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# Statistics 110

### Mtsac intercession chapters 4-6

Question | Answer |
---|---|

Binomial probability distribution 2 qualifications: | events are independent and each outcome is either success or failure |

Describe p, q and p+q in binomial probability | p(success)= p p(failure)=q p+q=1 |

Can p and q change in the events? | No, they must remain the same in events |

Describe x in binomial distribution | x is the number of successes, n – x is number of failures in n trials |

P(x) = nCx * | p^x*q^(n-x) |

When to use Binomialcdf() | when we have n,p and x n number of trials, p probability of success, x number of successes |

N = 100, p =.6 what is prob of exactly 50? | Binompdf(100,.6,50) |

N = 100, p =.6 what is prob of at most 50? | Binomcdf(100,.6,50) |

N=100,p=.6 what is prob of at least 50? | 1- binomcdf(100,.6,49) |

N=100,p=.6 what is prob of between 50 and 60? | binomcdf(100,.6,60)-binomcdf(100,.6,49) |

Use this chart x|p(x) find probability that x =6 | just add all probabilities and subtract from 1 |

Use this chart x|p(x) to draw histogram | just use x values as midpoints |

Use this chart x|p(x) to find µ | just plug into calculator |

Use this chart x|p(x) to find σ | just plug into calculator |

Only in the case of binomial probability µ | np |

Variance σ^2= | npq |

Continuous prob distribution we use | continuous random variables |

Types of continuous prob distribution | uniform prob distribution: standard normal and normal |

When p(x=#) = when dealing with a rectangle | p(x)=0 |

Find P(4.2<x<6.5) if we are between 2 and 18 | 6.5-4.2 mulitplied by 1/(18-2) |

Find k such that P(x>k)= .1 0 to 12 | (12-k)(1/12)= .1 because there is only 10%to the right of k |

uniform probability distribution means | you are working with a rectangle |

Standard normal distribution graph is | bell shaped |

Standard normal distribution : mean is equal to | mean= meadian = mode |

µ=0 and σ=1 when graph is___________ and | bell shaped and we are working with standard normal distributujtion |

p(-1<z<1) = | normalcdf(-1,1,0,1) |

find k such that P(z<k) = 0.845 | k= invNorm(.845,0,1) |

with inverse we always work with the | left value |

Zα is | α is the area of the right tail so |

For normally distributed we can also use µ and σ other than 1 and 0 and simply | input those two values in order followed by the other sides |

What do you use for infinity? | E99 |

The middle 90% from the rest (normally distributed) | 1-.90= .1, .1/2 gives P5 k1= p5, k2= P95, x1= invNorm(.05,µ,σ) |

z is related to | standard normal distribution; mu= 0 omega= 1 |

x is related to | normal distribution; mu and omega dont have to be 0,1 |

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