91Ó°ÊÓ

Determine whether a probability distribution is given. If a probability distribution is given, find its mean and standard deviation. If a probability distribution is not given, identify the requirements that are not satisfied. In a survey, cell phone users were asked which ear they use to hear their cell phone, and the table is based on their responses (based on data from "Hemispheric Dominance and Cell Phone Use," by Seidman et al., JAMA Orolaryngology-Head \& Neck Surgery, VoL. 139, No. 5). $$\begin{array}{|l|l|} \hline & P(x) \\ \hline \text { Left } & 0.636 \\ \hline \text { Right } & 0.304 \\ \hline \begin{array}{l} \text { No } \\ \text { preference } \end{array} & 0.060 \\ \hline \end{array}$$

Step by step solution

01

Verify if the table is a Probability Distribution

To confirm if the given table is a probability distribution, the following requirements must be satisfied: 1. The probabilities must be between 0 and 1.2. The sum of the probabilities must be 1. First check each value in the table to ensure they are between 0 and 1.

02

Check values of P(x)

\[ P(\text{Left}) = 0.636 \] \[ P(\text{Right}) = 0.304 \] \[ P(\text{No preference}) = 0.060 \] Each probability value lies between 0 and 1.

03

Sum the probabilities

Sum the probabilities: \[ 0.636 + 0.304 + 0.060 = 1.000 \] Since the sum is 1, the table satisfies the requirement for a probability distribution.

04

Calculate the Mean

The mean (expected value) of a probability distribution is calculated as: \[ \mu = \sum [x \times P(x)] \] Use values: 0 (Left), 1 (Right), and 2 (No preference) for x: \[ \mu = 0 \times 0.636 + 1 \times 0.304 + 2 \times 0.060 = 0.424 \]

05

Calculate the Variance

The variance is calculated as: \[ \sigma^2 = \sum [(x - \mu)^2 \times P(x)] \] Use the mean \( \mu = 0.424 \): \[ \sigma^2 = (0 - 0.424)^2 \times 0.636 + (1 - 0.424)^2 \times 0.304 + (2 - 0.424)^2 \times 0.060 \] \[ \sigma^2 = 0.179776 \times 0.636 + 0.331776 \times 0.304 + 2.471616 \times 0.060 = 0.114376416 \]

06

Calculate the Standard Deviation

The standard deviation is the square root of the variance: \[ \sigma = \sqrt{0.114376416} \approx 0.338 \]

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Key Concepts

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

Mean of Probability Distribution

To understand the mean of a probability distribution, think of it as the average value you'd expect if you repeated a random event many times. It's also known as the expected value and is denoted by \( \mu \). The mean is calculated by multiplying each possible outcome (x) by its probability (P(x)), and then summing all these products:

\( \mu = \sum [x \times P(x)] \)

In the given example, users either prefer their left ear (0), right ear (1), or have no preference (2). We multiply each x value by its probability and sum them up:

\( \mu = 0 \times 0.636 + 1 \times 0.304 + 2 \times 0.060 = 0.424\)

So, the expected mean or average number of ear preference is 0.424, rounding to three decimal places.

Standard Deviation Calculation

Standard deviation helps to understand the spread of values in a probability distribution. It represents how much the values in the distribution deviate from the mean. The first step to calculating the standard deviation is to find the variance.

After finding the variance, the standard deviation is simply the square root of the variance:

\( \sigma = \sqrt{\sigma^2} \)

Following the example, we found the variance to be \sigma^2 = 0.1144. Taking the square root of this value gives us:

\( \sigma = \sqrt{0.1144} \approx 0.338 \)

So, the standard deviation is approximately 0.338.

Variance in Probability Distribution

Variance describes how far the values in a probability distribution are from the mean. It gives us an idea of the spread of the values. The formula for variance, usually denoted \( \sigma^2 \), is:

\( \sigma^2 = \sum [(x - \mu)^2 \times P(x)] \)

Using the previously calculated mean \( \mu = 0.424 \, \sigma^2 = (0 - 0.424)^2 \times 0.636 + (1 - 0.424)^2 \times 0.304 + (2 - 0.424)^2 \times 0.060 = 0.1144 \)

This value of 0.1144 is the variance, showing that the data points are spread moderately around the mean.

Conditions for Probability Distribution

For a table to qualify as a probability distribution, it must satisfy two main conditions:

• All probabilities must be between 0 and 1.
• The sum of all probabilities must equal 1.

In our given example, we checked each probability value:

• \( P(\text{Left}) = 0.636 \)
• \( P(\text{Right}) = 0.304 \)
• \( P(\text{No preference}) = 0.060 \)
No value exceeds 1, and none are less than 0.

Next, we need to ensure the probabilities sum up to 1:

\( 0.636 + 0.304 + 0.060 = 1.000 \)

Since both conditions are met, the table represents a valid probability distribution.