Section 6.2 (p.320)

2. Let X denote the number of defects, the p.f. of X is

Therefore,

where f(3|1) = 0.0613 and f(3|1.5) = 0.1255.

Therefore, x(1|3) = 0.2456 and x(1.5|3) = 1-0.2456 = 0.7544.

6. x(q|x) µ q³ (1-q)⁵ 2 (1-q) = 2 q³ (1-q)⁶

Compare above with Eq. (1) of Sec 5.10. We know the posterior distribution is Beta distribution with parameters a = 4 and b = 7.

8. By Eq. (15) and (16), x(q|x₁,x₂) µ f(x₁|q) f(x₂|q) x(q).

Therefore by induction we can show that x(q|x₁, …, x_n) µ f(x₁|q) … f(x_n|q) x(q) (see exercise 7)

Therefore, x(q|x) µ q³ (1-q)⁵, which indicates that it’s a Beta distribution with parameters a = 4 and b= 6.

10.

Now min(x_1,…, x₆) = 10.9 and max(x_1,…, x₆) = 11.7, we get

x(q|x) µ positive constant for 11.2 < q < 11.4

Therefore, the posterior distribution of q is a uniform distribution on (11.2, 11.4).