Original practice · fully worked
Original variant: batch survival above a load threshold
Component strengths are normal with mean 100 and standard deviation 10. One hundred independent components are tested at load 90. Find the probability that at least 78 components survive.
- A
0.84 - B
0.90 - C
0.93 - D
0.96 - E
0.99
Answer and calculation
One component survives with probability Φ(1)=0.84134. The Binomial(100,0.84134) upper tail from 78 is 0.9606, choice D.
Setup
Setup
A component survives when its strength exceeds 90, which is one standard deviation below the mean. The one-component survival probability is approximately 0.841345.
p=Pr(S>90)=Φ(1)=0.841345 Model
Model
Independence makes the number of survivors among 100 components binomial with that success probability.
K∼Binomial(100,p) Compute
Compute
The exact binomial upper tail beginning at 78 survivors is approximately 0.960595.
Pr(K≥78)=k=78∑100(k100)pk(1−p)100−k=0.960595 Answer
Answer
The batch-survival probability rounds to 0.96.
0.96(D) 