A history professor decides to give a 12-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 0.10. What score should be set as the lowest passing grade? Group of answer choices

Respuesta :

Answer:

we can set the 9 as a benchmark to be the score for the passing grade so that probability of passing a student who guesses every question is less than 0.10

Step-by-step explanation:

From the given information;

Sample  size n = 12

the probability of passing a student who guesses on every question is less than 0.10

In a alternative - response question (true/false) question, the probability of answering  a question correctly  = 1/2 = 0.5

Let X be the random variable that is represent number of correct answers out of 12.

The X [tex]\sim[/tex] BInomial (12, 0.5)

The probability mass function :

[tex]P(X = k) = \dfrac{n!}{k!(n-k)!} \times p^k\times (1-p)^{n-k}[/tex]

[tex]P(X = 12) = \dfrac{12!}{12!(12-12)!} \times 0.5^{12}\times (1-0.5)^{12-12}[/tex]

P(X = 12) = 2.44 × 10⁻⁴

[tex]P(X = 11) = \dfrac{12!}{11!(12-11)!} \times 0.5^{11}\times (1-0.5)^{12-11}[/tex]

P(X =11 ) = 0.00293

[tex]P(X = 10) = \dfrac{12!}{10!(12-10)!} \times 0.5^{10}\times (1-0.5)^{12-10}[/tex]

P(X = 10) = 0.01611

[tex]P(X = 9) = \dfrac{12!}{9!(12-9)!} \times 0.5^{19}\times (1-0.5)^{12-9}[/tex]

P(X = 9) = 0.0537

[tex]P(X = 8) = \dfrac{12!}{8!(12-8)!} \times 0.5^{8}\times (1-0.5)^{12-8}[/tex]

P(X = 8)  = 0.12085

[tex]P(X = 7) = \dfrac{12!}{7!(12-7)!} \times 0.5^{7}\times (1-0.5)^{12-7}[/tex]

P(X = 7) = 0.19335

.........

We can see that,a t P(X = 9) , the probability is 0.0537 which less than 0.10 but starting from P(X = 8) downwards the probability is more than 0.01

As such, we can set the 9 as a benchmark to be the score for the passing grade so that probability of passing a student who guesses every question is less than 0.10