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Math Help - Binomial Trials, Probability, Please Help, I will love you.

  1. #1
    IHateMyProf
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    Binomial Trials, Probability, Please Help, I will love you.

    Alright here is the break down.....I would really appreciate someone helping me realize how to do this problem. I am so lost, my prof is a ****.



    Rosemary works in phone sales for a car insurance company. She has a success rate of 35% of getting a caller to buy car insurance from her company. The sales manager tells the phone sales staff that anyone who makes 10 sales that day will get a $500 bonus. If we take each call Rosemary handles as a binomial trial then:

    A) How many call must she take to insure a 95% probability that she will get the bonus?

    B) On average, how many sales will she amke if she takes the number of calls you found in part a above?

    C) Find the avreage number of calls she must take for each sale.



    Thank you soooo much for any help.

    -IHateMyProf
    Last edited by ThePerfectHacker; April 24th 2006 at 06:34 PM.
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  2. #2
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    Quote Originally Posted by IHateMyProf

    Rosemary works in phone sales for a car insurance company. She has a success rate of 35% of getting a caller to buy car insurance from her company. The sales manager tells the phone sales staff that anyone who makes 10 sales that day will get a $500 bonus. If we take each call Rosemary handles as a binomial trial then:

    A) How many call must she take to insure a 95% probability that she will get the bonus?

    Because by the binomial probability you need at least 10 out of n. Thus, you need,
    \sum^n_{k=10}\frac{n!}{k!(n-k)!}(.35)^k(.65)^{n-k}\geq .95

    Which is just a trail an error problem. You need one mighty program do to this. I seriouslly do not recommend doing this by hand unless you are Euler, anyways it seems that the answer is the "ultimate answer" which is 42.

    There might be more simpler ways of doing this, I am not an expert in probability the other mod is he might be able to help you out.
    Last edited by ThePerfectHacker; April 24th 2006 at 08:26 PM.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by IHateMyProf
    Alright here is the break down.....I would really appreciate someone helping me realize how to do this problem. I am so lost, my prof is a ****.



    Rosemary works in phone sales for a car insurance company. She has a success rate of 35% of getting a caller to buy car insurance from her company. The sales manager tells the phone sales staff that anyone who makes 10 sales that day will get a $500 bonus. If we take each call Rosemary handles as a binomial trial then:

    A) How many call must she take to insure a 95% probability that she will get the bonus?
    Mr Hacker has given one method of solving this, but the method I suspect
    you are supposed to employ is to use the Normal approximation to the
    Binomial distribution.

    Under the Normal approximation the number of sales made has approximatly
    a Normal distribution with mean 0.35 n, where n is the number
    of calls taken, and standard deviation \sqrt{0.35 \times 0.65\times n}}

    Then looking up the critical value of the Standard Normal distribution for 5\%
    probability of getting a value less than the critical value gives \sim  1.65 (this
    corresponds to a 95\% probability of exceeding the critical value), so:

    <br />
9.5 \approx 0.35 n -1.65 \sqrt{0.35 \times 0.65\times n}}<br />

    In this last equation the LHS is 9.5 rather than 10 as a continuity
    correction.

    This is a quadratic in \sqrt n which may be solved using the quadratic formula.
    This gives two solutions one of which is unphysical and can be discarded,
    and the other is \sqrt n \approx 6.4541 so n \approx 41.66, but as n must be an integer
    we round it up to 42.

    RonL
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by IHateMyProf
    B) On average, how many sales will she amke if she takes the number of calls you found in part a above?
    The avaerge number of sales made is the number of calls times the probability
    of making a sale on any one sale (this is virtualy the frequentist definition
    of this probability), so the average number of sales from 42 calls is:

    42 \times 0.35 = 14.7

    C) Find the avreage number of calls she must take for each sale.
    In the long run for every 42 calls she will make 14.7 sales so the average
    number of calls per sale is:

    42/14.7=2.86
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  5. #5
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    Quote Originally Posted by CaptainBlack
    Mr Hacker has given one method of solving this, but the method I suspect
    you are supposed to employ is to use the Normal approximation to the
    Binomial distribution.
    I suspected that the problem was solved differently from the mess I proposed. May you give a quick theory about what you said to him, the only probability I took is High School thus, I was not able to answer his question efficiently.

    Basically, as I understand you want to approximate the binomial distribution (that part I understand) as a normal distribution (you mean f(x)=e^{x^2}?).
    But how do you find this function, give a quick theory.
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker
    I suspected that the problem was solved differently from the mess I proposed. May you give a quick theory about what you said to him, the only probability I took is High School thus, I was not able to answer his question efficiently.

    Basically, as I understand you want to approximate the binomial distribution (that part I understand) as a normal distribution (you mean f(x)=e^{x^2}?).
    But how do you find this function, give a quick theory.
    You just replace the binomial distribution by a normal with the same mean
    and variance (or standard deviation).

    The mean and variance of a binomial distribution with probability of success
    on a single trial of p, and n trials are:

    <br />
\mu=n.p<br />
    <br />
\sigma^2=n.p.(1-p)<br />

    The theory about how the convergence works you will have to look up (but
    it has its peculiarities because you are approximating a discrete with a
    continuous distribution).

    RonL

    RonL
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