Dr. Clark claims to possess extrasensory perception (ESP). An experiment is cond

Question

Dr. Clark claims to possess extrasensory perception (ESP). An experiment is conducted in which a person in one room picks one of the integers 1, 2, 3, 4, or 5 at random and concentrates on it for one minute. In another room, Dr. Clark identifies the number she believes was picked. The experiment is done with three trials. After the third trial, the random numbers are compared with Dr. Clarks predictions.

Does Dr. Clark have ESP or is she merely guessing the number? To answer this, consider the following questions:

a) If she is truly guessing, what is the probability of a correct answer? (That is, what is the probability of a success?)

b) Let X = number of correct guesses in three trials. What is the sample space of X?

c) X has a binomial distribution, X ~ Bin(n,p). What are the values of n and p?

d) What is the probability of each event in the sample space? Write your result in table form where you list the outcomes and the corresponding probabilities. Verify the probabilities sum to one.

e) From (d), what is P(X = 2)?

f) You can also use the Binomial Distribution calculator on StatCrunch to verify this value. Open the StatCrunch website and go to Stat/Calculator/Binomial. Enter the correct values for n and p. Choose “Prob(X = 2)”. Press compute.

g) Dr. Clark got the correct result twice. Do you think Dr. Clark has ESP? Why or why not? What level and/or type of evidence would you require to be convinced that she has ESP?

Explanation / Answer

Dr. Clark claims to possess extrasensory perception (ESP). An experiment is conducted in which a person in one room picks one of the integers 1, 2, 3, 4, or 5 at random and concentrates on it for one minute. In another room, Dr. Clark identifies the number she believes was picked. The experiment is done with three trials. After the third trial, the random numbers are compared with Dr. Clarks predictions.

Does Dr. Clark have ESP or is she merely guessing the number? To answer this, consider the following questions:

a) If she is truly guessing, what is the probability of a correct answer? (That is, what is the probability of a success?)

The probability of a correct answer = p= 1/5 = 0.20

b) Let X = number of correct guesses in three trials. What is the sample space of X?

sample space (S) = {0,1,2,3}

c) X has a binomial distribution, X ~ Bin(n,p). What are the values of n and p?

X ~ Binomial(n = 3, p = 0.20)

n = 3 and p = 0.20

The pmf of X=x is,

P(X=x) = (3 C x) * 0.20x * (1-0.20)(3-x) , x = 0,1,2,3

P(X=0) = (3 C 0) * 0.200* (1-0.20)(3-0) = 0.512

P(X=1) = (3 C 1) * 0.201 * (1-0.20)(3-1) = 0.384

P(X=2) = (3 C 2) * 0.202 * (1-0.20)(3-2) = 0.096

P(X=3) = (3 C 3) * 0.203 * (1-0.20)(3-3) = 0.008

Total = 0.512+0.384+0.096+0.008 = 1

e) From (d), what is P(X = 2)?

P(X=2) = 0.096

f)ou can also use the Binomial Distribution calculator on StatCrunch to verify this value. Open the StatCrunch website and go to Stat/Calculator/Binomial. Enter the correct values for n and p. Choose “Prob(X = 2)”. Press compute.

1. Select Stats, highlight Calculators, select Binomial.
2. Enter the number of trials, n, and probability of success, p. In the pull-down menu, decide if you wish to compute P(X < x), P(X < x), and so on. Finally, enter the value of x. Click Compute.

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