# 5.11: Chapter 5 Solution (Practice + Homework)

1.

Uniform Distribution

3.

Normal Distribution

5.

$$P(6 < x < 7)$$

7.

one

9.

zero

11.

one

13.

0.625

15.

The probability is equal to the area from $$x = \frac{3}{2}$$ to $$x = 4$$ above the x-axis and up to $$f(x) = \frac{1}{3}$$.

17.

It means that the value of $$x$$ is just as likely to be any number between 1.5 and 4.5.

19.

$$1.5 \leq x \leq 4.5$$

21.

0.3333

23.

zero

24.

0.6

26.

$$b$$ is 12, and it represents the highest value of $$x$$.

28.

six

30.

Figure 5.38

33.

$$X =$$ The age (in years) of cars in the staff parking lot

35.

0.5 to 9.5

37.

$$f(x) = \frac{1}{9}$$ where x is between 0.5 and 9.5, inclusive.

39.

$$\mu = 5$$

41.

1. Check student’s solution.
2. $$\frac{3.5}{7}$$

43.

1. Check student's solution.
2. $$k = 7.25$$
3. 7.25

45.

No, outcomes are not equally likely. In this distribution, more people require a little bit of time, and fewer people require a lot of time, so it is more likely that someone will require less time.

47.

five

49.

$$f(x)=0.2 e^{-0.2 x}$$

51.

0.5350

53.

6.02

55.

$$f(x) = 0.75e^{-0.75x}$$

57.

Figure 5.39

59.

0.4756

61.

The mean is larger. The mean is $$\frac{1}{m}=\frac{1}{0.75} \approx 1.33$$, which is greater than 0.9242.

63.

continuous

65.

$$m = 0.000121$$

67.

1. Check student's solution
2. $$P(x < 5,730) = 0.5001$$

69.

1. Check student's solution.
2. $$k = 2947.73$$

71.

Age is a measurement, regardless of the accuracy used.

73.

1. Check student’s solution.
2. $$f(x)=\frac{1}{8} \text { where } 1 \leq x \leq 9$$
3. five
4. 2.3
5. $$\frac{15}{32}$$
6. $$\frac{333}{800}$$
7. $$\frac{2}{3}$$

75.

1. $$X$$ represents the length of time a commuter must wait for a train to arrive on the Red Line.
2. Graph the probability distribution.
3. $$f(x)=\frac{1}{8} \text { where } 0 \leq x \leq 8$$
4. four
5. 2.31
6. $$\frac{1}{8}$$
7. $$\frac{1}{8}$$

77.

d

78.

b

80.

1. The probability density function of $$X$$ is $$\frac{1}{25-16}=\frac{1}{9}$$.
$$P(X>19)=(25-19)\left(\frac{1}{9}\right)=\frac{6}{9}=\frac{2}{3}$$.
2. $$P(19<22)> 3. 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, $$(X=0)=\frac{3^{0} e^{-3}}{0 !}=e^{-3} \approx 0.0498$$

NOTE

You could let $$T =$$ duration of time between no-hitters. Since the time is exponential and there are 3 no-hitters per season, then the time between no-hitters is $$\frac{1}{3}$$ season. For the exponential, $$\mu = \frac{1}{3}$$.
Therefore, $$m = \frac{1}{\mu} = 3$$ and $$T \sim E x p(3)$$

The desired probability is $$P(T>1)=1-P(T<1)=1-\left(1-\mathrm{e}^{-3}\right)=\mathrm{e}^{-3} \approx 0.0498$$.

1. Let $$T =$$ duration of time between no-hitters. We find $$P(T > 2|T > 1)$$, and by the memoryless property this is simply $$P(T > 1)$$, which we found to be 0.0498 in part a.
2. Let $$X =$$ the number of no-hitters is a season. Assume that $$X$$ is Poisson with mean $$\lambda = 3$$. Then $$P(X > 3) = 1 – P(X \leq 3) = 0.3528$$.

94.

1. $$\frac{100}{9}=11.11$$
2. $$P(X>10)=1-P(X \leq 10)=1-\text { Poissoncdf }(11.11,10) \approx 0.5532$$.
3. The number of people with Type B blood encountered roughly follows the Poisson distribution, so the number of people X who arrive between successive Type B arrivals is roughly exponential with mean $$\mu = 9$$ and $$m = \frac{1}{9}$$. The cumulative distribution function of $$X$$ is $$P(X<x)>20)=1-P(X \leq 20)=1-\left(1-e^{-\frac{20}{9}}\right) \approx 0.1084$$.

NOTE

We could also deduce that each person arriving has a $$8/9$$ chance of not having Type B blood. So the probability that none of the first 20 people arrive have Type B blood is $$\left(\frac{8}{9}\right)^{20} \approx 0.0948$$. (The geometric distribution is more appropriate than the exponential because the number of people between Type B people is discrete instead of continuous.)

96.

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$$.