**17. Correct. The answer is true**

(a) The mode corresponds to the point where the density f(z) is a maximum. The relative maximum of f(z) occur where the derivative of the density function is zero:

df/dz[4z(9-z^{2})/81] = (36-12z^{2})/81 =0.

Then z* = 3^{0.5}
≈ 1.73 (US$ 1730) is the mode. Check that the second derivative, -24z/81
is negative at z* = 3^{1/2}. So, z* defines a maximum.

(b) The median is the value at which P(Z≤c) = 1/2. For 0<c<3:

P(Z≤c) = 4/ 81 ∫_{0}^{c}
z(9-z^{ 2})dz

P(Z≤c) = 4/
81(9c^{2}/2 – c^{4}/40

Setting this equal to 1/2:

2c^{4}-36c^{2}+81=0

c^{2} = 9_{}(9/2)*2^{1/2}

Since c must
be between 0 and 3, c^{2} = 9- (9/2)*2^{1/2}

c ≈ 1.62 (US$ 1620).

(c) E(Z) =4/81 ∫_{0}^{3} z^{2}
(9-z^{ 2})dz

E(Z) = 4/81 (3z^{ }– z^{5}
/ 5│_{0}^{3}

E(Z) = 1.60 (US$ 1600).

Mean < Median < Mode.