Al Bundy is evaluating a new advertising program that could increase shoe sales. Possible outcomes and probabilities of the outcomesare shown next.Possible OutcomesIneffective campaignNormal responseExtremely effectiveAdditionalSales in Units40190370Coefficient of variationProbabilities0.200.300.50Compute the coefficient of variation. (Do not round intermediate calculations. Round your answer to 3 decimal places.)

The formula of coefficient of variation is,From the information, given the probability distribution…

Answered: Al Bundy is evaluating a new…

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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Doctors are testing a new antidepressant. A group of patients, all with similar characteristics, take part in the study. Some of the patients receive the new drug, while others receive the traditional drug. During the study, a number of patients complain about insomnia. The data are shown in the contingency table below. What is the relative risk of insomnia for those who receive the new drug? Round to two decimal places. New drug Total Insomnia No insomnia Total Traditional drug 36 Provide your answer below: 52 88 226 295 521 278 331 609
Describe in detail the results achieved on the company’s test. Ignore other variables for now, just describe the test results. If the company is considering training for the lowest 10% of employees what score would they need to achieve to be exempted from training? Test Result 163 148 142 173 168 155 163 194 119 124 149 171 160 104 165 154 122 160 148 138 158 112 189 163 203 165 176 152 74 129 166 134 124 169 187 163 172 195 149 107 156 183 133 83 148 127 173 95 139 179 136 153 141 155 179 152 129 155 145 113 142 147 171 139 174 143 96 199 143 172 123 146 128 187 138 163 178 154 199 164 123 201 205 135 123 138 195 184 165 196 154 176 82 173 156 149 152 116 174 141 111 74 123 182 157 141 167 162 172 107 149 182 112 135 177 159 138 163 148 198…
The table below gives the percent of U.S. seniors (aged 65+ years) who used the internet in selected years. Year 2012 2014 2016 2018 2019 % of U.S. seniors using internet 54 57 64 66 73 Use a line of best-fit to predict the percent of U.S. seniors who will use the internet in the year 2022. A. 74 % B.76 % C.77 % D.79 % E.81 %
Lesson 7Q8 Identify the well-known consistent estimators. A.Sample mean B.Sample variance C.Sample mean and sample variance D.Population mean and population variance
Q1
5 The income of one policy holder is 69 ($1,000) The predicted life insurance policy is _______ ($1,000) a 263.578 b 274.561 c 286.001 d 297.917
Researchers are wondering whether they can predict college GPA with the number of hours studying per week. To test this, they randomly select 8 college students, record their GPA and the number of hours spent studying per week. GPA 2.5 2.7 3.4 3.7 2.7 1.9 2.0 4.0 ID 1 2 3 4 5 6 7 8 Hours studying per week 25.0 27.6 30.2 35.0 31.6 25.6 30.0 40.0 Based on the information in the table above, the researchers compute the regression equation: Y = -0.970 + 0.125 X. Do the researchers have enough evidence to claim that the number of hours studying predicts students' GPA? Based on the analysis, what would be the predicted GPA for someone who studies 30 hours per week? O2.00 O Cannot be predicted. O 1.25 2.78
2 part question done in Excel. 1. Data are collected in a clinical trial evaluating a new compound designed to improve wound healing in trauma patients. The new compound is compared against a placebo. After treatment for 5 days with the new compound or placebo, the extent of the wound healing is measured and the date are show in the table below. Is there a difference in the extent of wound healing by treatment (Hint: Are treatment and the percent wound healing independent?) Run the appropriate tes at a 5% level of significance. 2. Use the data in Problem 1 and pool the data across the treatments into one sample of size n=250. Use the pooled data to test whether the distribution of the percent wound healing is approximately normal. Specifically, use the following distributions: 30%, 40%, 20%, and 10%. Run the appropriate test at a = 0.05.
IQ tests supposedly measure our intelligence. They typically are ‘normed’ so that they have a mean of 100 and a standard deviation of 15. (Higher scores show higher intelligence.) Use the graph and use the z-score and percents from columns b and c in Table A, when relevant. If you scored 132, what is your z-score? (Interpret.) Use table A X, z(a), (b), and (c) X z (a) (b) (c) and use the graph! Write down or save your response to use in Part 2 of this quiz. Group of answer choices -2.53 z. I'm at the 98.34 percentile rank and 1.66% of people scored smarter than me. 2.13 z. I'm at the 98.34 percentile rank and 1.66% of people scored smarter than me. -2.13 z. I'm at the 1.66 percentile rank and 98.34% of people scored smarter than me. 2.53 z. I'm at the 0.57 percentile rank and 99.43% of people scored smarter than me. 2.13 z. I'm at the 1.66 percentile rank and 98.34% of people scored smarter than me. 2.53 z. I'm at the 99.43 percentile rank…
z stat please

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