Respuesta :
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;
[tex]P(X =x) = \binom{n}{r}\times p^{r}\times (1-p)^{n-r} ; x = 0,1,2,3,......[/tex]
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 = [tex]\frac{1}{4}[/tex] = 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 [tex]\geq[/tex] 18)
Because 60% of 30 = 18
P(X [tex]\geq[/tex] 18) = P(X = 18) + P(X = 19) +...........+ P(X = 29) + P(X = 30)
= [tex]\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}[/tex]
= 0.0000501
(b) The expected number of correct guesses is given by;
Mean of the binomial distribution, E(X) = [tex]n \times p[/tex]
= [tex]30 \times 0.25[/tex] = 7.5
(c) The standard deviation of the binomial distribution is given by;
S.D.(X) = [tex]\sqrt{n \times p \times (1-p)}[/tex]
= [tex]\sqrt{30 \times 0.25 \times (1-0.25)}[/tex]
= [tex]\sqrt{5.625}[/tex] = 2.372
