Question

An exam is given to all senior high school students, the standard deviation for all the students is given as 30. The average score for 40 students on the exam is 250. Assume the means to be measured to any degree of accuracy (assume that scores are Gaussian distributed). (a) Find the 95% confidence interval (CI) for the average score of all students taking the exam. (b) Find the prediction interval for the score of the next student chosen at random (not one of the original 40 students).

Answer 1

Here standard deviation given as a = 30 we know population distribution is normal and the population standard deviation. ich A 100(1-2)% confidence interval for u is (x-E, 7+E) where Margin error E-ZĄ x. za o in à = sample mean ne sample size d = 1- confidence level. Here given that 95% confidence inte. -rual so C.L=0.95, n =40, x = 250 '. d=1-0.95 -0.05 = 20.05 Zo.025 그슬 2 - 1 96 E = 22 xa 58.8 - 1.96 x 30 40 40 9.297

The 95% c. I. for the aug. score of all students taking the exam 18 (X-E ,x+E) (250-9.297 250+9 297) = (240.70, 259.297) In this problem, the population distribution is normal and the population std deviation (o) is known : The 100(1-2) 7 Prediection interoval for the score of the next student choosen at random is P.I (x-Zaxo Sitta to sinta we already calculated - Zo025 = |26 - Za 2 x 250 - ( 250 1.96 X 30x 250 +1.96x30x ( 190.46, 309 53)