An eyeglass manufacturer knows from past experience that 16% of manufactured lenses are defective dueto scratches. Suppose that the manufacturer examined seventeen eyeglasses. Assume that the binomialdistribution applies.(a) What are the number of trials n and the probability of "success" p in this case?n = 170, p = 0.16 0Use technology to determine the following probabilities, rounding each to 4 decimal places.(b) Probability that none of the lenses selected have scratches.Answer = 5.16 X(c) Probability that all of the lenses selected have scratches.Answer = 0.1362 X(d) Probability that at least three lenses have scratches.Answer = 52.66 X(e) Probability that at most three lenses have scratches.Answer = 71.59 X(f) Is it unusual for at least five lenses to be scratched?Probability of at least five lenses to be scratched = 12.24 XBecause this probability is greater than or equal toleast five lenses scratched.o 0.05, it is not unusualo to have at

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Answered: An eyeglass manufacturer knows from…

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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Use the scenario below to determine the correct values of n, p, q and x of the binomial distribution. Suppose that in a certain video game there is a 5.1% item drop rate of wild meat after defeating a Beast Master. Assume 14 Beast Masters are defeated. What is the probability that 5 wild meats will drop? n = p = q = X =
Suppose the number of cell phones in a household has a binomial distribution with parameters n = 15 and p = 85%. Find the probability of a household having: (a) 5 or 12 cell phones
Based on a poll, 70% of Internet users are more careful about personal information when using a public Wi-Fi hotspot. What is the probability that among threethree randomly selected Internet users, at least one is more careful about personal information when using a public Wi-Fi hotspot? How is the result affected by the additional information that the survey subjects volunteered to respond? The probability that at least one of them is careful about personal information is? (Round to three decimal places as needed.) How is the result affected by the additional information that the survey subjects volunteered to respond? A. It is very possible that the result is not valid because not everyone uses public Wi-Fi. B. The result should not be impacted by this because volunteers are likely to have the most relevant responses. C. The result should not be impacted by this because there is no reason to lie when responding. D. It is very possible that…
A survey showed that 83% of adults need correction (eyeglasses, contacts, surgery, etc.) for their eyesight. If 18 adults are randomly selected, find the probability that at least 17 of them need correction for their eyesight. Is 17 a significantly high number of adults requiring eyesight correction? ... The probability that at least 17 of the 18 adults require eyesight correction is (Round to three decimal places as needed.)
A concern for many adults is that computers and the Internet are reducing privacy. A survey conducted by Peter D. Hart Research Associates for the Shell Poll was reported in the USA Today. According to the survey, 37% of adults are concerned that employers are monitoring phone calls. Use the binomial distribution formula to calculate the probability that out of five adults, exactly three are concerned that employers are monitoring their phone calls. Select one: O a. 1.85 Ob. 0.799 Oc. 0.201 d. 3.15
About 69% of people who are murdered knew the person who committed the murder. Suppose that Detective Williams's case load currently has 49 unsolved murders. Use a nomal distribution approximation to the binomial distribution to find the following probabilities: a. What is the probability that at least 30 of the victims among Detective Williams's case files knew their murderer? b. What is the probability that between 30 and 39 of the victims knew their murderer?
Of all the students, 44% own an iPhone, with a standard deviation of .03 For that same sample of 400 students, what is the probability that at least 42% of them own an iPhone?
I. Boston Children’s Hospital estimates that 90% of Americans have had chickenpoxby the time they reach adulthood.32(a) Suppose we take a random sample of 100 American adults. Is the use of the binomial distributionappropriate for calculating the probability that exactly 97 out of 100 randomly sampled Americanadults had chickenpox during childhood? Explain.(b) Calculate the probability that exactly 97 out of 100 randomly sampled American adults had chickenpoxduring childhood.(c) What is the probability that exactly 3 out of a new sample of 100 American adults have not hadchickenpox in their childhood?(d) What is the probability that at least 1 out of 10 randomly sampled American adults have had chickenpox?(e) What is the probability that at most 3 out of 10 randomly sampled American adults have not hadchickenpox?
According to a study done by Nick Wilson of Otago University Wellington, the probability a randomly selected individual will not cover his or her mouth when sneezing is 0.267. Suppose you s bench in a mall and observe people's habits as they sneeze. Complete parts (a) through (c). (a) Using the binomial distribution, what is the probability that among 10 randomly observed individuals, exactly 8 do not cover their mouth when sneezing? The probability is (Round to four decimal places as needed.). (b) Using the binomial distribution, what is the probability that among 10 randomly observed individuals, fewer than 4 do not cover their mouth when sneezing? The probability is (Round to four decimal places as needed.) (c) Using the binomial distribution, would you be surprised if, after observing 10 individuals, fewer than half covered their mouth when sneezing? Why? it (Round to four decimal places as needed.) be surprising, because the probability is, which is C 0.05.
A recent study showed that 71% of adults need some type of corrective lens (glasses or contacts). A. If I grab a 6 random people what is the probability that fewer than 4 need corrective lenses? B. If I take a random sample of 2000 people what is the mean and standard deviation of the number of people who need corrective lens? C. Use the normal approximation to the binomial to find the probability that I get 1375 or fewer people who need corrective lenses out of a random sample of 2000. Illustrate what you are finding on a sketch of the normal curve and show your work.

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