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Determine whether or not the table is a valid probability distribution of a discrete random variable. Explain fully. a. \(\frac{m}{P(x)} \mid-100 \frac{4}{0.30 .50 .20 .1}\) b. \begin{tabular}{c|ccc} & 0.5 & 0.25 & 0.25 \\ \hline\(P(x)\) & -0.4 & 0.6 & 0.8 \end{tabular} C. \begin{tabular}{c|ccccc} & 1.1 & 2.5 & 4.1 & 4.6 & 5.3 \\ \hline\(P(x)\) & 0.16 & 0.14 & 0.11 & 0.87 & 0.22 \end{tabular}

Short Answer

Expert verified
Table (a) is valid. Table (b) is invalid due to negative probability. Table (c) is invalid as probabilities sum to more than 1.

Step by step solution

01

Understanding a Probability Distribution

A probability distribution for a discrete random variable assigns a probability to each possible outcome. The two main criteria are: 1) Each probability must be between 0 and 1, inclusive; 2) The sum of all probabilities must equal 1.
02

Evaluate Table (a)

Check the validity of table (a) by ensuring that each probability value is between 0 and 1, and then sum them up. The probability values provided are 0.10, 0.20, 0.50, and 0.30. Clearly, all these values are within the allowed range of 0 to 1. Adding them: 0.10 + 0.20 + 0.50 + 0.30 = 1.0. Thus, table (a) is a valid probability distribution.
03

Evaluate Table (b)

For table (b), check if each probability is between 0 and 1. The values given are -0.4, 0.6, and 0.8. Since -0.4 is less than 0, this makes table (b) invalid as a probability distribution because probabilities must be non-negative.
04

Evaluate Table (c)

Sum the probabilities in table (c) and ensure each is between 0 and 1. The provided probabilities are 0.16, 0.14, 0.11, 0.87, and 0.22. Check they are all within the allowed range. Calculate the sum: 0.16 + 0.14 + 0.11 + 0.87 + 0.22 = 1.5. The sum is greater than 1, making table (c) invalid as a probability distribution.

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Key Concepts

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

Discrete Random Variable
A discrete random variable is a type of random variable that can take on only a specific set of distinct values. Unlike continuous random variables, which can assume any value within a given range, discrete random variables are countable. This distinct nature of discrete random variables means they often arise in situations where the outcomes can be counted, such as the number of heads in a series of coin flips or the sum of rolled dice.

Discrete random variables are crucial because they provide a way to critically analyze real-world processes. For instance:
  • The number of students attending a class can be considered as a discrete random variable.
  • The number of faulty items in a production line is another example.
  • Election results, such as the number of votes for a candidate, fit the description perfectly.
To navigate these values, a probability distribution is employed, where each outcome is associated with a probability that sums up to 1.
Valid Probability Distribution
Creating a valid probability distribution for a discrete random variable involves meeting specific criteria. Essentially, it's all about ensuring that the probabilities assigned to each possible outcome add up correctly and logically.

For a probability distribution to be valid, it needs to follow these key conditions:
  • Each probability assigned to an outcome must be between 0 and 1, inclusive.
  • The sum of all possible outcome probabilities must equal 1, indicating that one of these outcomes will definitely occur.
If even one of these conditions is violated, like assigning a negative probability or having a total sum greater or less than 1, the distribution is considered invalid.

Let's consider a table demonstrating a probability distribution. If it contains an outcome with a probability like -0.4, it immediately becomes invalid due to negative probability. Additionally, if after calculating the total sum, it ends up greater than 1, it renders the distribution meaningless, as seen in an example where the sum is 1.5.
Probability Criteria
To properly assess whether a set of probabilities forms a valid distribution, understanding the criteria behind the probabilities is essential. This involves a straightforward checklist that quantifies the validity of probability distributions.

#### Core Criteria
  • The range for each probability is strictly from 0 to 1: No probability should ever be negative or exceed 1.
  • The cumulative sum of all probabilities must be exactly 1: This ensures that all possible outcomes are considered and that the entire set represents absolute certainty (i.e., one outcome from the set must occur).

This criteria can be tested by straightforward arithmetic: adding up all probability values and double-checking each one for adherence to the rules. When analyzing a table of probabilities, an initial scan should flag any immediate errors, such as a probability being outside the acceptable range.

For instance, in a sequence, if you notice that the sum of probabilities equals 0.9 or 1.1, this immediately indicates a problem. Similarly, spotting a single probability of 1.2 or -0.5 flags a fundamental rule breach, leading to an invalid probability distribution.

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Most popular questions from this chapter

When dropped on a hard surface a thumbtack lands with its sharp point touching the surface with probability \(2 / 3 ;\) it lands with its sharp point directed up into the air with probability \(1 / 3\). The tack is dropped and its landing position observed 15 times. a. Find the probability that it lands with its point in the air at least 7 times. b. If the experiment of dropping the tack 15 times is done repeatedly, what is the average number of times it lands with its point in the air?

Tybalt receives in the mail an offer to enter a national sweepstakes. The prizes and chances of winning are listed in the offer as: \(\$ 5\) million, one chance in 65 million; \(\$ 150,000,\) one chance in 6.5 million; \(\$ 5,000\), one chance in 650,000 ; and \(\$ 1,000,\) one chance in 65,000 . If it costs Tybalt 44 cents to mail his entry, what is the expected value of the sweepstakes to him?

An insurance company will sell a \(\$ 90,000\) one-year term life insurance policy to an individual in a particular risk group for a premium of \(\$ 478\). Find the expected value to the company of a single policy if a person in this risk group has a \(99.62 \%\) chance of surviving one year.

Investigators need to determine which of 600 adults have a medical condition that affects \(2 \%\) of the adult population. A blood sample is taken from each of the individuals. a. Show that the expected number of diseased individuals in the group of 600 is 12 individuals. b. Instead of testing all 600 blood samples to find the expected 12 diseased individuals, investigators group the samples into 60 groups of 10 each, mix a little of the blood from each of the 10 samples in each group, and test each of the 60 mixtures. Show that the probability that any such mixture will contain the blood of at least one diseased person, hence test positive, is about 0.18 . c. Based on the result in (b), show that the expected number of mixtures that test positive is about 11. (Supposing that indeed 11 of the 60 mixtures test positive, then we know that none of the 490 persons whose blood was in the remaining 49 samples that tested negative has the disease. We have eliminated 490 persons from our search while performing only 60 tests.)

In a hamster breeder's experience the number \(X\) of live pups in a litter of a female not over twelve months in age who has not borne a litter in the past six weeks has the probability distribution \begin{tabular}{c|ccccccc} \(x\) & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline\(P(x)\) & 0.04 & 0.10 & 0.36 & 0.31 & 0.22 & 0.05 & 0.02 \end{tabular} a. Find the probability that the next litter will produce five to seven live pups. b. Find the probability that the next litter will produce at least six live pups. c. Compute the mean and standard deviation of \(X\). Interpret the mean in the context of the problem.

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