Mathematics 243
March 16 Homework Solutions
(1.1)
1. We need to check that the function f is non-negative (which it obviously is) and has area 1 underneath the
curve (again obvious).
2. The area of the shaded triangle (left below) is 1/4.
p
3. The area of the shaded triangle (middle below) is 2m2 . We want m such that 2m2 = .5. So m = 1/ (2).
R1
R1
4. The mean of the random variable is 0 xf (x) dx = 0 x2x dx = 2x2 |10 = 2/3.
Density
0.000
0
0
0.005
0.010
0.015
2
y
y
1
2m
2
Histogram of time
0
1 2
1
0
x
m
1
0
20
x
40
60
80
100
120
140
time
(1.2)
1. A histogram (above) of the times between scores looks like the exponential might fit it.
2. Since the sample mean is 36.85 and the mean of the exponential is 1/λ, it is reasonable to choose λ = 1/36.85.
The resulting density is included on the histogram above.
3. A comparison of the model and the data is below.
> hist(time,prob=T)
> lambda=1/mean(time)
> x=seq(0,140,.01)
> lines(x,dexp(x,lambda))
> pexp(10,lambda) ; pexp(20,lambda) ; pexp(30,lambda) ; pexp(40,lambda)
[1] 0.2376659
[1] 0.4188468
[1] 0.5569671
[1] 0.662261
> sum(time<=10)/60 ; sum(time<=20)/60 ; sum(time<=30)/60 ; sum(time<=40)/60
[1] 0.06666667
[1] 0.3666667
[1] 0.4833333
[1] 0.65