Pat is taking an economics course. Pat\'s exam strategy is to rely no luck for t
ID: 3292876 • Letter: P
Question
Pat is taking an economics course. Pat's exam strategy is to rely no luck for the next exam. The exam consists of n multiple-choice quest ions. Each question has four possible answers, only one of which is correct. Pat plans to guess the answer to each question without reading it. If a grade oil the exam is more than 60%. Pat will pass the exam. (a) When n = 2, find the probability that Pat will pass the exam. (b) When n = 10, find the probability that Pat will pass the exam. (c) When n = 20, find the probability that Pat will pass the exam. (d) When n = 100, find the probability that Pat will pass the exam.Explanation / Answer
Solution:
The question is based on binomial distribution with p = 1/4 = 0.25
The Binomial distribution is given by:-
P (X = x) = nCx (0.25)^x (1 - 0.25)^n-x
a. When n = 2, he has to answer more than 0.60 x 2 = 1.2 ~ 1 correct to pass the exam.
P (X > 1) = 1 - P (X 1)
P (X > 1) = 1 - P (X = 0)
P (X > 1) = 1 - 2C0 (0.25)^0 (0.75)^2
P (X > 1) = 1 - 0.5625
P (X > 1) = 0.4375
b. When n = 6, he has to answer more than 0.60 x 6 = 3.6 ~ 4 correct to pass the exam.
P (X > 4) = 1 - P (X 4)
P (X > 4) = 1 - [P (X = 0) + P (X= 1) + P (X = 2) + P (X = 3) + P (X = 4)]
P (X > 4) = 1 - [0.178 + 0.356 + 0.297 + 0.132 + 0.033]
P (X > 4) = 0.005
c. When n = 20, he has to answer more than 20 x 0.60 = 12 correct to pass the exam.
P (X > 12) = P (X = 13) + P (X = 14) + .........+ P (X = 20)
P (X > 12) = 0.000154 + 0.0000257 + 0.000 + 0.000 + 0.000 + 0.000 + 0.000 + 0.000
P (X > 12) = 0.000184
d. When n = 100, he has to answer more than 100 x 0.60 = 60 correct to pass the exam. Since n is very large, np = 100 x 0.25 = 25 and n (1 - p) = 100 x (1 - 0.25) = 75 both are greater than 5, we use normal approximation to binomial.
Mean, µ = np = 100*0.25 = 25
Standard deviation, = np(1 – p) = 100*0.25*0.75 = 4.330
Using continuity correction, we have
P (X > 60) = P (X > 60.5)
The respective Z-score with X = 60.5 is
Z = (X - µ)/
Z = (60.5 – 25)/4.33
Z = 8.2
Using Z-tables, the probability is
P [Z > 8.2 ] = 0.000