Question

1.)

Assume that a procedure yields a binomial distribution with a
trial repeated n=5n=5 times. Use technology to find the probability
distribution given the probability p=0.702p=0.702 of success on a
single trial.

*(Report answers accurate to 4 decimal places.)*

k |
P(X = k) |
---|---|

0 | |

1 | |

2 | |

3 | |

4 | |

5 |

2.)

Assume that a procedure yields a binomial distribution with a
trial repeated n=5n=5 times. Find the probability distribution
given the probability p=0.799p=0.799 of success on a single trial.
Report answers accurate to 4 decimal places.

k |
P(X = k) |
---|---|

0 | |

1 | |

2 | |

3 | |

4 | |

5 |

Answer #1

Solution:

Given in the question

P(success)= 0.702

P(failure)= 0.298

No. Of trails =5

P(X=0)= 5C0*(0.702)^0*(0.298)^5 =0.0024

P(X=1)= 5C1*(0.702)^1*(0.298)^4= 0.0277

P(X=2)= 5C2*(0.702)^2*(0.298)^3= 0.1304

P(X=3)= 5C3*(0.702)^3*(0.298)^2= 0.3072

P(X=4)=5C4*(0.702)^4*(0.298)^1= 0.3619

P(X=5)= 5C5*(0.702)^5*(0.298)^0= 0.1705

Solution(b)

Given in the question

P(success)= 0.799

P(failure)= 0.201

No. Of trials = 5

P(X=0)= 5C0*(0.799)^0*(0.201)^5 = 0.0003

P(X=1)= 5C1*(0.799)^1*(0.201)^4 = 0.0065

P(X=2)= 5C2*(0.799)^2*(0.201)^3 = 0.0518

P(X=3)= 5C3*(0.799)^3*(0.201)^2 = 0.2061

P(X=4)= 5C4*(0.799)^4*(0.201)^1 = 0.4096

P(X=5)= 5C5*(0.799)^5*(0.201)^0 = 0.3256

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(Report answers accurate to 4 decimal places.)
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