# 5.11: Chapter Solution (Practice + Homework)

Figure $$\PageIndex{41}$$

• This is a conditional probability question. $$P(x > 21 | x > 18)$$. You can do this two ways:
• 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.

1. $$P(X>650)=\frac{700-650}{700-300}=\frac{50}{400}=\frac{1}{8}=0.125$$.
2. $$P(400<650)> 84. 1. \(X =$$ the useful life of a particular car battery, measured in months.
2. $$X$$ is continuous.
3. 40 months
4. 360 months
5. 0.4066
6. 14.27

86.

1. $$X =$$ the time (in years) after reaching age 60 that it takes an individual to retire
2. $$X$$ is continuous.
3. five
4. five
5. Check student’s solution.
6. 0.1353
7. before
8. 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)> 1. \(P(T<2)=1-1-e^{-\frac{2}{7}} \approx 0.2485$$.
2. $$P(T>15)=1-P(T<15)=1-\left(1-e^{-\frac{15}{7}}\right) \approx e^{-\frac{15}{7}} \approx 0.1173$$.
3. $$P(T>15 | T>10)=P(T>5)=1-\left(1-e^{-\frac{5}{7}}\right)=e^{-\frac{5}{7}} \approx 0.4895$$.
4. Let $$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$$.