A survey of 82 students found that 29% were in favor of raising tuition to pave new parking lots. The standard deviation of the sample proportion is 6.6%. How large a sample (to the nearest person) would be required to reduce this standard deviation to 3.7%? Answer: 48

In the 1992 presidential election, Alaska's 40 election districts averaged 1980 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.PLEASE ANSWER C, D AND E AND ROUND TO THE 4 DECIMALc. Find the probability that a randomly selected district had fewer than 2062 votes for President Clinton. d. Find the probability that a randomly selected district had between 2052 and 2347 votes for President Clinton. e. Find the first quartile for votes for President Clinton. Round your answer to the nearest whole number.

Calculate the standard deviation for the following set of scores: 18, 22, 26, 22, 34, 7, 18.  (Use n-1 as the denominator in your calculations.)  The standard deviation for this data is: Group of answer choices

You have a sample of 48 randomly selected Vancouverites. On average (the sample mean), the people in your sample spent 425 hours eating food in 2020. The least time a person in the sample spent eating food in 2020 is 367 hours and the most time a person in the sample spent eating food in 2020 is 484 hours. What would be your best guess about the population standard deviation of the time a Vancouverite spent eating food in 2020? Give the number and explain (including assumptions you make about the population distribution).