Integrating the probability density function over the stated interval gives

P (T < 8) = \int_{0}^{8} \frac{1}{25} e^{- t /25} d t = 1 - e^{- 8/25} = 0.274 (three significant figures) .

ii

Solution

P (T > t) = e^{- t /25}

for

t \geq 0

. Therefore

P (T > 20) = \int_{20}^{\infty} \frac{1}{25} e^{- t /25} d t = e^{- 20/25} = 0.449 (three significant figures) .

Subtract the two values of the survival function, or integrate the probability density function directly:

P (14 < T < 30) = \int_{14}^{30} \frac{1}{25} e^{- t /25} d t = e^{- 14/25} - e^{- 30/25} = 0.270 (three significant figures) .

The final zero records the requested third significant figure.

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