binomial distribution 42% of visitors at the zoo buy a grape juice drink at the concession stands. Assuming a binomialdistribution, 17 visitors are selected at random, what is the probability at least 5 bought a grape juicedrink?0.84680.78790.09490.21210.9051

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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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Subgroup of Participants Supine SBP Mean (SD) Standing SBP Mean (SD) Percent with > 5.0 Decrease Percent with > 5.0 Increase White women (N = 301) 120.4 (18.4) 120.0 (18.8) 32.6% 32.2% African-American women (N = 107) 122.7 (19.3) 122.8 (19.4) 30.2% 31.9% White men (N = 339) 121.2 (18.9) 120.2 (19.9) 39.5% 25.6% African-American men (N = 95) 124.7 (20.4) 125.9 (21.6) 13.7% 51.3% Total (N = 842) 121.5 (19.0) 121.1 (19.5) 29.8% 31.7% Would it be possible to analyze data in table 2 with a chi-square test? if yes, specify how
Hypothesis Test Consider the information above and in question 1. It is conjectured that the mean number of people attending all 2017 Veterans Day celebrations is 145, and of interest is to test this conjecture versus the alternative that the mean number of people attending all 2017 Veterans Day celebrations is different from 145 people To test the hypothesis, a simple random sample of 41 Veterans Day celebrations was selected and the number of people attending each was recorded. The mean number of people attending for this sample of 41 celebrations was 157, with a standard deviation of 23 people. The distribution of the data was bimodal. If appropriate, use this information to test the hypotheses stated in question 3 at the a = .05 level of significance. Is the population standard deviation known? a) Yes b) No Based on the p-value and using a significance level α = 0.05, which of the following is the correct decision? Based on the p-value and using a…

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42% of visitors at the zoo buy a grape juice drink at the concession stands. Assuming a binomial distribution, 17 visitors are selected at random, what is the probability at least 5 bought a grape juice drink? 0.8468 0.7879 O0.0949 0.2121 0.9051