The waiting time at which 10 percent of the people would continue to hold is given as 0.316
We have to solve for X ~ Exponential(λ).
then E(X) = 1/λ = 3,
= 0.3333
Remember that the cumulative distribution function of X is F(x) = 1 - e^(-λx). ; x is equal to the time in over case
For 10 percent of the people we would have a probability of
10/100 = 0.1
we are to find
P(X ≤ t)
= 1 - e^(0.3333)(t) = 0.1
Our concern is the value of t
Then we take the like terms
1-0.1 = e^(0.3333)(t)
1/0.9 = e^(0.3333)(t)
t = 3 * ln(1/0.9)
= 0.3157
~0.316
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