# continuous uniform distributions

• Apr 7th 2010, 12:06 PM
Tweety
continuous uniform distributions
Quote:

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
• Apr 7th 2010, 02:12 PM
awkward
Quote:

Originally Posted by Tweety
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.
• Apr 7th 2010, 02:20 PM
Tweety
Quote:

Originally Posted by awkward
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.
• Apr 7th 2010, 04:49 PM
awkward
Quote:

Originally Posted by Tweety
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?