Help me answer this 1.) A large corporation employs 37685 individuals. The average income of all employees is $69665, with a standard deviation of $19652 and is skewed to the right. Consider this to be the population distribution. You are given a data set consisting of the incomes of 60 randomly selected employees. The average of the sample is x = 63669 x = 63669 . Since the sample size is large, the Central Limit Theorem tells us that the sampling distribution for x x is approximately Normal. The sampling distribution of the sample mean is centered at mean = The sampling distribution has a standard deviation of SD=standard deviation/ square root of n . Round to two decimal places The z-score for the sample mean is z= ( x bar - mean )/ SD = . Round to two decimal places. Since the z-score is (between -1.96 and 1.96 not between -1.96 and 196) this sample mean is (typicalunusual) 2.) A consumer watchdog organization estimates the mean weight of 1-ounce "Fun-Size" candy bars to see if customers are getting full value for their money. A random sample of 31 bars is selected and weighed, and the organization reports that a 90% confidence interval for the true mean weight of the candy bars is 0.99 to 1 ounces. What is the point estimator used? ____ What is the point estimate? _____ What is the margin of error? ____ 3. If n=13, x x (x-bar)=48, and s=9, find the margin of error at a 95% confidence level Give your answer to two decimal places. 4. ) Assume that a sample is used to estimate a population proportion . Find the margin of error M.E. that corresponds to a sample of size 77 with a mean of 15.4 and a standard deviation of 5.4 at a confidence level of 95%. Report ME accurate to one decimal place because the sample statistics are presented with this accuracy. M.E. = Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places. 5.) A political candidate has asked you to conduct a poll to determine what percentage of people support her. If the candidate only wants a 7% margin of error at a 99% confidence level, what size of sample is needed? Give your answer in whole people