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Thread: Binomial distribution problem

  1. #1
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    Binomial distribution problem

    Hi I'd like to check my answer to this what I believe is a binomial problem whose statement is in the below paragraph. so I calculated the probability of having 1% success rate or less which was 5 success in 12 trials so I said the price should be $150 =50 + 20x5
    Could someone tell me if my conclusion is correct because I'm pretty shakey on how this problem should be approached.
    The problem statement is below.


    an audio player uses a low quality hard drive. The initial cost of building the player is $50. The hard drive fails with probability
    1/12 with each month of use. The cost to repair the hard drive is $20. If a 1 year warranty is offered, how much should the manufacturer charge so that the chance of losing money on a player is 1% or less.
    Last edited by ineedhelplz; Apr 17th 2017 at 08:42 AM.
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  2. #2
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    Re: Binomial distribution problem

    Quote Originally Posted by ineedhelplz View Post
    Hi I'd like to check my answer to this what I believe is a binomial problem whose statement is in the below paragraph. so I calculated the probability of having 1% success rate or less which was 5 success in 12 trials so I said the price should be \$150 =50 + 20x5
    Could someone tell me if my conclusion is correct because I'm pretty shakey on how this problem should be approached.
    The problem statement is below.


    an audio player uses a low quality hard drive. The initial cost of building the player is \$50. The hard drive fails with probability
    1/12 with each month of use. The cost to repair the hard drive is \$20. If a 1 year warranty is offered, how much should the manufacturer charge so that the chance of losing money on a player is 1% or less.
    reposted to fix \$ escapes
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  3. #3
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    Re: Binomial distribution problem

    an audio player uses a low quality hard drive. The initial cost of building the player is \$50. The hard drive fails with probability [/FONT][/COLOR]1/12 with each month of use. The cost to repair the hard drive is \$20. If a 1 year warranty is offered, how much should the manufacturer charge so that the chance of losing money on a player is 1% or less.
    the probability of losing money is

    $P[\text{loss}] = P\left[P-50 <20k\right] = P\left[\dfrac{P-50}{20}<k\right] = 1 - P\left[k \leq \dfrac{P-50}{20}\right] \leq 0.01$

    where $k$ is the number of repairs needed during the warranty period

    $P\left[k \leq \dfrac{P-50}{20}\right] \geq 0.99$

    Using a table of the CDF of the $binomial\left(12,\frac{1}{12}\right)$ distribution we find that

    $k = 4$ is the smallest value which produces a probability greater than 0.99

    and thus we must plan for $4$ failures which leads to a retail price of

    $P=20k+50 = \$130$
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