This Exam P sample reference tests Binomial Tail with Normal Input. A cable survives the test with probability Φ(1.28)=0.89973. An exact Binomial(400,0.89973) upper tail from 349 is 0.9676, which matches choice D.
How to solve this Binomial Tail with Normal Input question
Setup
Setup
First convert the normal strength model into the survival probability for one cable. The load is 1.28 standard deviations below the mean, giving survival probability approximately 0.899727.
p=Pr(S>12400)=Φ(2512432−12400)=Φ(1.28)=0.899727
Model
Model
With independent cables, the number surviving among 400 is binomial with that survival probability.
K∼Binomial(400,p)
Compute
Compute
Evaluate the exact binomial upper tail beginning at 349 survivors. The result is approximately 0.967553.
Pr(K≥349)=k=349∑400(k400)pk(1−p)400−k=0.967553
Answer
Answer
The required probability rounds to 0.97.
0.97(D)
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EThe value 1.00 treats the large binomial tail as certain. The exact tail still leaves about 3.24% probability below 349 survivors.
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
Variant answer in brief
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.
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