# 5.10: Chapter Solution (Practice + Homework)

- Page ID
- 5570

Figure \(\PageIndex{41}\)

- Draw the graph where a is now 18 and b is still 25. The height is \(\frac{1}{(25-18)}=\frac{1}{7}\)

So, \(P(x>21 | x>18)=(25-21)\left(\frac{1}{7}\right)=4 / 7\). - Use the formula: \(P(x>21 | x>18)=\frac{P(x>21 \cap x>18)}{P(x>18)}\)

\(=\frac{P(x>21)}{P(x>18)}=\frac{(25-21)}{(25-18)}=\frac{4}{7}\).

**82**.

- \(P(X>650)=\frac{700-650}{700-300}=\frac{50}{400}=\frac{1}{8}=0.125\).
- \(P(400
<650)>

**84**.

- \(X =\) the useful life of a particular car battery, measured in months.
- \(X\) is continuous.
- 40 months
- 360 months
- 0.4066
- 14.27

**86**.

- \(X =\) the time (in years) after reaching age 60 that it takes an individual to retire
- \(X\) is continuous.
- five
- five
- Check student’s solution.
- 0.1353
- before
- 18.3

**88**.

a

**90**.

c

**92**.

Let \(X =\) the number of no-hitters throughout a season. Since the duration of time between no-hitters is exponential, the number of no-hitters per season is Poisson with mean \(\lambda = 3\).

Therefore, \((X=0)=\frac{3^{0} e^{-3}}{0 !}=e^{-3} \approx 0.0498\)

Let \(T =\) duration (in minutes) between successive visits. Since patients arrive at a rate of one patient every seven minutes, \(\mu = 7\) and the decay constant is \(m = \frac{1}{7}\). The cdf is \(P(T<t)>

- \(P(T<2)=1-1-e^{-\frac{2}{7}} \approx 0.2485\).
- \(P(T>15)=1-P(T<15)=1-\left(1-e^{-\frac{15}{7}}\right) \approx e^{-\frac{15}{7}} \approx 0.1173\).
- \(P(T>15 | T>10)=P(T>5)=1-\left(1-e^{-\frac{5}{7}}\right)=e^{-\frac{5}{7}} \approx 0.4895\).
- Let \(X = #\) of patients arriving during a half-hour period. Then \(X\) has the Poisson distribution with a mean of \(\frac{30}{7}\), \(X \sim \text{Poisson}(\frac{30}{7})\). Find \(P(X>8)=1-P(X \leq 8) \approx 0.0311\).