Question

A literature professor decides to give a 10-question true-false quiz. She wants to choose the passing grade such that the probability of passing a student who guesses on every question is less than .10. What score should be set as the lowest passing grade?

Answer 1

Answer:

Step-by-step explanation:

7

Answer 2

Using the binomial distribution, it is found that the minimum passing grade should be set as 8.

For each question, there are only two possible outcomes. Either the student guesses the correct answer, or the student does not. The probability of a student guessing the correct answer of a question is independent of any other question, which means that the binomial distribution is used to solve this question.

Binomial probability distribution  

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$  

$C_{n,x} = \frac{n!}{x!(n-x)!}$

• x is the number of successes.
• n is the number of trials.

• p is the probability of a success on a single trial.

In this problem:

• 10 questions, thus $n = 10$
• Guesses on true-false questions, which have 2 options, thus $p = \frac{1}{2} = 0.5$.
• The minimum passing grade is the minimum value of x for which: $P(X \geq x) \leq 0.1$

Starting at 10:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{10,10}.(0.5)^{10}.(0.5)^{0} = 0.001$

Thus:

$P(X \geq 10) = P(X = 10) = 0.001$

Testing 9:

$P(X = 9) = C_{10,9}.(0.5)^{9}.(0.5)^{1} = 0.0098$

Thus:

$P(X \geq 9) = P(X = 9) + P(X \geq 10) = 0.0098 + 0.001 = 0.0108$

Testing 8:

$P(X = 8) = C_{10,8}.(0.5)^{8}.(0.5)^{2} = 0.0439$

Thus:

$P(X \geq 8) = P(X = 8) + P(X \geq 9) = 0.0439 + 0.0108 = 0.0547$

Testing 7:

$P(X = 7) = C_{10,7}.(0.5)^{7}.(0.5)^{3} = 0.1172$

Thus:

$P(X \geq 7) = P(X = 7) + P(X \geq 8) = 0.1172 + 0.0547 = 0.1719$

Since $P(X \geq 7) > 0.1$ and $P(X \geq 8) < 0.1$, the lowest passing grade should be set as 8.

A similar problem is given at https://brainly.com/question/24756209