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Math Help - continuous uniform distributions

  1. #1
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    continuous uniform distributions

    3. In an old computer game a white square representing a ball appears at random at the top of the
    playing area, which is 24 cm wide, and moves down the screen. The continuous random
    variable X represents the distance, in centimetres, of the dot from the left-hand edge of the
    screen when it appears. The distribution of X is rectangular over the interval
    [4, 28].
    (a) Find the mean and variance of X.
    (b) Find P( | X − 16 | < 3).

    During a single game, a player receives 12 “balls”.
    (c) Find the probability that the ball appears within 3 cm of the middle of the top edge of
    the playing area more than four times in a single game.
    I have found the mean and variance which are 16 and 48 respectively.

    However I need help/explanation on how to do part b and c. I know to work out the probability, you just find the area but how do I find it out on this one, when its has an absolute value sign and less than three is not in the interval?.

    thank you
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    Quote Originally Posted by Tweety View Post
    I have found the mean and variance which are 16 and 48 respectively.

    However I need help/explanation on how to do part b and c. I know to work out the probability, you just find the area but how do I find it out on this one, when its has an absolute value sign and less than three is not in the interval?.

    thank you
    b) |X - 16| < 3 if and only if 13 < X < 19.

    c) Hint: use the Binomial distribution.
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  3. #3
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    Quote Originally Posted by awkward View Post
    b) |X - 16| < 3 if and only if 13 < X < 19.

    c) Hint: use the Binomial distribution.

    Thank you,


    but I am not sure how you even got 19 and 13? Could you explain your thought process?

    Thanks.
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    Quote Originally Posted by Tweety View Post
    Thank you,


    but I am not sure how you even got 19 and 13? Could you explain your thought process?

    Thanks.
    Suppose |X - 16| < 3.

    Case 1. Suppose X - 16 < 0. Then |X - 16| = -(X - 16) and
    -(X - 16) < 3 <=> X - 16 > -3 <=> X > 13. So 13 < X < 16.

    Case 2. Suppose X - 16 > 0. Then |X -16| = X - 16 and
    X - 16 < 3 <=> X < 19. So 16 < X < 19.

    Combining 1 and 2, if |X - 16| < 3, then 13 < X < 19.

    Going the other way, we can show that if 13 < X < 19 then |X -16| < 3. The steps I leave up to you.

    So |X - 16| < 3 if and only if 13 < X < 19.

    So that's the proof. But you should be able to "see" this without a proof. Imagine a band of width 6 (i.e., plus or minus 3) centered at 16. Where is the left end of the band? Where is the right end?
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