Answer:
The probability that a defective rod can be salvaged = 0.50
Step-by-step explanation:
Given that:
A machine shop produces heavy duty high endurance 20-inch rods
On occasion, the machine malfunctions and produces a groove or a chisel cut mark somewhere on the rod.
If such defective rods can be cut so that there is at least 15 consecutive inches without a groove.
Then; The defective rod can be salvaged if the groove lies on the rod between 0 and 5 inches i.e ( 20 - 15 )inches
Now:
P(X ≤ 5) = 
= 0.25
P(X ≥ 15) = 
= 0.25
The probability that a defective rod can be salvaged = P(X ≤ 5) + P(X ≥ 15)
= 0.25+0.25
= 0.50
∴ The probability that a defective rod can be salvaged = 0.50
Answer:
2.) 0.10 (3.) 0.10 (4.) 2.43
Step-by-step explanation:
Given that:
x p(x)
0 0.12
1 0.18
2 0.30
3 0.15
4
5 0.10
6 0.05
X : __0__ 1 ___ 2 ___ 3 _____ 4 ____ 5 ____ 6
p(x):0.12_0.18_0.30_0.15__0.10___0.10 ___0.05
Σ of p(x) = 1
(0.12 + 0.18 + 0.30 + 0.15 + x + 0.10 + 0.05) = 1
0.9 + x = 1
x = 1 - 0.9
x = 0.1
2.)
P(x = 4) = 0.10
3.)
P(x = 5) = 0.10
4.)
Σ(x * p(x)) :
(0*0.12) + (1*0.18) + (2*0.3) + (3*0.15) + (4*0.1) + (5*0.1) + (6*0.05) = 2.43
Answer:
*reward
Step-by-step explanation:
i has helped your welcome
Answer:
x = 0
Step-by-step explanation:
.......
......
0=2x
x = 0