A report found that 60% of Americans of insurance fraud are young adults. Assuming this is...
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report found that 60% of Americans of insurance fraud are young
adults. Assuming this is...
it was found that 80 percent of americans victims of healthcare fraud are senior citizens. if...
it was found that 80 percent of americans victims of healthcare fraud are senior citizens. if 9 victims are randomly selected, find the probability that 5 are senior citizens. round your answers to three decimal places. P(exactly 5)=
A survey found that 60% of American victims of health care fraud are senior citizens. If...
A survey found that 60% of American victims of health care fraud are senior citizens. If 10 victims are selected at random, find (A) the mean number of victims who are senior citizens. (B) the probability that exactly 3 victims are senior citizens. (C) the probability that at most 6 victims are senior citizens. (D) the probability that all but one victim are senior citizens.
A survey found that 60% of American victims of health care fraud are senior citizens. If...
A survey found that 60% of American victims of health care fraud are senior citizens. If 10 victims are selected at random, find (A) the mean number of victims who are senior citizens. (B) the probability that exactly 3 victims are senior citizens. (C) the probability that at most 6 victims are senior citizens. (D) the probability that all but one victim are senior citizens. in words, what is the success and failure part of this binomial problem? also can...
A survey found that 60% of American victims of health care fraud are senior citizens. If...
A survey found that 60% of American victims of health care fraud are senior citizens. If 10 victims are selected at random, find (A) the mean number of victims who are senior citizens. (B) the probability that exactly 3 victims are senior citizens. (C) the probability that at most 6 victims are senior citizens. (D) the probability that all but one victim are senior citizens. I only need help with A,C, and D. I know that B is 0.0425. If...
A survey found that 60% of American victims of health care fraud are senior citizens. If...
A survey found that 60% of American victims of health care fraud are senior citizens. If 10 victims are selected at random, find (A) the mean number of victims who are senior citizens. ( C) the probability that at most 6 victims are senior citizens. (D) the probability that all but one victim are senior citizens. D is the one I need most help in, I have gotten the answer "0.0016" "0.0403" and "0.999895" im very confused with it. also...
A survey found that 60% of American victims of health care fraud are senior citizens. If...
A survey found that 60% of American victims of health care fraud are senior citizens. If 10 victims are selected at random, find (A) the mean number of victims who are senior citizens. (C) the probability that at most 6 victims are senior citizens. (D) the probability that all but one victim are senior citizens. D is the one I need most help in, I have gotten the answer "0.0016" "0.0403" and "0.999895" im very confused with it. Also can...
Based on a? poll, among adults who regret getting? tattoos, 14 14?% say that they were...
Based on a? poll, among adults who regret getting? tattoos, 14 14?% say that they were too young when they got their tattoos. Assume that five five adults who regret getting tattoos are randomly? selected, and find the indicated probability. Complete parts? (a) through? (d) below. a. Find the probability that none of the selected adults say that they were too young to get tattoos. nothing ?(Round to four decimal places as? needed.) b. Find the probability that exactly one...
Based on a poll, among adults who regret getting tattoos, 15 15% say that they were...
Based on a poll, among adults who regret getting tattoos, 15 15% say that they were too young when they got their tattoos. Assume that eight eight adults who regret getting tattoos are randomly selected, and find the indicated probability. Complete parts (a) through (d) below. a. Find the probability that none of the selected adults say that they were too young to get tattoos. nothing (Round to four decimal places as needed.) b. Find the probability that exactly one...
6)A survey of adults in the United States found that 60% ate at a restaurant at...
6)A survey of adults in the United States found that 60% ate at a restaurant at least once in the past week. You randomly select 28 adults and ask them whether they ate at a restaurant at least once in the past week (a) Verify that a normal distribution can be used to approximate the binomial distribution. (b) Find the probability that at most 13 adults ate at a restaurant at least once in the past week.
5. In Salinas, 68% of adults have health insurance. If 400 adults were randomly selected. a)...
5. In Salinas, 68% of adults have health insurance. If 400 adults were randomly selected. a) Find the probability that at least 230 of them have health insurance. (Use binomial approximation) b) Find the probability that exactly 230 of them have health insurance. (Use binomial approximation)
5. In Salinas, 68% of adults have health insurance. If 400 adults were randomly selected. a) Find the probability that at least 230 of them have health insurance. (Use binomial approximation) b) Find the probability that exactly 230 of them have health insurance. (Use binomial approximation)
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