Question

An exam consists of 64 independent multiple-choice questions. For each question, there are five choices, with only one being correct. Suppose a student randomly guesses the correct answer to each question. (a) What are the mean and standard deviation of the number of successful guesses? (b) Use the normal approximation to find the probability that the student will correctly guess at most 10 correct answers.

Answer 1

Let X be the random vari able follows a binomial di stribution. X-B(n.p) The probability mass function of the binomial distribution is P(X = x) = P(x) (*)p*q**, x = 0,1,2,.. 10 otherwise The random variable X has binomial distribution with parameters n = 64), and p = 0.20 p=0.20 = probability of guessing the correct answer on a question q=1-p = 1-0.20 = 0.80 q=probability of not guessing the correct answer on a question q=0.80 Let X be the random vari able follows a binomial distribution. X-B(n = 64, p = 0.20) --------------- -------------------------- --- The mean of the number of successful guesses is F(X)= ap => 64(0.20) 12.8 E(X)= 12.8 Therefore, the mean of the number of successful guesses is E(X)= 12.8) - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - The standard deviation of the number of sccessful guesses is o= Vapa = 64(0.20) (0.80) =V10.24 >3.2 T* 3.2) Therefore, the standard deviation of the number of sccessful guesses is 083.21 - - - - - - - - -

(b) The conditions of the Normal approximation to the Binomial distribution is appropriate is as follows. If the first condition is np 25 = np = 64(0.20) = 12.8 mp = 12.825 If the second condition is ng 25. = 64(0.80) = 51.2 ng = 51.225 Hence, the two conditions of the Normal approximation is used to calculate the Binomial probability. - - - - - - - - - - The Normal distribution as an approximation to the Binomial distribution with n = 154 and p = 0.20 The mean of the distribution is E(X) = 4 = np = 64(0.20) = 12.8 4 = 12.8 The standard deviation of the distribution is u= apa = (64)(0.20)(0.80) = V10.24 = 3.2 ox 3.2 The normal approximation to find the probability that the student will correctly guess at most 10 correct answers is P(x <10)= P(X <a+0.5) = P(z< 195020) =P[<33) =P(z<-0.71875) Since, the normal probability function of the Excel 2010 is 0.236147 (=normsdist (2) P(X510) - 10.2361 Therefore, the probability that the student will correctly guess at most 10 correct answers is P(X <10) 10.2361