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Based on its analysis of the future demand for its products, the financial department at Tipper Corporation has determined that there is a \(.17\) probability that the company will lose \(\$ 1.2\) million during the next year, a \(.21\) probability that it will lose \(\$ .7\) million, a \(.37\) probability that it will make a profit of \(\$ .9\) million, and a \(.25\) probability that it will make a profit of \(\$ 2.3\) million. a. Let \(x\) be a random variable that denotes the profit earned by this corporation during the next year. Write the probability distribution of \(x\). b. Find the mean and standard deviation of the probability distribution of part a. Give a brief interpretation of the value of the mean.

Short Answer

Expert verified
The probability distribution of the random variable \(x\) representing the profit earned by the corporation: For \(x = -1.2\), \(p(x) = 0.17\); For \(x = -0.7\), \(p(x) = 0.21\); For \(x = 0.9\), \(p(x) = 0.37\); For \(x = 2.3\), \(p(x) = 0.25\). The mean \(\mu\) of the distribution is 0.557 million, and the standard deviation \(\sigma\) is 1.287 million. This implies that on average, the corporation would earn 0.557 million per year, with a standard deviation of 1.287 million indicating the dispersion.

Step by step solution

01

Write the probability distribution of x

For this step, the values of \(x\) and their corresponding probability \(p(x)\) are listed. From the exercise: \(x = -1.2, p(x) = 0.17\) \(x = -0.7, p(x) = 0.21\) \(x = 0.9, p(x) = 0.37\) \(x = 2.3, p(x) = 0.25\)
02

Find the mean of the probability distribution

The mean (\(mu\)) of a probability distribution is calculated using the formula \(mu = \sum x * p(x)\). Plugging in the given values:\(mu = -1.2*0.17 + -0.7*0.21 + 0.9*0.37 + 2.3*0.25 = -0.204 + -0.147 + 0.333 + 0.575 = 0.557 million \)
03

Find the standard deviation of the probability distribution

The standard deviation (\(sigma\)) is calculated using the formula \(sigma = \sqrt{\sum (x - mu)^2 * p(x)}\). First calculate \(\sum (x - mu)^2 * p(x)\):\(-1.2 -0.557 = -1.757; (-1.757)^2 * 0.17 = 0.526; -0.7 - 0.557 = -1.257; (-1.257)^2 * 0.21 = 0.333; 0.9 - 0.557 = 0.343; (0.343)^2 * 0.37 = 0.043; 2.3 - 0.557 = 1.743; (1.743)^2 * 0.25 = 0.755;Thus \(sigma = \sqrt{(0.526 + 0.333 + 0.043 + 0.755)} = \sqrt{1.657} = 1.287 million \)
04

Interpret the mean

The mean of 0.557 million implies that if the possible outcomes were to repeat over several years, the average profit earned by the corporation would be approximately 0.557 million.

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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 concept that helps in working with uncertain or random phenomena. In simple terms, it's a variable whose value depends on the outcomes of a random event. For Tipper Corporation, the random variable \( x \) represents their profit or loss in the next year. The values \( -1.2 \), \( -0.7 \), \( 0.9 \), and \( 2.3 \) millions denote specific possible outcomes from this uncertain situation.
Understanding that random variables can take on any value from a set of possibilities, we express them with probabilities. These probabilities describe the likelihood of each outcome occurring:
  • \( x = -1.2 \) million with probability \( 0.17 \),
  • \( x = -0.7 \) million with probability \( 0.21 \),
  • \( x = 0.9 \) million with probability \( 0.37 \),
  • \( x = 2.3 \) million with probability \( 0.25 \).
Knowing the values and associated probabilities forms what is called a probability distribution, paving the way for deeper analysis.
Mean of Probability Distribution
The mean of a probability distribution, often called the expected value, gives a measure of the "center" of the distribution. It is computed by summing up the products of each possible outcome with its probability. This process is concisely captured in the formula:
\[\mu = \sum x * p(x)\]
For Tipper Corporation, the calculation follows:
  • \(-1.2 \times 0.17 = -0.204\),
  • \(-0.7 \times 0.21 = -0.147\),
  • \(0.9 \times 0.37 = 0.333\),
  • \(2.3 \times 0.25 = 0.575\).

Adding these together gives a mean profit of \(0.557\) million dollars. The mean suggests that if the company were to experience these probabilities repeatedly over many years, the average annual profit would settle around \(0.557\) million dollars.
Standard Deviation of Probability Distribution
The standard deviation in a probability distribution quantifies the amount of variation or spread in a set of values. It gives a sense of how much the values might deviate from the mean. To calculate it, we use:
\[\sigma = \sqrt{\sum (x - \mu)^2 * p(x)}\]
This formula involves calculating the squared differences between each value and the mean, weighting them by their probabilities, and then taking the square root.
For the exercise:
  • \((-1.2 - 0.557)^2 \times 0.17 = 0.526\),
  • \((-0.7 - 0.557)^2 \times 0.21 = 0.333\),
  • \((0.9 - 0.557)^2 \times 0.37 = 0.043\),
  • \((2.3 - 0.557)^2 \times 0.25 = 0.755\).

Adding up these values gives \(1.657\), and taking the square root results in a standard deviation of approximately \(1.287\) million dollars. This tells us that, although the mean profit is \(0.557\) million, real outcomes will vary around this figure by \(1.287\) million with each annual occurrence.

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