Question

prob 24

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. Using the Distribution tool below, find each of the probabilities requested for a person who has no ESP and is just guessing. Standard Normal Distribution Mean 0.0 Standard Deviation 1.0 5000 5000 0.0 0,0000 1.0 What is the probability of correctly predicting exactly 20 cards in a series of 100 trials? What is the probability of correctly predicting 10 or more cards in a series of 36 trials? What is the probability of correctly predicting more than 16 cards in a series of 64 trials?

Answer 1

Solution:-

Let X represent number of success

a.)X follow binomial distribution with p = 1/5=0.20 (as when np > 5 ,n(1-p) > 5)

So,

Mean(μ)=np

=100*0.2=20

Standard Deviation=sqrt(npq)

=sqrt(100*0.2*0.8)

=4

Binomial Distribution is given by:-

P(X = k) = 100Ck * (0.2)^k *(1-0.2)^(100-k)

P(X = 20) = 100C20 (0.2)^20* (0.8)^80

=0.0993

b.)

n=36, p=0.2

Mean((μ)=np=0.2*36=7.2 nq=36*0.80=28.8

as np>5 and nq>5 , we can use normal approximation

sd = sqrt(npq)=sqrt(36*0.2*0.8)=2.4

so, μ = 7.2 and σ = 2.4 For X = 36

P(X>=10)=1-P(X<10)

=1-P(Z<(10-7.2)/2.4)

=1-P(Z<1.17)

=1-0.879

=0.121

c.)

n=64, p=0.2

as np>5 and nq>5 , we can use normal approximation

μ = np=64*0.2=12.8

and σ = sqrt(npq)=sqrt(64*0.2*0.8)=3.2

X=16

P(X>=16)=1-P(X<16)

=1-P(Z<(16-12.8)/3.2)

=1-P(Z<1)

=1-0.8413 (using Z table)

=0.1587