Binomial Distribution

1.About 1% of the population has a particular genetic mutation. 1000 people are randomly selected.

Find the mean for the number of people with the genetic mutation in such groups of 1000.

- About 1% of the population has a particular genetic mutation. 400 people are randomly selected.

Find the standard deviation for the number of people with the genetic mutation in such groups of 400. Round your answer to three decimal places

- Assume that a procedure yields a binomial distribution with a trial repeated n=5n=5times. Use some form of technology like Excel or StatDisk to find the probability distribution given the probability p=0.373p=0.373of success on a single trial.

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

k |
P(X = k) |

0 | |

1 | |

2 | |

3 | |

4 | |

5 |

- After being rejected for employment, Kim Kelly learns that the Bellevue Credit Company has hired only two women among the last 22 new employees. She also learns that the pool of applicants is very large, with an approximately equal number of qualified men as qualified women.

Help her address the charge of gender discrimination by finding the probability of getting two or fewer women when 22 people are hired, assuming that there is no discrimination based on gender.

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

*P*(at most two) =

Because this is a serious claim, we will use a stricter cutoff value for unusual events. We will use 0.5% as the cutoff value (1 in 200 chance of happening by chance). With this in mind, does the resulting probability really support such a charge?

- yes, this supports a charge of gender discrimination
- no, this does not support a charge of gender discrimination

- A poll is given, showing 80% are in favor of a new building project.

If 9 people are chosen at random, what is the probability that exactly 2 of them favor the new building project?

- A manufacturing machine has a 1% defect rate.

If 7 items are chosen at random, what is the probability that at least one will have a defect?

- The television show
*Green’s Anatomy*has been successful for many years. That show recently had a share of 27, which means, that among the TV sets in use, 27% were tuned to*Green’s Anatomy*. An advertiser wants to verify that 27% share value by conducting its own survey, and a pilot survey begins with 11 households have TV sets in use at the time of a*Green’s Anatomy*broadcast.

Find the probability that none of the households are tuned to*Green’s Anatomy*.

*P*(none) =

Find the probability that at least one household is tuned to*Green’s Anatomy*.

*P*(at least one) =

Find the probability that at most one household is tuned to*Green’s Anatomy*.

*P*(at most one) =

If at most one household is tuned to*Green’s Anatomy*, does it appear that the 27% share value is wrong? (Hint: Is the occurrence of at most one household tuned to*Green’s Anatomy*unusual?)

- no, it is not wrong
- yes, it is wrong

- A high school baseball player has a 0.316 batting average. In one game, he gets 7 at bats. What is the probability he will get at least 5 hits in the game?

**THIS LINKS MAY BE USEFUL FOR THIS.**

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http://onlinestatbook.com/2/calculators/binomial_dist.html

https://www.youtube.com/watch?v=bvRP2CzSAtc&feature=youtu.be

http://onlinestatbook.com/2/probability/binomial.html

https://www.learner.org/series/against-all-odds-inside-statistics/binomial-distributions/

https://www.youtube.com/watch?v=oQzCO5itsvU&feature=youtu.be