Suppose scores on exams in statistics are normally distributed with an unknown population mean and a population standard deviation of three points. a random sample of 36 scores is taken and gives a sample mean (sample mean score) of 68. find a confidence interval estimate for the population mean exam score (the mean score on all exams). find a 90% confidence interval for the true (population) mean of statistics exam scores.

( 67.18, 68.82)

Step-by-step explanation:

Let be the true (population) mean of statistics exam scores. We have a large random sample of n = 36 scores with a sample mean (sample mean score) of . We know that the population standard deviation is . A pivotal quantity is which is approximately normally distributed and where , and so, . Therefore the 90% confidence interval is (68-(1/2)(1.6449), 68+(1/2)(1.6449)), i.e., ( 67.18, 68.82)

i don't know, but that looks pretty bad. my school passed us unless we failed in all the subjects, but idk abt ur school's rules.

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