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Use a normal approximation to find the probability ofthe indicated number of voters. In this case, assume that 156 eligible voters aged 18-24 are randomly selected. Suppose a previous study showed that among eligible voters aged 18-24, 22% of them voted. Probability that fewer than 39 voted The probability that fewer than 39 of 156 eligible voters voted is (Round to four decimal places as needed.) In a survey of 1200 people, 790 people said they voted in a recent presidential election. Voting records show that 63% of eligible voters actually did vote. Given that 63% of eligible voters actually did vote, (a) nd the probability that among 1200 randomly selected voters, at least 790 actually did vote. (b) What do the results from part (a) suggest? (a) P(X 2 790) = (Round to four decimal places as needed.) The overhead reach distances of adult females are normally distributed with a mean 01 200 cm and a standard deviation of 8.9 cm. a. Find the probability that an individual distance is greater than 209.30 cm. b. Find the probability that the mean for 15 randomly selected distances is greater than 198.20 cm. c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30? E) a. The probability is D. (Round to four decimal places as needed.) If a procedure meets all of the conditions of a binomial distribution except the number of trials is not xed, then the geometric distribution can be used. The probability of getting the first success on the xth trial is given by P(x) = p(1 - p)x_1, where p is the probability of success on any one trial. Subjects are randomly selected for a health survey. The probability that someone is a universal donor (with group O and type Rh negative blood) is 0.04. Find the probability that the rst subject to be a universal blood donor is the eighth person selected. The probability is . (Round to four decimal places as needed.) If we sample from a small nite population without replacement, the binomial distribution should not be used because the events are not independent. If sampling is done without replacement and the outcomes belong to one of two types, we can use the hypergeometric distribution. If a population has A objects of one type, while the remaining B objects are of the other type, and if n objects are sampled without replacement, then the probability of getting x objects of type A and n - x objects of type B under the hypergeometric distribution is given by the following formula. In a lottery game, a bettor selects six numbers from 1 to 53 (without repetition), and a winning six-number combination is later randomly selected. Find the probabilities of getting exactly two winning numbers with one ticket. (Hint: Use A = 6, B = 47, n = 6, and x = 2.) A! B! (A+ B)! p(X) = . + (A - x)!x! (B - n + x)!(n x)! (A + B - n)!n! P(2) = (Round to four decimal places as needed.) The waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between 0 and 7 minutes. Find the probability that a randomly selected passenger has a waiting time greater than 3.25 minutes. Find the probability that a randomly selected passenger has a waiting time greater than 3.25 minutes. (Simplify your answer. Round to three decimal places as needed.) A statistics professor plans classes so carefully that the lengths of her classes are uniformly distributed between 46.0 and 56.0 minutes. Find the probability that a given class period runs between 51.25 and 51.75 minutes. Find the probability of selecting a class that runs between 51.25 and 51.75 minutes. (Round to three decimal places as needed.)