Question

Find the indicated probabilities using the geometric distribution, the Poisson distribution, or the binomial distribution. Then determine if the events are unusual. If convenient, use the appropriate probability table or technology to find the probabilities. Fifty dash two percent of adults say that they have cheated on a test or exam before. You randomly select six adults. Find the probability that the number of adults who say that they have cheated on a test or exam before is (a) exactly four, (b) more than two, and (c) at most five. (a) Upper P left parenthesis 4 right parenthesisequals nothing (Round to three decimal places as needed.) (b) Upper P left parenthesis more than two right parenthesisequals nothing (Round to three decimal places as needed.) (c) Upper P left parenthesis at most five right parenthesisequals nothing (Round to three decimal places as needed.) Which of the events are unusual? Select all that apply. A. The event Upper P left parenthesis 4 right parenthesis is unusual. B. The event Upper P left parenthesis more than two right parenthesis is unusual. C. The event Upper P left parenthesis at most five right parenthesis is unusual. D. None of the events are unusual.

Answer #1

Ans:

Use binomial distribution with n=6 and p=0.52

P(x=k)=^{6}C_{k}*0.52*(1-0.52)^{6-k}

x | P(x) |

0 | 0.0122 |

1 | 0.0795 |

2 | 0.2153 |

3 | 0.3110 |

4 | 0.2527 |

5 | 0.1095 |

6 | 0.0198 |

a)

P(x=4)=^{6}C_{4}*0.52^4*0.48^2=0.2527

b)

P(x>2)=1-P(x=0)-P(x=1)-P(x=2)

=1-(1-0.52)^6-^{6}C_{1}*0.52*(1-0.52)^5-^{6}C_{2}*0.52^2*(1-0.52)^4

=0.6930

c)

P(at most 5)=P(x<=5)=1-P(x=6)

=1-0.52^6**=**0.9802

Option D is correct.

**None of the events are unusual.**

(as all probabilities are greater than 0.05)

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