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 p(x)=frac{1}{sqrt{2 pi}^{sigma}}e^{-(x-mu)^{2}}/(2 sigma^{2}) where pi = 3.14159265 . . . and sigma and mu are constants called the standard deviation and the mean, respectively. Its graph(text{for} sigma=1 text{and} mu=2)is shown in the figure. With sigma = 5 text{and} mu = 0, approximate int_0^{+infty} p(x) dx.

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 p(x)=frac{1}{sqrt{2 pi}^{sigma}}e^{-(x-mu)^{2}}/(2 sigma^{2}) where pi = 3.14159265 . . . and sigma and mu are constants called the standard deviation and the mean, respectively. Its graph(text{for} sigma=1 text{and} mu=2)is shown in the figure. With sigma = 5 text{and} mu = 0, approximate int_0^{+infty} p(x) dx.

Question
Modeling data distributions
asked 2021-01-31
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 \(p(x)=\frac{1}{\sqrt{2 \pi}^{\sigma}}e^{-(x-\mu)^{2}}/(2 \sigma^{2})\) where \(\pi = 3.14159265 . . .\) and sigma and mu are constants called the standard deviation and the mean, respectively. Its graph\((\text{for}\ \sigma=1\ \text{and}\ \mu=2)\)is shown in the figure. With \(\sigma = 5 \text{and} \mu = 0\), approximate \(\int_0^{+\infty}\ p(x)\ dx.\)

Answers (1)

2021-02-01
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Relevant Questions

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? NKS 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.) NKS (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.
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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
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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.
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asked 2020-10-23
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