want to obtain a sample to estimate how much parents spend on their kids birthday parties.  I believe the population standard deviation is approximately 45.1 dollars.  I would like to be 98% confident that I estimate is within 2 dollars of average spending on the birthday parties.  How many parents do I have to sample?

The sample mean amount of savings is $

In a recent survey of 621 working Americans ages 25-34, the average weekly amount spent on lunch was $44.61 with a standard deviation $3.03. The weekly amounts are approximately bell-shaped. Estimate the percentage of amounts that were greater than $47.64. Round the answer to one decimal place.

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You want to obtain a sample to estimate how much parents spend on their kids birthday parties. Based on previous study, you believe the population standard deviation is approximately ở = 69.9 dollars. You would like to be 95% confident that your estimate is within 3 dollar(s) of average spending on the birthday parties. How many parents do you have to sample? instructor

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In the 1992 presidential election, Alaska's 40 election districts averaged 2038 votes per district for President Clinton. The standard deviation was 576. (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 2038 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 1842 votes for President Clinton. d. Find the probability that a randomly selected district had between 1997 and 2099 votes for President Clinton. e. Find the first quartile for votes for President Clinton. Round your answer to the nearest whole number.