MATH 511, Meade HW Solutions
3.2 –3 (omit c), 5, 14, 18, **3c, **8 2/13/04
3.
a. f(x) = x/10, x = 1,2,3,4
E(x) = (1) (1/10) + (2) (2/10) + (3) (3/10) + (4) (4/10)
= (1 + 4 + 9 + 16 )/10 = 30/10 = 3
b. f(x) = x/55 , x = 1, 2, 3, 4 ,5 , 6, 7, 8, 9, 10
E(x) = x=1∑10 x2/55 = [(10)(11)(21)/(6)]/55 = 385/55 = 7
**c. f(x) = 3(1/4)x , x = 1, 2, 3, …
E(x) = x=1∑∞ 3x (1/4)x = x=1∑∞ 3x/4x ≤ x=1∑∞ 3x/4x = 3
This assures that the series does indeed converge. The series, x=1∑∞ x/4x ,
can now be approximated. After about ten approximations it is clear that
the series converges to 0.44444… which is 4/9.
So the E(X) = 3(4/9) = (4/3)
d. f(x) = (1/30)(x+1)2 , x = 0, 1, 2, 3
E(x) = (1/30) [ 0 + 4 + 18 + 48] = 70/30 = 7/3
e. f(x) = 2/n(n+1)x , x = 1, 2, 3, …, n
E(x) =2/n(n+1) i=1∑n x2 = [2/n(n+1)] [n(n+1)(2n+1)/6)] = (2n+1)/3
5. E(x) = [2,987,994(0) + 12,000(25) + 4(10,000) + 1(50,000) + 1(200,000)] - 0.50
3,000,000
= 19 ⅔ cents – 50 cents = - 30⅓ cents
**8. f(x) = 6/( π2x2), x = 1, 2, 3, …
E(x) = (6/ π2) x=1∑∞ x/x2 = (6/ π2) x=1∑∞ 1/x ; this is the harmonic series, which
diverges. Therefore E(x) does not exist. One way to show this is the
integral test for infinite series.
14.
a. [16(25) + 3(100) + 1(300)]/20 = 1,000/20 = 50. Note that this makes sense:
1000 students divided into 20 classes; the average class size must be
1000/20=50.
b. f(x) = 0.4, x = 25
= 0.3, x = 100
= 0.3, x = 300. Note that here you are selecting a random student, not
a random class.
c. E(x) = 25(0.4) + 100(0.3) + 300(0.3) = 130
18.
a. E[(X - µ) / σ] = (1/σ) E(X - µ) = (1/σ) [E(X) – E(µ)] =(1/σ) [µ – µ] = 0
b. E{[(X – µ) / σ]2} = (1/σ2) (E[(X – µ)2]) = (1/σ2) (E(X2) – 2µE(X) + µ2)
= (1/σ2) (E(X2) – 2µ2 + µ2) = (1/σ2) (E(X2) – µ2)
= (1/σ2) σ2 = 1
