Question

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

Modeling data distributions
ANSWERED
asked 2020-11-30

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

Expert Answers (1)

2020-12-01

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.

15
 
Best answer

expert advice

Have a similar question?
We can deal with it in 3 hours

Relevant Questions

asked 2021-08-04

This exercise requires the use of a graphing calculator or computer programmed to do numerical integration. The normal distribution curve, which models the distributions of data in a wide range of applications, is given by the function
\(\displaystyle{p}{\left({x}\right)}={\frac{{{1}}}{{\sqrt{{{2}\pi^{\sigma}}}}}}{e}-\frac{{\left({x}-\mu\right)}^{{2}}}{{{2}\sigma^{{2}}}}\)
where \(\pi\) = 3,14159265 ... and \(\sigma\) and \(\mu\) are constants called the standard deviation and the mean, respectively. Its graph (for \(\sigma=1\) and \(\mu=2)\) is shown in the figure.
With \(\displaystyle\sigma={\color{red}{{5}}}\) and \(\mu=0\), approximate \(\displaystyle{\int_{{{0}}}^{{+\infty}}}{p}{\left({x}\right)}{\left.{d}{x}\right.}\).(Round your answer to four decimal places.)

asked 2021-05-05


A random sample of \( n_1 = 14 \) winter days in Denver gave a sample mean pollution index \( x_1 = 43 \).
Previous studies show that \( \sigma_1 = 19 \).
For Englewood (a suburb of Denver), a random sample of \( n_2 = 12 \) winter days gave a sample mean pollution index of \( x_2 = 37 \).
Previous studies show that \( \sigma_2 = 13 \).
Assume the pollution index is normally distributed in both Englewood and Denver.
(a) State the null and alternate hypotheses.
\( H_0:\mu_1=\mu_2.\mu_1>\mu_2 \)
\( H_0:\mu_1<\mu_2.\mu_1=\mu_2 \)
\( H_0:\mu_1=\mu_2.\mu_1<\mu_2 \)
\( H_0:\mu_1=\mu_2.\mu_1\neq\mu_2 \)
(b) What sampling distribution will you use? What assumptions are you making?

The Student's t. We assume that both population distributions are approximately normal with known standard deviations.
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations.
The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.
(c) What is the value of the sample test statistic? Compute the corresponding z or t value as appropriate.
(Test the difference \( \mu_1 - \mu_2 \). Round your answer to two decimal places.)

(d) Find (or estimate) the P-value. (Round your answer to four decimal places.)
(e) Based on your answers in parts (i)−(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level \alpha?
At the \( \alpha = 0.01 \) level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
At the \( \alpha = 0.01 \) level, we reject the null hypothesis and conclude the data are statistically significant.
At the \( \alpha = 0.01 \) level, we fail to reject the null hypothesis and conclude the data are statistically significant.
At the \( \alpha = 0.01 \) level, we reject the null hypothesis and conclude the data are not statistically significant.
(f) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver. (g) Find a 99% confidence interval for
\( \mu_1 - \mu_2 \).
(Round your answers to two decimal places.)
lower limit
upper limit
(h) Explain the meaning of the confidence interval in the context of the problem.
Because the interval contains only positive numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, we can not say that the mean population pollution index for Englewood is different than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains only negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is less than that of Denver.

asked 2021-06-09

The following table represents the Frequency Distribution and Cumulative Distributions for this data set: 12, 13, 17, 18, 18, 24, 26, 27, 27, 30, 30, 35, 37, 41, 42, 43, 44, 46, 53, 58

\(\begin{array}{|c|c|} \hline \text{Class}&\text{Frequency}&\text{Relative Frequency}&\text{Cumulative Frequency}\\ \hline \text{10 but les than 20}&5\\ \hline \text{20 but les than 30}&4\\ \hline \text{30 but les than 40}&4\\ \hline \text{40 but les than 50}&5\\ \hline \text{50 but les than 60}&2\\ \hline \text{TOTAL}\\ \hline \end{array}\)

What is the Relative Frequency for the class: 20 but less than 30? State you answer as a value with exactly two digits after the decimal. for example 0.30 or 0.35

asked 2021-08-08

Presenting data in the form of table. For the data set shown by the table, Solve,
a) Create a scatter plot for the data.
b) Use the scatter plot to determine whether an exponential function, a logarithmic function, or a linear function is the best choice for modeling the data. (If applicable, you will use your graphing utility to obtain these functions.)
\(\begin{array}{|c|c|}\hline \text{Intensity (wattd per}\ meter^{2}) & \text{Loudness Level (decibels)} \\ \hline 0.1\text{(loud thunder)} & 110 \\ \hline 1\text{(rock concert, 2 yd from speakers)} & 120 \\ \hline 10 \text{(jackhammer)} & 130 \\ \hline 100 \text{(jet take off, 40 yd away)} & 140 \\ \hline \end{array}\)

...