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 50 students on the exam is 540. ASS the means to be measured to any degree of accuracy (assume that scores are Gaussian distributed). (a) Find the 90% 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 50 students! TTTT Paragraf Arial 3 (120) T. 3136

Answer 1

1 Solcetion: Given data: a) tet u and 6 Jenetz Jupestively the standard deviation of and population mean the Scores Here We Scores are cessume that the distributed and given that 6=30 normally Here The population distribution is normal population standard devication is and the known А. 100 (1-0)4. C.I for l (T-€, 5 +E), where E Za x 6 न sample mean n > sample Size E = margin of cerror.

d = I - confidence Level Given that confidence Level = 0 96, n=50, = 540 . d = 1-0.90 0.10 .. Z Za Zo 10 2 20:05 1.645 E- zo 05 X 음 VO = 1.645 x 30 850 -6.98 . The 904. C.1 for the average score of ou students taking the excem (T - E, šte) (540-6.98,540 +6.98) ( 1533 62. 46 989

3 b) In this problem. The population distribution and the population standard normal deviation (6) is known The 100 (1-2)t. prediction internal CP.I) for the score The next studend chosen at random ł - za 4 x 6 I + Zulx odith for the score of the next . The 904 P.I steedent chosen at random is Cał- 70.05 * ET THEO at t zo.05 X 6 - تا بار ء (540-1.645 30x Vitt 60 540+1.645x30 Vito (490.16, 589.84)

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