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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
a) X is a random variable indicating profit, with possible values -1.2, -0.7, 0.9 & 2.3. The accompanying probabilities are 0.17, 0.21, 0.37 & 0.25. The probability distribution table connects each profit value (X) with its corresponding probability. b) The mean of the probability distribution, which is the expected profit, is -0.84 million dollars; it's a loss expected by the company. The standard deviation is 1.19, which describes how widely profits are spread around the expected value.

Step by step solution

01

Construct the Probability Distribution

A Probability Distribution can be written as a two-column table with all possible outcomes in one column and their corresponding probabilities in the other column.\n\nSo, from the problem, we have the following Probability Distribution : \n\nx = [-1.2, -0.7, 0.9, 2.3] \nand \nP(x) = [0.17, 0.21, 0.37, 0.25] respectively.
02

Compute the Mean

The mean, or expected value, of a random variable X (E(X)) is calculated by multiplying each outcome 'x' with its corresponding probability 'P(x)' and summing them all up.\n\nTherefore, the Mean (E(X)) = \(Summation (x * P(x))\)\n= (-1.2 * 0.17) + (-0.7 * 0.21) + (0.9 * 0.37) + (2.3 * 0.25)\n= -0.84
03

Compute the Variance

The Variance of a random variable X is the expectation of the squared deviation of a random variable from its mean. That is, it measures how far each number in the set is from the mean. \n\nVariance (Var(X)) = E((X−E(X))^2) = \(Summation ((x-E(X))^2 * P(x))\)\n= ((-1.2 - (-0.84))^2 * 0.17) + ((-0.7 - (-0.84))^2 * 0.21) + ((0.9 - (-0.84))^2 * 0.37) + ((2.3 - (-0.84))^2 * 0.25)\n = 1.42
04

Compute the Standard Deviation

The Standard Deviation is the square root of the variance.\n\nTherefore, Standard Deviation = \(sqrt(Var(X))\) = sqrt(1.42) = 1.19

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Key Concepts

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

Expected Value
Understanding the expected value, often called the mean, is a vital part of probability distribution. It's like finding the average of a set of random outcomes weighted by their probabilities. This provides a central value that represents the distribution. For our example, the financial fortunes of Tipper Corporation can range from a loss to a profit, and the expected value aids in summarizing these different scenarios into one comprehensive number.

To compute the expected value of a random variable, multiply each outcome by its probability and sum up all these products. In mathematical terms, it’s expressed as: \[ E(X) = \sum x_i P(x_i) \] where \( x_i \) are the outcomes and \( P(x_i) \) their respective probabilities.

For Tipper Corporation, this addition results in an expected value of \-0.84 million (indicating an average expected loss). This means that over a long period, the corporation can anticipate a slight loss, considering all possibilities and their probabilities.
Mean and Standard Deviation
The mean provides a middle point of the data, while the standard deviation measures how spread out the numbers in the probability distribution are around this mean point. Together, they give a picture of the data’s central tendency and variability.

Computing the mean in this setting involves the same calculation as the expected value. Whereas the standard deviation requires an additional step: first, calculate the variance, then find the square root. The formula for the standard deviation is: \[ \text{Standard Deviation} = \sqrt{\text{Variance}} \] This helps in understanding how much the actual earnings deviate from the expected value, or mean, over time.

For Tipper Corporation, their calculated standard deviation of \(1.19\) million suggests that their profits and losses can veer away from the expected \-0.84 million by about 1.19 million on average.
Random Variable
A random variable is a concept used to model uncertainty. It represents the outcomes of a probabilistic event, usually denoted by symbols like \( X \). In our context, the random variable \( X \) represents the various profits and losses Tipper Corporation might experience.

It’s important to note that a random variable is not a variable in the traditional math sense, but a function that assigns numerical outcomes to every possible occurrence of a random process.

Suppose \( x \) is defined as
  • \(-1.2 \) (loss of \\(1.2 million)
  • \(-0.7 \) (loss of \\)0.7 million)
  • \(0.9 \) (profit of \\(0.9 million)
  • \(2.3 \) (profit of \\)2.3 million)
These numerical outcomes are part of constructing the probability distribution for \( X \), representing the comprehensive view of all likely financial results.
Variance
Variance measures how much the set of numbers (or outcomes of a random variable) are scattered around the mean. It gives insight into the stability and risk intrinsic to the probability distribution.

More technically, variance is calculated as the expectation of the squared deviation from the mean. This formula for variance introduces the concept of squaring the differences from the mean, ensuring that they all contribute positively to the final result: \[ \text{Variance} = E((X - E(X))^2) = \sum (x_i - E(X))^2 P(x_i) \] For Tipper Corporation, the variance is computed to be \(1.42\). This numerical value showcases the extent of variability in profit and loss figures, implying how spread out the outcomes are from the average expected loss of \-0.84 million. A higher variance would suggest a wide dispersion and, therefore, a higher risk.

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Most popular questions from this chapter

Many of you probably played the game "Rock, Paper, Scissors" as a child. Consider the following variation of that game. Instead of two players, suppose three players play this game, and let us call these players \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\). Each player selects one of these three items- Rock, Paper, or Scissors-independent of each other. Player A will win the game if all three players select the same item, for example, rock. Player B will win the game if exactly two of the three players select the same item and the third player selects a different item. Player \(\mathrm{C}\) will win the game if every player selects a different item. If Player B wins the game, he or she will be paid \(\$ 1\). If Player \(\mathrm{C}\) wins the game, he or she will be paid \(\$ 3\). Assuming that the expected winnings should be the same for each player to make this a fair game, how much should Player A be paid if he or she wins the game?

Let \(x\) be a Poisson random variable. Using the Poisson probabilities table, write the probability distribution of \(x\) for each of the following. Find the mean, variance, and standard deviation for each of these probability distributions. Draw a graph for each of these probability distributions. a. \(\lambda=1.3\) b. \(\lambda=2.1\)

Briefly explain the concept of the mean and standard deviation of a discrete random variable.

An instant lottery ticket costs \(\$ 2\). Out of a total of 10,000 tickets printed for this lottery, 1000 tickets contain a prize of \(\$ 5\) each, 100 tickets have a prize of \(\$ 10\) each, 5 tickets have a prize of \(\$ 1000\) each, and 1 ticket has a prize of \(\$ 5000\). Let \(x\) be the random variable that denotes the net amount a player wins by playing this lottery. Write the probability distribution of \(x\). Determine the mean and standard deviation of \(x\). How will you interpret the values of the mean and standard deviation of \(x\) ?

A review of emergency room records at rural Millard Fellmore Memorial Hospital was performed to determine the probability distribution of the number of patients entering the emergency room during a 1 -hour period. The following table lists this probability distribution. $$ \begin{array}{l|ccccccc} \hline \begin{array}{l} \text { Patients } \\ \text { per hour } \end{array} & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text { Probability } & .2725 & .3543 & .2303 & .0998 & .0324 & .0084 & .0023 \\ \hline \end{array} $$ a. Make a histogram for this probability distribution. b. Determine the probability that the number of patients entering the emergency room during a randomly selected 1 -hour period is i. 2 or more ii. exactly 5 iii. fewer than 3 iv. at most 1

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