Question

Suppose your statistics instructor gave six examinations during the semester. You received the following exam scores (percent correct): 83, 76, 82, 84, 89, and 73. To compute your final course grade, the instructor decided to randomly select two exam scores, compute their mean, and use this score to determine your final course grade.

Compute the population mean. This is your average grade based on all of your grades. (Round your answer to 2 decimal places.) Compute the population standard deviation. (Round your answer to 2 decimal places.)

How many difference scores could be calculated if your instructor decided to random sample two of your exam scores?

List all possible samples of size 2 and compute the mean of each. (Round your answers to 1 decimal place.)

Compute the mean of the sample means and the standard error of the sample means. (Round your answers to 2 decimal places.)

Applying the central limit theorem, if the instructor randomly samples two of your exam scores to compute an average as your final course grade, what is the probability your final course grade will be less than 81.17? What is the probability that your final course grade will be more than 81.17? (Round your answers to 1 decimal places.)

Answer 1

Solution :- Given data is the 6 examination scores during the semister. di = 83,76,82,84,89,43 The sample size n=6 a) population mean : x = sai U8+ 83+46+62+4+89+ 33 a-81.17 5) Standard deviation: o=1 & cxi-aj? NA! (83-81-1734 676-81.17)+(82-81-19) + (84-8117) + (89-81.1934623-81." 6-1 3 / 3.3489 +06.4289 +0.6889 +8.0089 +61-3089+ 66.2489 166.6224 -> 438-34566 = 5.78

(c) Total number of scores n36 from toda! 69 Gelect 66. = random Šarriples 6! - 675x4! 0!(6-2)! ! xx? = 15 samples from select a random Therefore, 15 Ways to total 6 Samples man paccible ways 83,76 6+84 83,82 83,84 83,89 83,73 - 16,82 76,84 76189 76,73 82184 - 82,89 82,23 84,89 -- 80 , 73 89173 → 83+26 - 79.5 > (83+82)/2 = 82.5 > (83484)/2 = 83.5 my (83789)/286 (83+73)/2 = 78 46+82)/2 = 79 4 6+84)/2 = 80 A6+89)/2 = 82.5 €16+73)/2 = 74.5 (82+84)12 583 (82+89)/2 = 85.5 (82+73)/2 = 7705 (84+89)/2 = 86.5 - 04+03512278.5 → 089+93)/2=81

mean Sample means X1 182 6889 5776 Bondard crror : 00+ Exş?- Nles)? 6924 N 89 7056 7921 5329 139695-6 (81.17) u87 39695 162.5866 129.26 T50=5.22 ) © 1) probabilis less than 81:17: probability P18€81-19) = Plote = P(x231•19 – 2116 5.22 186 = P(Z < 0.01 2.13 = P(240.0004) (p<3 <814) = 0.675;

13) probability more than 81117; P/328113) - |- P(X< 1:19) -1-0.5.; /P(7 78117) = 0.5 therefore, PCR 81.17)=P(>81'17) = 0.5 ll Hey, if you are 'Okaywith this answer eive a thumps up, thank you