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

ID: 3334060 • Letter: J

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 = 12, 24, and 48 trials, respectively. In the three experiments, his numbers correct are 3, 6, and 12, respectively, for a total of 21 correct out of 84 trials. (a) For each experiment, what is the expected number correct if Joe is just guessing? for:

n = 12 the mean =

n = 24 the mean =

n = 48 the mean =

(b) Over all 84 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 = 12 P(x 3) =

n = 24 P(x 6) =

n = 48 P(x 12) =

(d) Out of the 84 trials overall, find the approximate probability that someone would get as many correct as Joe did (21), or more, if the person was just guessing. (Use the binomial distribution. Round the answer to four decimal places.)

P(x 21) =

(e) Compare your answers in parts (c) and (d). If Joe were trying to convince someone of his ability, would it be better to show the three experiments separately, or would it be better to show the combined data?

It would be better to show the combined data.

or

It would be better to show the three experiments separately.

Explanation / Answer

as we know that mean=np

hence n = 12 the mean = 12*0.2=2.4

n = 24 the mean =24*0.2=4.8

n = 48 the mean =48*0.2=9.6

c)

from binomial distribution:

n = 12 P(x 3) = 1-P(X<=2) =1-0.5583 =0.4417

n = 24 P(x 6) = 1-P(X<=5)=1-0.6559 =0.3441

n = 48 P(x 12) =1-P(X<=11)=1-0.7595 =0.2405

d)

n = 48 P(x 21) =1-P(X<=20)=1-0.8437 =0.1563

e)

It would be better to show the three experiments separately.

please revert for any clarification requiired

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