example, the discrete random variable is the yearly profit, denoted by \(X\). Since it can only assume specific values based on the model's success, it is considered discrete.
There are three potential outcomes for \(X\):
  • \(X = 15\) million dollars when the product is very successful.
  • \(X = 5\) million dollars when it is moderately successful.
  • \(X = -0.5\) million dollars when not successful.
Each of these outcomes is tied to its probability, which makes \(X\) a discrete random variable.
Expected Profit Calculation
The expected profit is an essential metric for evaluating potential business successes or risks. It's calculated by weighing each potential profit outcome with its associated probability, providing a statistical average of expected performance.
The formula for calculating expected profit, \(E(X)\), is:\[E(X) = \sum_{i} (x_i \cdot P(X = x_i))\]
In the exercise, we determine this as follows:
  • \(x_1 = 15\) million dollars with \(P(X = 15) = 0.6\)
  • \(x_2 = 5\) million dollars with \(P(X = 5) = 0.3\)
  • \(x_3 = -0.5\) million dollars with \(P(X = -0.5) = 0.1\)
Substituting into the formula, we have:\[E(X) = (15 \times 0.6) + (5 \times 0.3) + (-0.5 \times 0.1)\]This results in an expected profit which provides a clear quantitative measure of the model's financial potential.
Business Success Probability
Business success probability is a critical factor in decision-making processes. It quantifies the likelihood of various outcomes that influence overall business success or failure.
By analyzing the given probabilities:
  • Probability of a very successful launch: 0.6
  • Probability of moderate success: 0.3
  • Probability of no success: 0.1
This exercise helps in assessing the risk and preparing for different scenarios that a business may encounter when launching a new product.
The probabilities highlight not just potential gains, but also the associated risks, such as a 10% chance of experiencing losses. By understanding these probabilities, businesses can better strategize and allocate resources to maximize their profit and mitigate risks.

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Most popular questions from this chapter

The random variable \(X\) has a binomial distribution with \(n=10\) and \(p=0.5 .\) Sketch the probability mass function of \(X\). (a) What value of \(X\) is most likely? (b) What value(s) of \(X\) is(are) least likely?

The number of telephone calls that arrive at a phone exchange is often modeled as a Poisson random variable. Assume that on the average there are 10 calls per hour. (a) What is the probability that there are exactly five calls in one hour? (b) What is the probability that there are three or fewer calls in one hour? (c) What is the probability that there are exactly 15 calls in two hours? (d) What is the probability that there are exactly five calls in 30 minutes?

The random variable \(X\) has a binomial distribution with \(n=10\) and \(p=0.01 .\) Determine the following probabilities (a) \(P(X=5)\) (b) \(P(X \leq 2)\) (c) \(P(X \geq 9)\) (d) \(P(3 \leq X<5)\)

Messages that arrive at a service center for an information systems manufacturer have been classified on the basis of the number of keywords (used to help route messages) and the type of message, either e-mail or voice. Also, \(70 \%\) of the messages arrive via e-mail and the rest are voice. $$ \begin{array}{llllll} \text { number of keywords } & 0 & 1 & 2 & 3 & 4 \\ \text { e-mail } & 0.1 & 0.1 & 0.2 & 0.4 & 0.2 \\ \text { voice } & 0.3 & 0.4 & 0.2 & 0.1 & 0 \end{array} $$ Determine the probability mass function of the number of keywords in a message.

Data from www.centralhudsonlabs determined the mean number of insect fragments in 225 -gram chocolate bars was \(14.4,\) but three brands had insect contamination more than twice the average. See the U.S. Food and Drug Administration-Center for Food Safety and Applied Nutrition for Defect Action Levels for food products. Assume the number of fragments (contaminants) follows a Poisson distribution. (a) If you consume a 225-gram bar from a brand at the mean contamination level, what is the probability of no insect contaminants? (b) Suppose you consume a bar that is one-fifth the size tested (45 grams) from a brand at the mean contamination level. What is the probability of no insect contaminants? (c) If you consume seven 28.35-gram (one-ounce) bars this week from a brand at the mean contamination level, what is the probability that you consume one or more insect fragments in more than one bar? (d) Is the probability of contamination more than twice the mean of 14.4 unusual, or can it be considered typical variation? Explain.
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