A multiple choice quiz has 200 questions, each with 4 possible answers, out of which only one is correct. What is the probability (using normal approximation to binomial distribution) that sheer guesswork yields 25 to 30 correct answers for 80 problems (out of 200 problems) about which the student has no knowledge?

In this problem, we are given a multiple-choice quiz containing 200 questions, with 4 possible answers for each question, out of which only one is correct. We need to find the probability that a student who has no knowledge of the subject will answer 25 to 30 questions correctly out of 80 randomly chosen questions.

Since we have a large number of questions (200) and a small probability of getting the correct answer (1/4), we can approximate this problem using the normal distribution. To use the normal approximation to the binomial distribution, we need to check whether the conditions for the normal approximation are met or not. The conditions for normal approximation are:

1. The sample size should be large (np ≥ 10 and n(1-p) ≥ 10).
2. The samples should be independent.
3. The probability of success (p) should be constant for each trial.

Let's check these conditions:

n = 80 (sample size) p = 1/4 (probability of success) q = 1 - p = 3/4 (probability of failure)

np = 80 × 1/4 = 20 nq = 80 × 3/4 = 60

Both np and nq are greater than or equal to 10, so the sample size is large enough for the normal approximation. Each trial is independent since we are assuming that each question is answered randomly. The probability of success (p) is constant for each trial since we are assuming that each question has an equal chance of being answered correctly.

Now, let X be the number of questions answered correctly out of 80 randomly chosen questions. We want to find the probability that 25 ≤ X ≤ 30. We can use the normal distribution to approximate the binomial distribution as follows:

μ = np = 20 σ = sqrt(npq) = sqrt(80 × 1/4 × 3/4) ≈ 3.87

The probability of getting 25 to 30 questions correct is:

P(25 ≤ X ≤ 30) = P((25 - 20)/3.87 ≤ (X - 20)/3.87 ≤ (30 - 20)/3.87) = P(1.29 ≤ Z ≤ 2.58) [where Z = (X - μ)/σ]

We can use a standard normal distribution table or calculator to find the probability of Z lying between 1.29 and 2.58. This probability turns out to be 0.0537.

Therefore, the probability that sheer guesswork yields 25 to 30 correct answers for 80 problems (out of 200 problems) about which the student has no knowledge is approximately 0.0537.

It is important to note that this calculation assumes that the student is answering the questions completely randomly with no knowledge of the subject. In reality, students may have some knowledge of the subject or may use other strategies to guess the correct answer. Additionally, the normal approximation to the binomial distribution is an approximation and may not be perfectly accurate for small sample sizes or extreme values of p. 

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