Joe performs remarkably well on remote viewing ESP tests, which require the \"vi

Question

Joe performs remarkably well on remote viewing ESP tests, which require the "viewer" to draw a picture, then determine which of five possible photos was the intended "target". Because the target is randomly selected from among the five choices, the probability of a correct match by chance is 1/5, or 0.2. Joe participates in three experiments with n = 11, 33, and 55 trials, respectively. In the three experiments, his numbers correct are 2, 6, and 10, respectively, for a total of 18 correct out of 99 trials.

(a) For each experiment, what is the expected number correct if Joe is just guessing?

for n = 11 the mean =

for n = 33 the mean =

for n = 55 the mean =

(b) Over all 99 trials, what is the expected number correct, ?

=

(c) For each experiment separately, find the approximate probability, P that someone would get as many correct as Joe did, or more, if the person was just guessing. (Use the binomial distribution. Round all answers to four decimal places.

) for n = 11 P(x 2) =

for n = 33 P(x 6) =

for n = 55 P(x 10) =

(d) Out of the 99 trials overall, find the approximate probability that someone would get as many correct as Joe did (18), or more, if the person was just guessing. (Use the binomial distribution. Round the answer to four decimal places.)

P(x 18) =

Explanation / Answer

Given,

p = probability of correct match = 0.2

Let x be the number of correct match.

x follows Binomial distribution with parameter n and p .

So, Mean = n*p

a )

i ) n = 11 , p = 0.2

expected number correct if Joe is just guessing is,

E( x ) = n* p = 11*0.2 = 2.2

ii) n = 33 , p = 0.2

E(x) = 33*0.2 = 6.6

iii) n = 55

E(x) = 55*0.2 = 11

b ) n = 99 , p = 0.2

the expected number correct, = n*p = 99*0.2 = 19.8

c )

i ) n = 11

P( x >= 2 ) = 1 - P( x < 2 ) = 1 - P( x <= 1 )

Using Excel function, = BINOMDIST( X, N, P, 1 )

P( x <= 1 ) = BINOMDIST( 1, 11, 0.2 , 1 ) = 0.322123

P( x >= 2 ) = 1 - 0.322123 = 0.6779

ii) n = 33

P( x >= 6 ) = 1 - P( x < 6 ) = 1 - P( x <= 5 )

Using Excel function, = BINOMDIST( X, N, P, 1 )

P( x <= 5 ) = BINOMDIST( 5, 33, 0.2 , 1 ) = 0.32903

P( x >= 6 ) = 1 - 0.32903= 0.6710

iii) n = 55

P( x >= 10 ) = 1 - P( x < 10 ) = 1 - P( x <= 9 )

Using Excel function, = BINOMDIST( X, N, P, 1 )

P( x <= 5 ) = BINOMDIST( 9, 55, 0.2 , 1 ) = 0.315891

P( x >= 6 ) = 1 - 0.315891= 0.6841

d )

n = 99

P( x >= 18 ) = 1 - P( x <= 17 )

P( x <= 17 ) = BINOMDIST( 17 , 99, 0.2, 1 ) = 0.2876

P( x >= 18 ) = 1 - 0.2876 = 0.7124

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