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)

solution:

as, mean is defined as sum of all the observations divided by total number of observations.

and , median is mid value of the data whether arranged in ascending order or descending order.

consider, roadsters:

2,3,4,5,6,7,8,9

mean =

also, median= , because number of observations is even, so median is mean of two mid values.

now, coming to bandits

2,3,4,5,6

mean =  

median = 4, as number of observations is odd.

so, roadsters has greater median as well mean than bandits.

→option a is true about the repairs performed on 2 types of cars:

bandits have lower median and mean than roadsters.