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A group of college students were surveyed to learn how many times they had visited a dentist in the previous year. \({ }^{14}\) The probability distribution for \(Y,\) the number of visits, is given by the following table: $$ \begin{array}{|cc|} \hline Y \text { (No. Visits) } & \text { Probability } \\ \hline 0 & 0.15 \\ 1 & 0.50 \\ 2 & 0.35 \\ \hline \text { Total } & 1.00 \\ \hline \end{array} $$ Calculate the mean, \(\mu_{Y},\) of the number of visits.

Short Answer

Expert verified
\(\mu_Y = 1.20\) visits per year

Step by step solution

01

Understand the Concept of Mean for a Discrete Random Variable

The mean or expected value of a discrete random variable is the sum of the products of each possible value the variable can take and its corresponding probability. This is represented by the formula \(\mu_Y = \sum (Y \times P(Y))\), where Y is the random variable representing the number of visits, and P(Y) is the probability of Y occurring.
02

List the Possible Values and Corresponding Probabilities

From the table provided, list the possible values of the random variable Y and their corresponding probabilities: 0 visits (P = 0.15), 1 visit (P = 0.50), and 2 visits (P = 0.35).
03

Calculate the Products of Values and Probabilities

For each possible value of Y, multiply it by its corresponding probability: \(0 \times 0.15 = 0\), \(1 \times 0.50 = 0.50\), \(2 \times 0.35 = 0.70\).
04

Sum the Products to Find the Mean

Add up all the products to calculate the mean number of visits: \(\mu_Y = 0 + 0.50 + 0.70 = 1.20\).

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Key Concepts

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

Probability Distribution
When studying a certain phenomenon like how often students visit a dentist, we encounter a set of values each with a particular likelihood of occurring. This is known as a probability distribution, and it's essentially a mathematical description of outcomes. For a discrete random variable, such as the number of dental visits by college students in a year, a probability distribution assigns a probability to each possible value.

Picture this: Every outcome that the random variable can take on is listed, just like in a menu at a restaurant. The probability distribution tells you how 'popular' each menu item is, that is, how likely it is to be chosen. In our dental visit example, the probabilities are 0.15 for 0 visits, 0.50 for 1 visit, and 0.35 for 2 visits. Notice that the probabilities add up to 1, which means we've accounted for all possible outcomes — a fundamental property of probability distributions.
Expected Value
The expected value, often symbolized as \(\mu\) or \(E(X)\), goes by many names — mean, average, first moment. But regardless of the term you use, it serves as a measure of the 'center' of the distribution, giving us an idea of the typical value we expect from our random variable.

So, how to find it? Imagine our dentist visit scenario as a game of chance. If you were to play this game over and over, what would be your average score? To compute the expected value, we take each number of visits, called \(Y\), and weigh it by its probability, \(P(Y)\). Calculating \(\sum (Y \times P(Y))\) gives us the sum of these weighted visits, which in our case resulted in 1.20 visits. This average represents what you might expect over many instances of the scenario.
Discrete Random Variable
A discrete random variable is like a box of chocolates with a limited selection — you know exactly what pieces you might get. It can only take on a countable number of values, such as 0, 1, or 2 dentist visits. Each value has its own probability that is greater than or equal to 0 and all probabilities summed together equal 1.

Think of it in terms of a simple coin flip: heads or tails. There are two possible outcomes, making the variable 'discrete'. Now, apply this to our example of dental visits. Here, our discrete random variable is the number of visits, with each possibility (0, 1, or 2) happening with different probabilities. It is an excellent way to handle real-world processes that don't have infinite possibilities but a distinct set of outcomes.

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Most popular questions from this chapter

Neuroblastoma is a rare, serious, but treatable disease. A urine test, the VMA test, has been developed that gives a positive diagnosis in about \(70 \%\) of cases of neuroblastoma. \({ }^{22}\) It has been proposed that this test be used for large-scale screening of children. Assume that 300,000 children are to be tested, of whom 8 have the disease. We are interested in whether or not the test detects the disease in the 8 children who have the disease. Find the probability that (a) all eight cases will be detected. (b) only one case will be missed. (c) two or more cases will be missed. [Hint: Use parts (a) and (b) to answer part (c).]

The following table shows the distribution of ages of Americans. \({ }^{3}\) Age distribution in reference population $$ \begin{array}{|cc|} \hline \text { Age } & \text { Proportion } \\ \hline 0-19 & 0.27 \\ 20-29 & 0.14 \\ 30-39 & 0.13 \\ 40-49 & 0.14 \\ 50-64 & 0.19 \\ 65+ & 0.13 \\ \hline \end{array} $$ Find the probability that the age of a randomly chosen American (a) is less than 20 . (b) is between 20 and \(49 .\) (c) is greater than \(49 .\) (d) is greater than \(29 .\)

In a certain college, \(55 \%\) of the students are women. Suppose we take a sample of two students. Use a probability tree to find the probability (a) that both chosen students are women. (b) that at least one of the two students is a woman.

Suppose that a disease is inherited via a sex-linked mode of inheritance so that a male offspring has a \(50 \%\) chance of inheriting the disease, but a female offspring has no chance of inheriting the disease. Further suppose that \(51.3 \%\) of births are male. What is the probability that a randomly chosen child will be affected by the disease?

In a certain population of the freshwater sculpin, Cottus rotheus, the distribution of the number of tail vertehrae is as shown in the table. $$ \begin{array}{|cr|} \hline \text { No. of vertebrae } & \text { Percent of fish } \\ \hline 20 & 3 \\ 21 & 51 \\ 22 & 40 \\ 23 & 6 \\ \hline \text { Total } & 100 \\ \hline \end{array} $$ Find the probability that the number of tail vertebrae in a fish randomly chosen from the population (a) equals 21 . (b) is less than or equal to 22 (c) is greater than 21 . (d) is no more than 21 .

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