Attempts: Score: 3 13. Chapter 6, Problem 25 AaAa One test for claims of extrase

Question

Attempts: Score: 3 13. Chapter 6, Problem 25 AaAa One test for claims of extrasensory perception (ESP) involves using Zener cards. Each card shows one of five different symbols (square, circle, star, cross, wavy lines), and the person being tested has to predict the shape on each card before it is selected. Find each of the probabilities requested for a person who has no ESP and is just guessing. To enable you to select a z-score to two decimal places on the Standard Normal Distribution tools, only a portion of the range of z-scores can be shown at a time. Click the button to see the Standard Normal Distribution tool for the range of z-scores stated. -2.4

Explanation / Answer

Here probability of success (p) = 1/5

number of trilas (n) = 100

np = 1/5*100 = 20 > 10

nq = 100 * (1 - 1/5) = 80 > 10

So we can use here binomial approximation to the normal distribution.

mean = np = 20

se = sqrt(p*q)/n = sqrt((1/5) * (4/5) / 100) = 0.04

a) What is the probability that correctly predicting 20 cards in a series of 100 trials.

Here we have to find P(X = 20)

Convert x = 20 into z-score.

z = + or - 0.13

Now we have to find P(-0.13 < Z < 0.13)

P(-0.13 < Z < 0.13) = P(Z < 0.13) - P(Z < -0.13)

This probability we can find in EXCEl.

syntax :

=NORMSDIST(z)

where z is z-score.

P(-0.13 < Z < 0.13) = 0.5517 - 0.4483 = 0.1034

b) What is the probability that correctly predicting more than 30 cards ina series of 100 trials.

Here we have to find P(X > 30.5)

z-score = 2.63

P(Z > 2.63) = 1 - P(Z < 2.63) = 1 - 0.9957 = 0.0043

c) What is the probability of correctly predicting 50 or more cards in a series of 200 trials

Here mean will be 40 and standard deviation will be 5.66

And we have to find P(X > 49.5)

z-score = 1.68

Now we have to find P(Z > 1.68)

P(Z > 1.68) = 1- P(Z < 1.68) = 1 - 0.9535 = 0.0465

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