In a test for ESP (extrasensory perception), the experimenter looks at cards tha

Question

In a test for ESP (extrasensory perception), the experimenter looks at cards that are hidden from the subject. Each card contains either a star, a circle, a wave, a square, or a cross. As the experimenter looks at each of 10 cards in turn, the subject names the shape on each card. The experimenter then reveals the hidden card. If a subject simply guesses the shape on each card, what is the probability of a successful guess on a single card? Suppose the experimenter is drawing cards from a deck of only 10 cards, with exactly 2 cards with each shape. The subject also knows the makeup of the deck. Why is the binomial distribution not a good model for this study? Now suppose that the cards are drawn from a very large, fully shuffled deck of cards, so the trials may now be assumed to be independent. A binomial distribution with 10 trials is a good model for this distribution. What is the chance that the subject (who is guessing) guesses at least 5 shapes correctly? In many repetitions of this study, for a subject who is guessing, how1 many cards will he guess correctly on the average? What is the standard deviation of the number of correct guesses?

Explanation / Answer

a)

There are 5 possibilities, so

P = 1/5 = 0.2 [ANSWER]

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b)

It is because the probability of success gets updated (not constant) on every turn, as the experimenter reveals the card each time.

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c)

Note that P(at least x) = 1 - P(at most x - 1).          
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    10      
p = the probability of a success =    0.2      
x = our critical value of successes =    5      
          
Then the cumulative probability of P(at most x - 1) from a table/technology is          
          
P(at most   4   ) =    0.967206502
          
Thus, the probability of at least   5   successes is  
          
P(at least   5   ) =    0.032793498 [ANSWER]

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d)

u = mean = np =    2 [ANSWER, AVERAGE]
  
s = standard deviation = sqrt(np(1-p)) =    1.264911064 [ANSWER, STANDARD DEVIATION]

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