Question

Suppose you take a 30-question, multiple-choice test, in which each question contains 4 choices: A, B, C, and D. If you randomly guess on all 30 questions, what is the probability you pass the exam (correctly guess on 60% or more of the questions)? Assume none of the questions have more than one correct answer (hint: this assumption of only 1 correct choice out of 4 makes the distribution of X, the number of correct guesses, binomial). What is the expected number of correct guesses, from problem #19? What is the standard deviation, ? (Remember that X is a binomial random variable!) What would be considered an unusual number of correct guesses on the test mention in problem number 19 using ?

Answer 1

Answer:

(a) The probability you pass the exam is 0.0000501.

(b) The expected number of correct guesses is 7.5.

(c) The standard deviation is 2.372.

Step-by-step explanation:

We are given that you take a 30-question, multiple-choice test, in which each question contains 4 choices: A, B, C, and D. And you randomly guess on all 30 questions.

Since there is an assumption of only 1 correct choice out of 4 which means the above situation can be represented through binomial distribution;

$P(X =x) = \binom{n}{r}\times p^{r}\times (1-p)^{n-r} ; x = 0,1,2,3,......$

where, n = number of trials (samples) taken = 30

           r = number of success = at least 60%

           p = probbaility of success which in our question is the probability

                 of a correct answer, i.e; p = $\frac{1}{4}$ = 0.25

Let X = Number of questions that are correct

So, X ~ Binom(n = 30 , p = 0.25)

(a) The probability you pass the exam is given by = P(X $\geq$ 18)

Because 60% of 30 = 18

P(X $\geq$ 18) = P(X = 18) + P(X = 19) +...........+ P(X = 29) + P(X = 30)

= $\binom{30}{18}\times 0.25^{18}\times (1-0.25)^{30-18} + \binom{30}{19}\times 0.25^{19}\times (1-0.25)^{30-19} +.......+ \binom{30}{29}\times 0.25^{29}\times (1-0.25)^{30-29} + \binom{30}{30}\times 0.25^{30}\times (1-0.25)^{30-30}$

= 0.0000501

(b) The expected number of correct guesses is given by;

  Mean of the binomial distribution, E(X) =  $n \times p$

                                                                =  $30 \times 0.25$ = 7.5

(c) The standard deviation of the binomial distribution is given by;

      S.D.(X)  =  $\sqrt{n \times p \times (1-p)}$

                    =  $\sqrt{30 \times 0.25 \times (1-0.25)}$

                    =  $\sqrt{5.625}$  =  2.372