Problem 1(15 pts) A student is studying for the test which has 40 questions. Eac

Question

Problem 1(15 pts) A student is studying for the test which has 40 questions. Each question is a multiple choice question with 6 options. The student studied 25% of the content of the test. The rest of the test he will randomly pick the answer.

(A)   (1pt)   Find P(Score = 5)   Is this 1 or 0? Since we expect the student to answer at least 10 questions correctly

(B)   (2pts) Find P(Score = 30)   (1^10)*(()^20)*(()^10)

(C)   (2pts) Find P(Score = k)   (1^10)*(()^(k-10))*(()^(40-k))

(D)   (3pts) What should student expect for his score?   10+()*30 = 15

(E)     (5pts) The student would like to be 95% confident in his score result. Compute the 95% CI for the student.

(F)     (2pts) The student needs 13 points on the test to get B. Can he be 95% confident that he will get B on the test? Explain!!!!

Explanation / Answer

(A)   (1pt)   Find P(Score = 5)   Is this 1 or 0? Since we expect the student to answer at least 10 questions correctly

so it should be 0 as we are talking about exact probability here if it is asked that P(score>5) then it would be 1.

(B)   (2pts) Find P(Score = 30) = it would be when student will get 20 marks out of 30 unknown questions.

so P( score = 30) = 30C20 * (1/6)20 * (5/6)10 this value is less than 0.0001 so it is about to be 0.

(C)   (2pts) Find P(Score = k)   =  30Ck-10 * (1/6)k-10 * (5/6)40-k

(D)   (3pts) What should student expect for his score?   10+()*30 = 15

(E)     (5pts) The student would like to be 95% confident in his score result. Compute the 95% CI for the student.

Here mean score = 10 + np =15

Here 10 is the fixed part and 5 is the variable part so confidence interval will be applied on variable part only

so number of marks in unknown questions are Here n = 30 and p = 1/6

SEo = sqrt [ 1/6 * 5/6 * 30] = 2.0412

so 95% confidence interval for unknown questions = 5 +- Z0.05 * sqrt [ n * p * (1-p) ]

= 5 +- 1.96 * sqrt [ 30 * 1/6 * 5/6] = 5 + 1.96 * 2.04124 = ( 1.00, 9.00)

so 95 % confidence interval is = 10 + (1.00, 9.00) = ( 11, 19)

so it is 95 % chance that he will score between 11 and 19

(F)     (2pts) The student needs 13 points on the test to get B. Can he be 95% confident that he will get B on the test? Explain!!!!

Probability of him getting marks above 95% = Pr ( Marks >= 13; 15; 2.0412)

Z = ( 13 - 15)/ 2.0412 = - 0.98 => p- value from Z - table = 0.1635

so Probability of him getting marks above 95% = 1 - 0.1635 = 0.8365

so, no he can not be 95% confident that he will get B on the test.

This can also be detemined by answer of question (E). One can be 95% sure to get 11 or more marks and as marks are increasing, his probability of getting that marks will be lower than 95%.

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