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Math Help - Need STATS and PROBABILITY HELP!!!

  1. #1
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    Need STATS and PROBABILITY HELP!!!

    Given the independant variable with mean =ux=120 and uy=300 The St. Dev is Qx= 12, Qy=16. What is the mean and ST.D of each veriibles
    a) 0.8y+20 b)2x-100 c)x+2y

    Please explain how you did this and show steps so that i can understand how it was done!!

    #2. Lets say 13% of ppl are lift handed, if we randomly test 5 persons, find the probability of these:
    a) some lefties among 5 ppl
    b) there are exaclty three left handed ppl in the team
    c) there are at least three lefted hands in the team
    d)there are less than 3 lefties in the team

    Again please explain how you did this thanks!
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  2. #2
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    Re: Need STATS and PROBABILITY HELP!!!

    #1
    There are 4 rules to remember for transferring means and standard deviations
    Where A and B are constants
    If the mean of x is ux then the mean of Ax is Aux
    If the mean of y is uy then the mean of y+B is uy+B
    If the St dev of x is Qx then when all the values get increased by a factor of A to Ax then all the values get A times farther apart and the standard deviation gets A times larger. So the St dev of Ax is AQx
    If the St dev of y is Qy then when all the values get increased to y+B the values just get bigger they don't get farther apart since they were all increased by the same amount and so the standard deviation doesn't change.


    #2
    a) Find the probability that all are right handed. 1 minus this probability is the probability that not all are right handed (one or more are left handed)

    b) First find the probability that when you choose 5 people
    ABCDE
    the first 3 are left handed (L) and the last 2 are right handed (R)
    LLLRR
    Use the AND rule for this. Think of it as choosing Left and Left and Left and Right and Right

    Now there are many ways you can arrange the left and right handed people such as LRLLR or RLLLR. In fact there are 5C3 ways of 3 left handers from 5 people (this is the nCr function on your calculator)
    So the total probably of getting 3 left handers is [the probability of picking LLLRR] multiplied by 5C3

    c) Repeat what you did in b) but find the chance that there are exactly 4 left handers, and the chance there is exactly 5, you already know from b) that the chance that there are exactly 3
    The chance that there are 3 or more is [chance there are exactly 3]+[chance there are exactly 4]+[chance there are exactly 5]

    d) Hint: [chance there are less than 3] + [chance there are at least 3] = 1
    Last edited by Shakarri; March 8th 2013 at 11:08 AM.
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  3. #3
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    Re: Need STATS and PROBABILITY HELP!!!

    Thank you for replying but is it possible you can show me using the numbers. Im still confused to which number to be used with which thanks!!




    Quote Originally Posted by Shakarri View Post
    Given the independant variable with mean =ux=120 and uy=300 The St. Dev is Qx= 12, Qy=16. What is the mean and ST.D of each veriibles
    a) 0.8y+20 b)2x-100 c)x+2y

    There are 4 rules to remember for transferring means and standard deviations
    Where A and B are constants
    If the mean of x is ux then the mean of Ax is Aux
    If the mean of y is uy then the mean of y+B is uy+B
    If the St dev of x is Qx then when all the values get increased by a factor of A to Ax then all the values get A times farther apart and the standard deviation gets A times larger. So the St dev of Ax is AQx
    If the St dev of y is Qy then when all the values get increased to y+B the values just get bigger they don't get farther apart since they were all increased by the same amount and so the standard deviation doesn't change.


    #2
    a) Find the probability that all are right handed. 1 minus this probability is the probability that not all are right handed (one or more are left handed)

    b) First find the probability that when you choose 5 people
    ABCDE
    the first 3 are left handed (L) and the last 2 are right handed (R)
    LLLRR
    Use the AND rule for this. Think of it as choosing Left and Left and Left and Right and Right

    Now there are many ways you can arrange the left and right handed people such as LRLLR or RLLLR. In fact there are 5C3 ways of 3 left handers from 5 people (this is the nCr function on your calculator)
    So the total probably of getting 3 left handers is [the probability of picking LLLRR] multiplied by 5C3

    c) Repeat what you did in b) but find the chance that there are exactly 4 left handers, and the chance there is exactly 5, you already know from b) that the chance that there are exactly 3
    The chance that there are 3 or more is [chance there are exactly 3]+[chance there are exactly 4]+[chance there are exactly 5]

    d) Hint: [chance there are less than 3] + [chance there are at least 3] = 1
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