Note: if the population standard deviations o1 and oz, then we use the standard normal distribution.

asked 2021-08-04

\(\displaystyle{p}{\left({x}\right)}={\frac{{{1}}}{{\sqrt{{{2}\pi^{\sigma}}}}}}{e}-\frac{{\left({x}-\mu\right)}^{{2}}}{{{2}\sigma^{{2}}}}\)

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 2021-05-14

When σ is unknown and the sample size is \(\displaystyle{n}\geq{30}\), there are tow methods for computing confidence intervals for μμ. Method 1: Use the Student's t distribution with d.f. = n - 1. This is the method used in the text. It is widely employed in statistical studies. Also, most statistical software packages use this method. Method 2: When \(\displaystyle{n}\geq{30}\), use the sample standard deviation s as an estimate for σσ, and then use the standard normal distribution. This method is based on the fact that for large samples, s is a fairly good approximation for σσ. Also, for large n, the critical values for the Student's t distribution approach those of the standard normal distribution. Consider a random sample of size n = 31, with sample mean x¯=45.2 and sample standard deviation s = 5.3. (c) Compare intervals for the two methods. Would you say that confidence intervals using a Student's t distribution are more conservative in the sense that they tend to be longer than intervals based on the standard normal distribution?

asked 2021-08-21

The amount of time Ricardo spends brushing his teeth follows a Normal distribution with unknown mean and standard deviation.

Ricardo spends less than one minute brushing his teeth about 40% of the time.

He spends more than two minutes brushing his teeth 2% of the time.

Use this information to determine the mean and standard deviation of this distribution.

Ricardo spends less than one minute brushing his teeth about 40% of the time.

He spends more than two minutes brushing his teeth 2% of the time.

Use this information to determine the mean and standard deviation of this distribution.

asked 2021-06-23

The pathogen Phytophthora capsici causes bell pepper plants to wilt and die. A research project was designed to study the effect of soil water content and the spread of the disease in fields of bell peppers. It is thought that too much water helps spread the disease. The fields were divided into rows and quadrants. The soil water content (percent of water by volume of soil) was determined for each plot. An important first step in such a research project is to give a statistical description of the data.
Soil Water Content for Bell Pepper Study
1591510amp;14amp;15amp;12amp;10amp;14amp;12amp;9amp;10amp;14amp;9amp;10amp;11amp;13amp;10amp;9amp;9amp;12amp;7amp;9amp;amp;11amp;14amp;16amp;amp;11amp;13amp;16amp;amp;11amp;14amp;12amp;amp;11amp;8amp;10amp;amp;10amp;9amp;11amp;amp;11amp;8amp;11amp;amp;13amp;11amp;12amp;amp;16amp;13amp;15amp;amp;10amp;13amp;6
(a) Make a box-and-whisker plot of the data. Find the interquartile range.

asked 2021-05-01

Which distribution (standard normal or Student's t) did you use? Why? Do you need information about the soil water content distributions?

asked 2021-06-05

Use the following Normal Distribution table to calculate the area under the Normal Curve (Shaded area in the Figure) when \(Z=1.3\) and \(H=0.05\);

Assume that you do not have vales of the area beyond \(z=1.2\) in the table; i.e. you may need to use the extrapolation.

Check your calculated value and compare with the values in the table \([for\ z=1.3\ and\ H=0.05]\).

Calculate your percentage of error in the estimation.

How do I solve this problem using extrapolation?

\(\begin{array}{|c|c|}\hline Z+H & Prob. & Extrapolation \\ \hline 1.20000 & 0.38490 & Differences \\ \hline 1.21000 & 0.38690 & 0.00200 \\ \hline 1.22000 & 0.38880 & 0.00190 \\ \hline 1.23000 & 0.39070 & 0.00190 \\ \hline 1.24000 & 0.39250 & 0.00180 \\ \hline 1.25000 & 0.39440 & 0.00190 \\ \hline 1.26000 & 0.39620 & 0.00180 \\ \hline 1.27000 & 0.39800 & 0.00180 \\ \hline 1.28000 & 0.39970 & 0.00170 \\ \hline 1.29000 & 0.40150 & 0.00180 \\ \hline 1.30000 & 0.40320 & 0.00170 \\ \hline 1.31000 & 0.40490 & 0.00170 \\ \hline 1.32000 & 0.40660 & 0.00170 \\ \hline 1.33000 & 0.40830 & 0.00170 \\ \hline 1.34000 & 0.41010 & 0.00180 \\ \hline 1.35000 & 0.41190 & 0.00180 \\ \hline \end{array}\)

Assume that you do not have vales of the area beyond \(z=1.2\) in the table; i.e. you may need to use the extrapolation.

Check your calculated value and compare with the values in the table \([for\ z=1.3\ and\ H=0.05]\).

Calculate your percentage of error in the estimation.

How do I solve this problem using extrapolation?

\(\begin{array}{|c|c|}\hline Z+H & Prob. & Extrapolation \\ \hline 1.20000 & 0.38490 & Differences \\ \hline 1.21000 & 0.38690 & 0.00200 \\ \hline 1.22000 & 0.38880 & 0.00190 \\ \hline 1.23000 & 0.39070 & 0.00190 \\ \hline 1.24000 & 0.39250 & 0.00180 \\ \hline 1.25000 & 0.39440 & 0.00190 \\ \hline 1.26000 & 0.39620 & 0.00180 \\ \hline 1.27000 & 0.39800 & 0.00180 \\ \hline 1.28000 & 0.39970 & 0.00170 \\ \hline 1.29000 & 0.40150 & 0.00180 \\ \hline 1.30000 & 0.40320 & 0.00170 \\ \hline 1.31000 & 0.40490 & 0.00170 \\ \hline 1.32000 & 0.40660 & 0.00170 \\ \hline 1.33000 & 0.40830 & 0.00170 \\ \hline 1.34000 & 0.41010 & 0.00180 \\ \hline 1.35000 & 0.41190 & 0.00180 \\ \hline \end{array}\)

asked 2021-10-04

Assume that all the distributions are normal, average is always taken to be the arithmetic mean xˉ or μ. A college Physical Education Department offered an Advanced First Air course last semester. The scores on the comprehensive final exam were normally distributed, and the the z scores for some of the students are shown below:
Robert, 1.11
Juan, 1.66
Susan, –1.9
Joel, 0.00
Jan, –0.65
Linda, 1.46
(c) Which of these students scored below the mean?