The time that a butterfly lives after emerging from its chrysalis can be modelled by a random variable tt, the model here taking the probability that a butterfly survives for more than tt days as

The probabilities that either a) the butterfly dies within 7 days or b) the butterfly lives longer than 7 days must add up to one. We know the second probability (lives longer than 7 days)--it is given. So now set up an equation and solve: P(t > 7) = 36 / 13² = 36/169 --> 36/169 + x = 1 --> x = 1 - 36/169 = 133/169 ~ 0.7870 or 78.70% b) 7% is the probability (the expected value) that a butterfly will survive more than t days. Again, this formula is given: 36/(t + 6)² = 0.07 --> (t + 6)² = 36/.07 --> t = âš(36/.07) - 6 ~ 16.68 days This is the value of t that will give a probability of 7%. c) This one is a little trickier. Technically the mean lifetime should be calculated as: â«t * p(t)dt where p(t) is the probability distribution function But we have a cumulative probability distribution. I think we need to find the probability distribution from this cumulative one. Which is easy because the cumulative distribution function is an integral of the probability distribution function: P(t) = â«p(x)dx, from x = t, to x = âž Now hopefully it's easy to see that P(t) is an anti-derivative of p(x). Said the other way p(x) is the derivative of P(t): P'(t) = -72/(t + 6)Âł --> but p(t) shouldn't be negative--also you can convince yourself that it should be positive by using the above value (from t --> âž...you get -0 - a negative = a positive) p(t) = 72/(t + 6)Âł Now we can calculate the mean lifetime by integrating: 72â«t/(t + 6)Âłdt --> there's a very easy u-substitution to use here: u = t + 6 --> du = dt --> t = u - 6 --> â«(u - 6)/uÂłdu = â«(1/u² - 6/uÂł)du = -1/u + 3/u² + C --> now evaluate from t = 0 to t = âž or from u = 0 + 6 = 6 to u = âž 0 - (-1/6 + 3/36) = 1/6 - 1/12 = 1/12 --> multiply by the 72 72/12 = 6 days

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