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Random variables and distribution functions
14 original Exam P questions on random variables and distribution functions.
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Exam PRandom variables and distribution functionsCore
A loss X has CDF F(x)=1−(1+x)−3 for x>0. Find P(1<X≤4).
A0.0080
B0.1170
C0.1250
D0.8750
E0.9920
Solution
- Any interval probability is a difference of CDF values: P(a<X≤b)=F(b)−F(a).
- F(4)=1−5−3=1−0.008=0.992.
- F(1)=1−2−3=1−0.125=0.875.
- P=0.992−0.875=0.117, so about 11.7% of losses fall in that band.
- Because the variable is continuous, using strict or weak inequalities makes no difference to the answer.
Trap. Integrating the density from scratch when the CDF is already given - slower and an easy place to lose a constant.
Exam PRandom variables and distribution functionsExam level
A random variable is uniform on the integers 1 through 20. Find P(X>14).
A0.0500
B0.2500
C0.3000
D0.3500
E0.7000
Solution
- Each of the 20 integers has probability 1/20.
- 'Greater than 14' means the values 15 through 20 - six of them.
- P(X>14)=6/20.
- =0.30. Counting carefully matters: 'at least 14' would give seven values and 0.35.
Trap. Including 14 itself and answering 0.35.
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