Exercise 4 Let be a random sample of size from a probability distribution with probability density function a. (1.5 pts) Find an expression for the maximum likelihood estimator of . b. (0.5 pts) Given a sample of size , where , calculate the value of the MLE of

Fran would like to estimate the mean amount of time it takes people in her town to travel to work. the town's population is 150,000 and about 90,000 of those are working adults. which statements are true? select each correct answer. there is no way to infer the value of the mean. she must collect data for all 90,000 working adults and calculate the mean. a larger sample of working adults will provide a better estimate of the true mean than a smaller sample. the population of working adults is too large for using a sample to estimate the mean of the entire group. if she takes a random sample of the population of working adults in her town, the mean for that group is likely close to the mean for the entire group.

Recall that the owner of a local health food store recently started a new ad campaign to attract more business and wants to know if average daily sales have increased. historically average daily sales were approximately $2,700. the upper bound of the 95% range of likely sample means for this one-sided test is approximately $2,843.44. if the owner took a random sample of forty-five days and found that daily average sales were now $2,984, what can she conclude at the 95% confidence level?

36x2 + 49y2 = 1,764 the foci are located at: (-√13, 0) and (√13,0) (0, -√13) and (0,√13) (-1, 0) and (1, 0)edit: the answer is (- the square root of 13, 0) and (the square root of 13, 0)