St361HW4Sol
5.26 (a)
x is a discrete random variable, so  =
 xp(x)
= (0)(.08) + (1)(.15) + (2)(.45) + (3)(.27) + (4)(.05) =
x
2.06.
5.28
x is a discrete random variable, so
 p(x) must equal 1.
x
That is, p(0) + p(1) + ... + p(5) = 1. Using the formula for p(x), the equation becomes c(5-0) + c(5-1) + c(5-2)
+ c(5-3) + c(5-4) + c(5-5) = c(5+4+3+2+1+0) = 15c = 1, so c = 1/15.
3)
(a)
The total area under f(x)=1/2*(2*1)=1
(b) for less than 1 hour
P(0=<x<1)= area over the interval [0,1]=1/2*(0.5*1)=1/4=0.25
1.19
(a) The density curve forms a rectangle over the interval [4, 6]. For this reason, uniform densities
are also called rectangular densities by some authors. Areas under uniform densities are easy to find
(i.e., no calculus is needed) since they are just areas of rectangles. For example, the total area under this
density curve is 12 (6  4) = 1.
height = 1/(6-4) = 1/2
x
4
6
(b) The proportion of values between 4.5 and 5.5 is depicted (shaded) in the diagram below.
The area of this rectangle is 12 (5.5  4.5) = .5. Similarly, the proportion of x values that
exceed 4.5 would be 12 (6  4.5) = .75.
x
4
4.5
5.5
6
(c) The median of this distribution is 5 because exactly half the area under this density sits over
the interval [4,5].
1.26
(a) (ii) qualifies as a distribution because the probabilities add exactly to 1. On the other hand, the
probabilities in (i) add to .7, which disqualifies (i). In (iii), notice that the probability associated with
x = 4 is negative, which is impossible for a valid distribution.
(b) Using (ii), the proportion of cars having at most 2 (i.e., 1 or less) under-inflated tires is
p(0)+p(1)+p(2) = .4 + .1 +.1 = .6. Similarly, Proportion(x  1) = 1- Proportion(x =0) = 1-.4 = .6.