Maria claims that she can tell with 85% accuracy whether a pregnancy will result
ID: 2926017 • Letter: M
Question
Maria claims that she can tell with 85% accuracy whether a pregnancy will result in a boy or a girl simply by looking at the mother. You test this by introducing Maria to 10 random pregnant women, and later (once the babies are born) compairing her guesses with the actual gender of the child. For the purpose of this problem, assume that every pregnancy results in the birth of either a single boy or a single girl. You will believe her claim if she gets 7 or more of the baby’s genders right. How likely are you to believe her if she does have this ability? How likely are you to believe her if she, in fact, is just guessing randomly?Explain your answers as best you can. Please:)
Explanation / Answer
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 excueted
p = success probability
mean = 10 * 0.85
= 8.5
II.
variance = npq
where
n = total number of repetitions experiment is excueted
p = success probability
q = failure probability
variance = 10 * 0.85 * 0.15
= 1.275
III.
standard deviation = sqrt( variance ) = sqrt(1.275)
=1.12916
Maria claims that she can tell with 85% accuracy whether a pregnancy will result in a boy or a girl simply by looking at the mother
P( X < 7) = P(X=6) + P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)
= ( 10 6 ) * 0.85^6 * ( 1- 0.85 ) ^4 + ( 10 5 ) * 0.85^5 * ( 1- 0.85 ) ^5 + ( 10 4 ) * 0.85^4 * ( 1- 0.85 ) ^6 + ( 10 3 ) * 0.85^3 * ( 1- 0.85 ) ^7 + ( 10 2 ) * 0.85^2 * ( 1- 0.85 ) ^8 + ( 10 1 ) * 0.85^1 * ( 1- 0.85 ) ^9 + ( 10 0 ) * 0.85^0 * ( 1- 0.85 ) ^10
= 0.04997
P( X > = 7 ) = 1 - P( X < 7) = 0.95003
95.0% believe that she would have been guessing ramdomly