StateParolees (per 100,000 people)Florida28Illinois299Pennsylvania1,035Ohio193Georgia334Michigan239North Carolina130New Jersey214Virginia27Source: National Institute of Corrections, Correction Statistics by State, 2016.a. Assume that o = 226.83 for the entire population of 50 states. Calculate and interpretthe standard error. (Consider the formula for the standard error. Since we providedthe population standard deviation, calculating the standard error requires only minorcalculations.)b. Write a brief statement on the following: the standard error compared with thestandard deviation of the population, the shape of the sampling distribution, andsuggestions for reducing the standard error. lon?C12. The following table presents the number of parolees (per 100,000 people) for 12 of themost populous states as of July 2015.StateParolees (per 100,000 people)California292Texas556New York288

Given, State Parolees (per 100,000 people) California 292 Texas 556 New York 288 Florida…

Answered: a. Assume that o = 226.83 for the…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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A simple random sample of 100 observations was taken from a large population. The sample mean and the standard deviation were determined to be 81 and 9 respectively. The standard error of the mean (i.e. the standard error of the sampling distribution) is 8.10 0.90 O 0.81 O 0.27 Clear my choice
A primary measure of dispersion for sample data is the sample standard deviation, s. Choose the best option to explain what the meaning of the sample standard deviation is, in general. A. The distance between one observation and the mean. B. The average distance of an observation from the middle of the distribution. C. The average distance between observations in the sample. D. The distance between the third and first quartiles.
A population has a mean p= 79 and a standard deviation o = 12. Find the mean and standard deviation of a sampling distribution of sample means with sample size n = 36. %3D H=(Simplify your answer.) %3D (Simplify your answer)
In the 1992 presidential election, Alaska's 40 election districts averaged 1993 votes per district for President Clinton. The standard deviation was 584. (There are only 40 election districts in Alaska.) The distribution of the votes per district for President Clinton was bell-shaped. Let X = number of votes for President Clinton for an election district. (Source: The World Almanac and Book of Facts) Round all answers except part e. to 4 decimal places. a. What is the distribution of X? X ~ N( b. Is 1993 a population mean or a sample mean? Select an answer v C. Find the probability that a randomly selected district had fewer than 2053 votes for President Clinton. d. Find the probability that a randomly selected district had between 2074 and 2208 votes for President Clinton. e. Find the 95th percentile for votes for President Clinton. Round your answer to the nearest whole number.
A nurse at a local hospital is interested in estimating the birth weight of infants. How large a sample must she select if she desires to be 98% confident that the true mean is within 2 ounces of the sample mean? The population standard deviation of the birth weights is known to be 5 ounces. O A. 6 O B. 33 O C. 34 O D. 5 Click to select your answer. DELL EVD RW Op DELL
For the following situation, find the mean and standard deviation of the population. List all samples (with replacement) of the given size from that population and find the mean of each. Find the mean and standard deviation of the sampling distribution and compare them with the mean and standard deviation of the population. The word counts of five essays are 506, 623, 544, 614, and 587. Use a sample size of 2.
4. The variance of a data set is 3410.56. Find the standard deviation for the data set. This ample of
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A population has a mean p= 75 and a standard deviation o= 6. Find the mean and standard deviation of a sampling distribution of sample means with sample size n= 36. (Simplify your answer.)
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The average income in a state was $64 comma 64,000 per person per year. Suppose the standard deviation is $29 comma 29,000 and the distribution is right-skewed. Suppose we take a random sample of 100 residents of the state. Complete parts (a) and (b) below. a. What value should we expect for the sample mean? Why? b. What is the standard error for the sample mean?
On average, accountants earn approximately an average of $6500 per year in the United States. Assume that the distribution for electricians’ yearly earnings is normally distributed and that the standard deviation is $8680. Given a sample of 64 electricians, what is the standard deviation for the sampling distribution of the sample mean? a. 1205 b. 8680 c. 6000 d. 1085

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