The distribution of heights of adult American men is approximately normal with a mean of 69 inches and a standard deviation of 2.5 inches. a. What percent of these men are at least 71.5 inches tall? b. What percent of these men are between 64 and 74 inches tall?
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The distribution of heights of adult American men is approximately normal with a mean of 69 inches and a standard deviation of 2.5 inches. a. What percent of these men are at least 71.5 inches tall? b. What percent of these men are between 64 and 74 inches tall?
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- A bakery owner tabulated sales to 50 customers and obtained the distribution. Find the mean and standard deviation for the data. Sales in dollars Number of customers0.00 – 1, 99 18 2.00 – 3.99 164.00 – 5.99 9 6.00 – 7.99 48.00 – 9.99 3In the table below, the distribution of the height of 100 students is given. According to this; a. mean deviation b. Find the variance and standard deviation?In the 1992 presidential election, Alaska's 40 election districts averaged 1824 votes per district for President Clinton. The standard deviation was 573. (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 1824 a population mean or a sample mean? c. Find the probability that a randomly selected district had fewer than 1872 votes for President Clinton. d. Find the probability that a randomly selected district had between 1842 and 2091 votes for President Clinton. e. Find the first quartile for votes for President Clinton. Round your answer to the nearest whole number.
- The heights of Crepe Myrtle saplings at a local nursery are measured and the data recorded. The data is then plotted and the distribution is found to be bell-shaped. The mean of the set is 67 inches and the standard deviation is 3 inches. A graph of the situation is shown below: 58 61 64 67 70 73 76 a) What percentage of the saplings is more than 70 inches tall? b) What percentage of the saplings is between 61 and 73 inches tall?In a science test, the mean score is 42 and the standard deviation is 5. Assuming the scores are normally distributed, what percent of score is: a. greater than 48 b. less than 50 c. between 30 and 48In the 1992 presidential election, Alaska's 40 election districts averaged 1993 votes per district for President Clinton. The standard deviation was 584. (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 1993 a population mean or a sample mean? Select an answer v C. Find the probability that a randomly selected district had fewer than 2053 votes for President Clinton. d. Find the probability that a randomly selected district had between 2074 and 2208 votes for President Clinton. e. Find the 95th percentile for votes for President Clinton. Round your answer to the nearest whole number.
- Consider the following data set. You may assume that the mean of this data set is x¯ = 7. 8 10 3 5 7 5 11 a) Find the range of the data set. b) Find the standard deviation of the data set. c) How many of the seven data points lie within one standard deviation of the mean?According to Chebyshev's theorem, which of the following is the percentage of measurements in a data set that fall within three standard deviations of their mean? Select one: a. 75%. b. At least 75%. c. 89%. d. At least 89%. Clear my choiceIn the 1992 presidential election, Alaska's 40 election districts averaged 1898 votes per district for President Clinton. The standard deviation was 599. (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 1898 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 1904 votes for President Clinton. d. Find the probability that a randomly selected district had between 2006 and 2197 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 1989 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 1989 a population mean or a sample mean? Select an answer _______. c. Find the probability that a randomly selected district had fewer than 2087 votes for President Clinton. _____ d. Find the probability that a randomly selected district had between 1905 and 2194 votes for President Clinton.______ e. Find the first quartile for votes for President Clinton. Round your answer to the nearest whole number. ______A biologist found the wing span of a group of MONARY butterflies to be normally distributed with a mean of 53.7 mm and a standard deviation of 2.5 mm. What percent of the butterflies have the following wingspan round to one Decimal decimal place what percent is less than 48.5 mm What percent is between 50 and 56 mmThe average driver spends $39 at the gas station each week with a standard deviation of $11. Assuming that the amount a driver spends on gas follows a normal distribution; find the percentage of drivers who spend: 1. More than $67? 2. Less than $25? 3. Between $25 and $67
