Answers: Continuous Probability
Distributions
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1. There are very few continuous distributions that we
can do much with without a lot of math, but the simple
uniform distribution is one of them. A uniform distribution
is one that is just thatuniform. If we imagine any
possible number between 0 and 2 as equally likely, we have
an infinite possible number of numbers, so the probability
of any particular number is close to zero. Instead of
focusing on particular numbers, we have to focus on
intervals.
The probability that some number between zero and two
will come up will be one. To find the probability of any
interval, we find the length of the interval and multiply by
.5. Hence, the probability of getting a number between zero
and one is .5.
For the following, draw the picture and compute the
probabilities. What is:
 a) P(0<x<.8)
 b) P(x>.8)
 c) P(.5<x<1.5)
 d) P(.5<x<1.2) or P(.8<x<1.3)
 e) compute the probability that we will not be in the
interval of .4 to 1.5.
 f) compute the middle interval in which we will be
90% of the time.
 g) compute the tails (outside intervals) that we will
be in 5% of the time
a) .8*.5 = .40
b) 1  .4 = .6
c) 1*.5 = .5
d) P(.5<x<1.3) = .8*.5 = .4
e) 1 P(.4<x<1.5) = 1  1.1*5 = 1  .55 = .45
f) 90% of 2 is 1.8, which centered will be from .1 to
1.9.
g) 95% of the time we will be in the interval .05 to 1.95.
5% of the time we will be between 0 and .05 or 1.95 and
2.
2. The picture below shows a continuous probability
distribution with the formula, y=.5x for the range, x=0 to
x=2. In this case, area is probability.
 To compute the probability of being between 0 and 1,
for example, you need to know y when x equals 1. The
formula says it is .5. Hence, to find the probability
that x is between 0 and 1 Prob(0<x<1), we compute
the area of the triangle with a length of 1 and a height
of .5. The answer is .25 because .5HW = .5*.5*1.
 a) Show that the area under the line is equal to
1.
 b) Find the following probabilities:
 Prob(0<x<.5)
 Prob(0<x<1.5)
 Prob(1.5<x<2)
 Prob(.5<x<1.5)
 Prob(x<.5 or x>1.5)
 c) Suppose that we draw twice from this
distribution and that probability of the second draw is
independent of the first. What is the probability of
getting greater than 1.5 on both draws? What is the
probability of getting less than 1.5 on both draws? What
is the probability of getting a result greater than 1.5
on one draw and less than 1.5 on the other draw?
a) Length*height*.5 is the
area of a triangle. This triangle has length of 2 and height
of 1, so 2*1*.5 = 1. The area is one, and for a continuous
distribution, area is probability.
b) Prob(0<x<.5)
 Prob(0<x<.5) =
.5*.5*.25 = .0625
 Prob(0<x<1.5)
=.5*1.5*.75 = .5625
 Prob(1.5<x<2) = 1
Prob(0<x<1.5) = 1  .5625 =
.4375.
 Prob(.5<x<1.5) =
.5625  .0625 = .5
 Prob(x<.5 or x>1.5)
= .0625 + .4375 = .5
 c) Both less than 1.5:
.5625*.5625 = .3164: both greater than 1.5: .4375*.4375 =
.1914; one greater and one less: 1  .3164  .1914 =
.4922.
3. Here is another continuous probability distribution
for which we can work the problems without using complex
math or tables. The range of possible outcomes is from zero
to 1. The probability density function is y=2x. The
probability histogram looks like this:
If you remember geometry, you can find the area of any
interval using the formula for a triangle, which is .5(H*W)
(one half height multiplied by width). We can find the
height at any spot with the formula y=2x, where y is the
height and x is the distance out on the xaxis. The total
area under the triangle is 1, which it must be.
 Find the following
probabilities:
 P(0<.5) = .5*.5*1 =
.25 (Since the formula is .5*x*2x, it becomes
x^{2}.)
P(.5>x>1) = 1  .25 = .75
P(.2<x<.7) = .7*.7  .2*.2 = .49  .04 = .45. (We
find the probability of being less than .7 and then
subtract from that the probability of being less than
.2)

 Find the interval so there is a 5% chance of being
below the interval and a 5% chance of being above
it.
.05 = x^{2}; x =
.2236;
.95 = x^{2}; x = .9747
4. A uniform distribution that has a range from 0 to 4
and that is continuous would look like the picture below.
Probability is area.
Find:
 (0 < P(x) < 4) =
1
 (P(x) > 1) =
.75
 (P(x) = 1) =
0
 (P(x) < 1) =
.25
What is the expected value of x? (You cannot do the math,
but you should be able to see it.)
2  it is the middle, the
balancing point.
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