As a group, the Dutch are among the tallest people in the world. The average Dutch man is 184 cm tall - just over 6 feet. The standard deviation of men's height is about 8 cm. Assuming the distribution is approximately Normal, use the 68-95-99.7 Rule to sketch a model for the heights of Dutch men. What fraction of Dutch men should be less than 190 cm tall? type your answer... % What proportion of Dutch men should be taller than 205 cm tall? type your answer... %
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- In a Statistics examination, the mean score is 82 and the standard deviation is 3.5. What is the Z- score of a particular student whose score is 90? a) 2.92 b) 2.29 c) 2.19 d) 2.91Use Table 13.10 to answer each question. Note: Round z-scores to the nearest hundredth and then find the required A values using Table 13.10. The resting heart rates of a group of healthy adult men were found to have a mean of 73.4 beats per minute with a standard deviation of 5.9 beats per minute. What percent of these men had a resting heart rate of: b. between 70 and 85 beats per minute?In the 1992 presidential election, Alaska's 40 election districts averaged 1856 votes per district for President Clinton. The standard deviation was 552. (There are only 40 election districts in Alaska.) The distribution of the votes per district for President Clinton was bell-shaped. Let X = number of votes for President Clinton for an election district. (Source: The World Almanac and Book of Facts) Round all answers except part e. to 4 decimal places.a. What is the distribution of X? X ~ N(,)b. Is 1856 a population mean or a sample mean? Select an answer Sample Mean Population Mean c. Find the probability that a randomly selected district had fewer than 1792 votes for President Clinton. d. Find the probability that a randomly selected district had between 1777 and 1881 votes for President Clinton. e. Find the third quartile for votes for President Clinton. Round your answer to the nearest whole number.
- A box plot is constructed using several different values. Which of the following values are included in a box plot: the standard deviation the 90th percentile the smallest value the third quartile O the second quartileIn the 1992 presidential election, Alaska's 40 election districts averaged 2071 votes per district for President Clinton. The standard deviation was 591. (There are only 40 election districts in Alaska.) The distribution of the votes per district for President Clinton was bell-shaped. Let X = number of votes for President Clinton for an election district. (Source: The World Almanac and Book of Facts) Round all answers except part e. to 4 decimal places.a. What is the distribution of X? X ~ N(,)b. Is 2071 a population mean or a sample mean? Select an answer Population Mean Sample Mean c. Find the probability that a randomly selected district had fewer than 2029 votes for President Clinton. d. Find the probability that a randomly selected district had between 2247 and 2359 votes for President Clinton. e. Find the first quartile for votes for President Clinton. Round your answer to the nearest whole number.C. H=34, o= 4.5, the score is 2.5 standard deviations above the mean
- Approximately 6% of American drivers d not text while they drive. Suppose you are working with a sample group of 300 people.a. Find the mean and standard deviationPlease solve for D,E, and FIn the 1992 presidential election, Alaska's 40 election districts averaged 2121 votes per district for President Clinton. The standard deviation was 553. (There are only 40 election districts in Alaska.) The distribution of the votes per district for President Clinton was bell-shaped. Let X = number of votes for President Clinton for an election district. (Source: The World Almanac and Book of Facts) Round all answers except part e. to 4 decimal places.a. What is the distribution of X? X ~ N(,)c. Find the probability that a randomly selected district had fewer than 1996 votes for President Clinton. d. Find the probability that a randomly selected district had between 2306 and 2476 votes for President Clinton. e. Find the third quartile for votes for President Clinton. Round your answer to the nearest whole number.
- In the 1992 presidential election, Alaska's 40 election districts averaged 1979 votes per district for President Clinton. The standard deviation was 568. (There are only 40 election districts in Alaska.) The distribution of the votes per district for President Clinton was bell-shaped. Let X = number of votes for President Clinton for an election district. (Source: The World Almanac and Book of Facts) Round all answers except part e. to 4 decimal places. a. What is the distribution of X? X - N b. Is 1979 a population mean or a sample mean? Select an answer v c. Find the probability that a randomly selected district had fewer than 1898 votes for President Clinton. d. Find the probability that a randomly selected district had between 1998 and 2201 votes for President Clinton. e. Find the first quartile for votes for President Clinton. Round your answer to the nearest whole number.Researchers wanted to know if living in schools protects minnows from predators. The researchers placed different numbers of minnows into separate aquaria. Then, the researchers counted how many times per minute predatory fish approached the schools. Calculate the sample means, ?¯,x¯, and sample standard deviations, ?,s, for the number of times per minute a predator approached each school size. Round ?¯x¯ to one decimal place. Round ?s to two decimal places. Minnow school size Trial 3 5 10 15 20 1 0.9 0.7 0.4 0.4 0.1 2 0.8 0.8 0.5 0.5 0.1 3 0.7 0.9 0.6 0.3 0.2 4 1.1 0.6 0.8 0.2 0.3 5 1.0 1.0 0.7 0.6 0.3Solve for the mean, median, mode, midrange, range and sample standard deviation of the following items. Write NONE if there is no possible answer. The ones in the parenthesis will say the concept of rounding off.