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Math Help - The mean and standard deviation

  1. #1
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    The mean and standard deviation

    A heavy equipment sales person can contact either one or two customers per day with probability 1/3 and 2/3 respectively. Each contact will result in no sale or a $50 000 sale with the probabilities .9 and .1. Give the probability distribution for daily sales. Find the mean and standard deviation of the daily sales.

    Here is how I started:

    Let Y denote # of daily sales. Then Range of Y = {0,1,2}

    Given: P(contacting 1) = 1/3
    P(contacting 2) = 2/3

    P(Sale) = 1/10
    P(No Sale) = 9/10

    Possible outcomes:

    Contacted 1 sold 0
    Contacted 1 sold 1
    Contacted 2 sold 0
    Contacted 2 sold 1
    Contacted 2 sold 2

    Now I have a problem in going further, how do I calculate these probabilities to create distribution of Y. For example how to calculate contacted 2 sold 1 or contacted 2 sold 2? Are the events independent?

    Should it be something like this

    P of SS(Sale Sale) 1/10 * 1/10 = 1/100
    P of SN(Sale no Sale) 1/10*9/10= 9/100
    P of NN(no sale no sale) 9/10 *9/10 = 81/100

    then use these values to calculate:

    P(Y= 0)= 1/3 * 81/100 + 2/3 * 81/100 = 81/100

    P(Y = 1) = 1/3 * 9/100 + 2/3 * 9/100 = 9/100

    P(Y = 2) = 2/3 * 1/100 + 1/3 * 1/100 = 1/100

    ?
    Last edited by somestudent2; February 17th 2008 at 11:50 AM. Reason: added more
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by somestudent2 View Post
    A heavy equipment sales person can contact either one or two customers per day with probability 1/3 and 2/3 respectively. Each contact will result in no sale or a $50 000 sale with the probabilities .9 and .1. Give the probability distribution for daily sales. Find the mean and standard deviation of the daily sales.

    Here is how I started:

    Let Y denote # of daily sales. Then Range of Y = {0,1,2}

    Given: P(contacting 1) = 1/3
    P(contacting 2) = 2/3

    P(Sale) = 1/10
    P(No Sale) = 9/10

    Possible outcomes:

    Contacted 1 sold 0
    Contacted 1 sold 1
    Contacted 2 sold 0
    Contacted 2 sold 1
    Contacted 2 sold 2

    Now I have a problem in going further, how do I calculate these probabilities to create distribution of Y. For example how to calculate contacted 2 sold 1 or contacted 2 sold 2? Are the events independent?

    Should it be something like this

    P of SS(Sale Sale) 1/10 * 1/10 = 1/100
    P of SN(Sale no Sale) 1/10*9/10= 9/100
    P of NN(no sale no sale) 9/10 *9/10 = 81/100

    then use these values to calculate:

    P(Y= 0)= 1/3 * 81/100 + 2/3 * 81/100 = 81/100

    P(Y = 1) = 1/3 * 9/100 + 2/3 * 9/100 = 9/100

    P(Y = 2) = 2/3 * 1/100 + 1/3 * 1/100 = 1/100

    ?

    Use a contingency tree similar to that shown in the attachment, where the leaves are labled with the contribution of that leaf to the expected sales.

    RonL
    Attached Thumbnails Attached Thumbnails The mean and standard deviation-gash.png  
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  3. #3
    Junior Member
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    Thank You

    is $50000 a really necessary piece of information here? I thought we needed to find the probability distribution of sales, so it can be omitted? So we have 2 different distribution tables (I was trying to put it all in one...)

    ===for 1 contact===

    y____P(y)

    0____1/3*9/10
    1____1/3*1/10



    ===for 2 contacts====

    y___P(y)
    0___2/3*1/9*1/9
    1___2/3*(1/10*9/10+9/10*1/10)
    2___2/3*1/100

    ?
    Last edited by somestudent2; February 17th 2008 at 08:18 PM.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by somestudent2 View Post
    Thank You

    is $50000 a really necessary piece of information here? I thought we needed to find the probability distribution of sales, so it can be omitted? So we have 2 different distribution tables (I was trying to put it all in one...)

    ===for 1 contact===

    y____P(y)

    0____1/3*9/10
    1____1/3*1/10



    ===for 2 contacts====

    y___P(y)
    0___2/3*1/9*1/9
    1___2/3*(1/10*9/10+9/10*1/10)
    2___2/3*1/100

    ?

    Each leaf has a pay-off times a probability, all the information you need is there, the
    probabilities of daily sales are:

    p(\$0)=(1/3)\times (0.9) +(2/3) \times (0.81) = 0.84

    p(\$50k) = (1/3)\times (0.1) +(2/3) \times (0.18) \approx 0.1533

    p(\$100k) = (2/3) \times 0.01 \approx 0.00667

    RonL
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