Question

0 0.0438 QUESTION 15 Use the normal distribution to approximate the desired probability. A coin is tossed 20 times. A person, who claims to have extrasensory perception, is asked to predict the outcome of each flip in advance. She predicts correctly on 14 tosses. What is the probability of being correct 14 or more times by guessing? Does this probability seem to verify her claim? 0.4418, yes 0.4418, no 0.0582. yes 0.0582, no

Answer 1

15

This is a binomial distribution question with
n = 20
p = 0.5
q = 1 - p = 0.5
where
$P(X = x) = \binom{n}{x} p^x q^{n-x}$

a)
$\\ P(X \ge 14), n = 20, p = 0.5 \\ P(X \ge 14) = P(14 \le X \le 20) \\ P(X \ge 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)$
$\\ P(X \ge 14) = \binom{20}{14} 0.5^{14} 0.5^{20-14} + \binom{20}{15} 0.5^{15} 0.5^{20-15} + \binom{20}{16} 0.5^{16} 0.5^{20-16} + \binom{20}{17} 0.5^{17} 0.5^{20-17} + \binom{20}{18} 0.5^{18} 0.5^{20-18} + \binom{20}{19} 0.5^{19} 0.5^{20-19} + \binom{20}{20} 0.5^{20} 0.5^{20-20}$
$\\ P(X \ge 14) = 38760.0 (0.5^{14}) (0.5^{6}) + 15504.0 (0.5^{15}) (0.5^{5}) + 4845.0 (0.5^{16}) (0.5^{4}) + 1140.0 (0.5^{17}) (0.5^{3}) + 190.0 (0.5^{18}) (0.5^{2}) + 20.0 (0.5^{19}) (0.5^{1}) + 1 (0.5^{20}) (0.5^{0})$
$\\ P(X \ge 14) = 38760.0 (0.0001) (0.0156) + 15504.0 (0.0) (0.0312) + 4845.0 (0.0) (0.0625) + 1140.0 (0.0) (0.125) + 190.0 (0.0) (0.25) + 20.0 (0.0) (0.5) + 1 (0.0) (1.0) \\ P(X \ge 14) = \textbf{0.0369644165} + \textbf{0.0147857666} + \textbf{0.0046205521} + \textbf{0.0010871887} + \textbf{0.0001811981} + \textbf{1.90735e-05} + \textbf{9.537e-07} \\ P(X \ge 14) = \textbf{0.057659149169921875} \\ P(X \ge 14) = \textbf{0.0577} \\ \par\noindent\rule{\textwidth}{0.4pt}$

OPTION D: 0.0582, NO

to correctly predict the outcome on 14th flip and to correctly predict the outcomes continuously till 14th flip are two different thing altogether