Question

Assume that random guesses are made for

ninenine

multiple choice questions on an SAT​ test, so that there are

nequals=99

​trials, each with probability of success​ (correct) given by

pequals=0.450.45.

Find the indicated probability for the number of correct answers.

Find the probability that the number x of correct answers is fewer than

P(X<4)=

Answer 1

BIONOMIAL DISTRIBUTION
pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where   
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial
I.
mean = np
where
n = total number of repetitions experiment is executed
p = success probability
mean = 9 * 0.45
= 4.05
II.
variance = npq
where
n = total number of repetitions experiment is executed
p = success probability
q = failure probability
variance = 9 * 0.45 * 0.55
= 2.2275
III.
standard deviation = sqrt( variance ) = sqrt(2.2275)
=1.4925

a.
the probability that the number x of correct answers is fewer than
P(X<4)
P( X < 4) = P(X=3) + P(X=2) + P(X=1) + P(X=0)
= ( 9 3 ) * 0.45^3 * ( 1- 0.45 ) ^6 + ( 9 2 ) * 0.45^2 * ( 1- 0.45 ) ^7 + ( 9 1 ) * 0.45^1 * ( 1- 0.45 ) ^8 + ( 9 0 ) * 0.45^0 * ( 1- 0.45 ) ^9
= 0.3614