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    smrtboy90's Avatar
    smrtboy90 Posts: 2, Reputation: 1
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    #1

    Oct 5, 2010, 11:52 AM
    Help solving binomial distributions
    Does anyone know the solutions to these distributions?

    A survey found that 25% of pet owners had their pets bathed professionally rather than do it themselves. If 18 pet owners are randomly selected, find the probability that exactly 5 people have their pets bathed professionally.

    Approximately 10.3% Of American High School Students Drop Out Before Graduation. Choose 10 Students Entering High School At Random. Find the probability that at least 6 will graduate

    Thanks
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #2

    Oct 6, 2010, 07:01 AM

    From your previous thread, I'll take this as a follow up.

    1. n = 18, p = 0.25, q = 0.75

    Let X be the event that a pet owner had their pets bathed professionally.

    X ~ B(18, 0.25)

    Find P(X = 5) using the equation I gave you in your previous thread.

    2. Practically the same thing here.

    n = 10, p = 0.103, q = 0.897

    Let X be the number who drop out.

    Then you need to find which means 6, 7, 8, 9 or 10 who graduate.

    Now, you if X represent those that drop out, and you need at least 6 graduates, it means that you need to have 4, 3, 2, 1 or 0 drop outs, which is



    Post what you get! :)

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