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A multiple-choice test consists of a series of questions, each with four possible answers.
(a) If there are 60 questions, estimate the probability that a student who guesses blindly at each question will get at least 30 questions right.
(b) How many questions are needed in order to be 99% confident that a student who guesses blindly at each question scores no more than 35% on the test

Sagot :

fichoh

Answer:

0.00001 ; 92

Step-by-step explanation:

Given that:

Number of possible answers / options per question = 4 ; correct answers per question = 1

P(choosing the correct answer) ; p = 1/4 = 0.25

Probability of getting atleast 30 questions right :

Mean, m = np = 60 * 0.25 = 15

Standard deviation, s = sqrt(np(1-p)) = sqrt(60 * 0.25 * 0.75) = 3.354

Using binomial approximation :

P(x ≥ 29.5) = (29.5 - 15) / 3.354 = 4.32

P(Z ≥ 4.32) = 0.00001 (Z probability calculator)

To get no more than 35%

P(x ≤ 0.35) ; normal approximation P(x ≤ 0.35n + 0.5)

m= n * 1/4 = n/4

Variance = np(1-p) = n * 1/4 * 3/4 = 3n/16

X ~(n/4, 3n/16)

P(x ≤ 0.35) = [(0.35n + 0.5 - 0.25n) / sqrt(0.1875n)] = Zcritical 99%

Zcritical at 99% = 2.326

0.1n + 0.5 / sqrt(0.1875n) = 2.326

0.1n + 0.5 = 2.326 * sqrt(0.1875n)

Square both sides :

(0.1n + 0.5)² = (2.326*sqrt0.1875n)²

Quadratic relation obtained :

0.01n² - 0.914427 + 0.25n = 0

Solving using the quadratic. Formula :

-b ± [sqrt(b² - 4ac) / 2a]

a = 0.01 ; b = - 0.914427 ; c = 0.25

Output = 91.17 or 0.27

Hence, n = 92

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