I have found the mean and variance which are 16 and 48 respectively.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
(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.
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?.
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?