Answer:
Step-by-step explanation:
There are total 5 batches and two boards are selected from each batch for inspection.
Let the boards are numbered from 1 to 5.
If the selected boards are 1 and 2, then it is represented in pair as (1, 2).
If the selected boards are 1 and 3, then it is represented in pair as (1, 3).
Similarly, other pairs can be obtained.
a) Let X be the number of defective boards observed among the two inspected.
If the boards 1 and 2 are the only defective boards in a lot of five, then
(1,2),x=2; (1,3),x=1; (1,4),x=1; (1,5),x=1;
(2,3),x=1; (2,4),x=1; (2,5),x=1;
(3,4),x=0;\ (3,5),x=0;\ (4,5),x=0.(3,4),x=0; (3,5),x=0; (4,5),x=0.
P(X=0)= 3/10 =0.3
P(X=1)={6 \over 10}=0.6P(X=1)= 6/10 =0.6
P(X=2)={1 \over 10}=0.1P(X=2)= 1/10 =0.1
b)
x 0 1 2
p(x) 0.3 0.6 0.1
μ X =0⋅0.3+1⋅0.6+2⋅0.1=0.8
σ X ² =(0−0.8)² ⋅0.3+(1−0.8)² ⋅0.6+(2−0.8)² ⋅0.1=0.36
μₓ=0.8, σX =0.6