Identifying Probability Distributions. In Exercises 7–14, determine whether a probability distribution is given. If a probability distribution is given, find its mean

ddaeeric 2020-11-30 Answered

Identifying Probability Distributions. In Exercises 7–14, 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. Cell Phone Use 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 Otolaryngology—Head & Neck Surgery , Vol. 139, No.

5). \(\begin{array}{|c|c|}& P(x)\\ Left & 0.636 \\ Right & 0.304 \\ No\ preference & 0.060\end{array}\)

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FieniChoonin
Answered 2020-12-01 Author has 102 answers

Requirements: The following requirements should be satisfied for the distribution to follow the probability distribution:

1. The given random variable (x) must be a numerical value and should have its own corresponding probabilities.

2. The sum of all probabilities must be equal to

1. That is, \(\displaystyle\sum{P}{\left({x}\right)}={1}\)P 3. The probability values must lie between 0 and 1, inclusive. That is, \(\displaystyle{0}\leq{P}{\left({x}\right)}\leq{1}\) The given data does not have a random variable (x) and the preferences to hear the cell phone can be left, right, or no preference. That is, there is no numerical value for the random variable. Hence, the requirement is not satisfied. The sum of all probabilities is, \(\displaystyle\sum{P}{\left({x}\right)}={0.636}+{0.304}+{0.060}={1}\)

Hence, the requirement 2 is satisfied. In the given distribution, all the values of P(x) for random variable x lie between 0 and 1. Hence, the requirement 3 is satisfied. Since, the requirement 1 is not satisfied. That is, the requirement “The given random variable (x) must be a numerical value” is not satisfied. Hence, the distribution preference of the ear used to hear the cell phone is not a probability distribution.

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