Probability of errorsA large accountancy firm finds that over the long run 10% of its statements forclients are in error in some way. A quality assurance officer at the firm investigatingthe source of the errors takes a sample of 30 statements produced by a singleemployee and classifies each as being "in error" or "not in error".Question 10 (2 points)If the variable X is used to represent the number of statements in the sample that arein error, and assuming the error rate is the firm's level of 10% then state:⚫ the type of distribution this variable has; (1 mark) andthe parameter/s of this distribution. (1 mark)

This is a binomial distribution since it satisfies the following conditions:Fixed number of trials:…

Answered: Probability of errors A large…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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Dataset: NES. Variables: spend8, nesw.) The NES dataset contains spend8, which records the number of government policy areas where respondents think spending should be increased. Scores range from 0 (the respondent does not want to increase spending on any of the policies) to 8 (the respondent wants to increase spending on all eight policies). The NES, of course, polls a random sample of U.S. adults. In this exercise you will analyze spend8 using the Analyze ▶ Compare Means ▶ One-Sample T Test procedure. You then will draw inferences about the population mean. Make sure you weight observations using the nesw variable so your results are nationally representative. A. The spend8 variable has a sample mean of ______.
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Determine if the following statements are true or false. a. Jane took a sample of n = 100 observations from a population with mean and standard deviation. Mike also took a sample of n = 200 observations from the same population. Then, Mike’s sample mean will be closer to the true population mean than Jane’s sample. b. Jane took a sample of n = 100 observations from a population with mean and standard deviation. Mike also took a sample of n = 200 observations from a population with mean and standard deviation 2. Then, the sample mean from Mike’s sample is more likely to be close to the true mean than the sample mean from Jane’s sample. That is, the probability that X is between 90% and 110% of mean is higher for Mike’s sample than for Jane’s sample.
BUS-173 (Applied Statistics)
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