5.11: Solutions

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Uniform Distribution

Normal Distribution

P(6 < x < 7)

one

zero

11.

one

13.

0.625

15.

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

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 ≤ x ≤ 4.5

21.

0.3333

23.

zero

25.

0.6

27.

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

29.

six

31.

This graph shows a uniform distribution. The horizontal axis ranges from 0 to 12. The distribution is modeled by a rectangle extending from x = 0 to x = 12. A region from x = 9 to x = 12 is shaded inside the rectangle.

Figure 5.52

33.

4.8

35.

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

37.

0.5 to 9.5

39.

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

41.

μ = 5

43.

Answers may vary.
$\frac{3.5}{7}$ $\frac{3.5}{7}$

45.

Answers may vary.
k = 7.25
7.25

47.

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.

49.

five

51.

f(x) = 0.2e^-0.2x

53.

0.5350

55.

6.02

57.

f(x) = 0.75e^-0.75x

59.

This graph shows an exponential distribution. The graph slopes downward. It begins at the point (0, 0.75) on the y-axis and approaches the x-axis at the right edge of the graph. The decay parameter, m, equals 0.75.

Figure 5.53

61.

0.4756

63.

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

65.

continuous

67.

m = 0.000121

69.

Answers may vary.
P(x < 5,730) = 0.5001

71.

Answers may vary.
k = 2947.73

73.

Age is a measurement, regardless of the accuracy used.

75.

X ~ U(1, 9)
Answers may vary.
$f (x) = \frac{1}{8}$ $f (x) = \frac{1}{8}$ where $1 \leq x \leq 9$ $1 \leq x \leq 9$
five
2.3
$\frac{15}{32}$ $\frac{15}{32}$
$\frac{333}{800}$ $\frac{333}{800}$
$\frac{2}{3}$ $\frac{2}{3}$
8.2

77.

X represents the length of time a commuter must wait for a train to arrive on the Red Line.
X ~ U(0, 8)
Graph the probability distribution.
$f (x) = \frac{1}{8}$ $f (x) = \frac{1}{8}$ where $0 \leq x \leq 8$ $0 \leq x \leq 8$
four
2.31
$\frac{1}{8}$ $\frac{1}{8}$
$\frac{1}{8}$ $\frac{1}{8}$
3.2

79.

81.

83.

The probability density function of X is $\frac{1}{25 - 16} = \frac{1}{9}$ $\frac{1}{25 - 16} = \frac{1}{9}$ .
P(X > 19) = (25 – 19) $(\frac{1}{9})$ $(\frac{1}{9})$ = $\frac{6}{9}$ $\frac{6}{9}$ = $\frac{2}{3}$ $\frac{2}{3}$ .
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Figure 5.54
P(19 < X < 22) = (22 – 19) $(\frac{1}{9})$ $(\frac{1}{9})$ = $\frac{3}{9}$ $\frac{3}{9}$ = $\frac{1}{3}$ $\frac{1}{3}$ .

Figure 5.55
The area must be 0.25, and 0.25 = (width) $(\frac{1}{9})$ $(\frac{1}{9})$ , so width = (0.25)(9) = 2.25. Thus, the value is 25 – 2.25 = 22.75.
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 $\frac{1}{(25 - 18)}$ $\frac{1}{(25 - 18)}$ = $\frac{1}{7}$ $\frac{1}{7}$
  So, P(x > 21|x > 18) = (25 – 21) $(\frac{1}{7})$ $(\frac{1}{7})$ = 4/7.
- Use the formula: P(x > 21|x > 18) = $\frac{P (x > 21 AND x > 18)}{P (x > 18)}$ $\frac{P (x > 21 AND x > 18)}{P (x > 18)}$
  = $\frac{P (x > 21)}{P (x > 18)}$ $\frac{P (x > 21)}{P (x > 18)}$ = $\frac{(25 - 21)}{(25 - 18)}$ $\frac{(25 - 21)}{(25 - 18)}$ = $\frac{4}{7}$ $\frac{4}{7}$ .

85.

P(X > 650) = $\frac{700 - 650}{700 - 300} = \frac{50}{400} = \frac{1}{8}$ $\frac{700 - 650}{700 - 300} = \frac{50}{400} = \frac{1}{8}$ = 0.125
P(400 < X < 650) = $\frac{650 - 400}{700 - 300} = \frac{250}{400}$ $\frac{650 - 400}{700 - 300} = \frac{250}{400}$ = 0.625
0.10 = $\frac{width}{700 - 300}$ $\frac{width}{700 - 300}$ , so width = 400(0.10) = 40. Since 700 – 40 = 660, the drivers travel at least 660 miles on the furthest 10% of days.

87.

X = the useful life of a particular car battery, measured in months.
X is continuous.
X ~ Exp(0.025)
40 months
360 months
0.4066
14.27

89.

X = the time (in years) after reaching age 60 that it takes an individual to retire
X is continuous.
X ~ Exp $(\frac{1}{5})$ $(\frac{1}{5})$
five
five
Answers may vary.
0.1353
before
18.3

91.

93.

95.

Let T = the life time of a light bulb.

The decay parameter is m = 1/8, and T ∼ Exp(1/8). The cumulative distribution function is $P (T < t) = 1 - e^{- \frac{t}{8}}$ $P (T < t) = 1 - e^{- \frac{t}{8}}$

Therefore, P(T < 1) = 1 – e^{$- \frac{1}{8}$ $- \frac{1}{8}$} ≈ 0.1175.
We want to find P(6 < t < 10).
To do this, P(6 < t < 10) – P(t < 6)
$= (1 - e^{- \frac{1}{8} * 10}) - (1 - e^{- \frac{1}{8} * 6})$ $= (1 - e^{- \frac{1}{8} * 10}) - (1 - e^{- \frac{1}{8} * 6})$ ≈ 0.7135 – 0.5276 = 0.1859

Figure 5.56
We want to find 0.70 $= P (T > t) = 1 - (1 - e^{- \frac{t}{8}}) = e^{- \frac{t}{8}} .$ $= P (T > t) = 1 - (1 - e^{- \frac{t}{8}}) = e^{- \frac{t}{8}} .$
Solving for t, e^{$- \frac{t}{8}$ $- \frac{t}{8}$} = 0.70, so $- \frac{t}{8}$ $- \frac{t}{8}$ = ln(0.70), and t = –8ln(0.70) ≈ 2.85 years.
Or use t = $\frac{l n (area_to_the_right)}{(- m)} = \frac{l n (0 .70)}{- \frac{1}{8}} \approx 2. 85 years$ $\frac{l n (area_to_the_right)}{(- m)} = \frac{l n (0 .70)}{- \frac{1}{8}} \approx 2. 85 years$ .
2.85) = 0.70."> 2.85) = 0.70." width="981" height="564" src="/apps/archive/20240130.204924/resources/443d2d74b21ce16f1b3033b0eaee2d8aa8fe7a3d">
Figure 5.57
We want to find 0.02 = P(T < t) = 1 – e^{$- \frac{t}{8}$ $- \frac{t}{8}$}.
Solving for t, e^{$- \frac{t}{8}$ $- \frac{t}{8}$} = 0.98, so $- \frac{t}{8}$ $- \frac{t}{8}$ = ln(0.98), and t = –8ln(0.98) ≈ 0.1616 years, or roughly two months.
The warranty should cover light bulbs that last less than 2 months.
Or use $\frac{\ln (area_to_the_right)}{(- m)} = \frac{\ln (1 - 0.2)}{- \frac{1}{8}}$ $\frac{\ln (area_to_the_right)}{(- m)} = \frac{\ln (1 - 0.2)}{- \frac{1}{8}}$ = 0.1616.
We must find P(T < 8|T > 7).
Notice that by the rule of complement events, P(T < 8|T > 7) = 1 – P(T > 8|T > 7).
By the memoryless property (P(X > r + t|X > r) = P(X > t)).
So P(T > 8|T > 7) = P(T > 1) = $1 - (1 - e^{- \frac{1}{8}}) = e^{- \frac{1}{8}} \approx 0.8825$ $1 - (1 - e^{- \frac{1}{8}}) = e^{- \frac{1}{8}} \approx 0.8825$
Therefore, P(T < 8|T > 7) = 1 – 0.8825 = 0.1175.

97.

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 λ = 3.
Therefore, (X = 0) = $\frac{3^{0} e^{- 3}}{0!}$ $\frac{3^{0} e^{- 3}}{0!}$ = e^–3 ≈ 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 $\frac{1}{3}$ $\frac{1}{3}$ season. For the exponential, µ = $\frac{1}{3}$ $\frac{1}{3}$ .
Therefore, m = $\frac{1}{μ}$ $\frac{1}{μ}$ = 3 and T ∼ Exp(3).

The desired probability is P(T > 1) = 1 – P(T < 1) = 1 – (1 – e^–3) = e^–3 ≈ 0.0498.
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.
Let X = the number of no-hitters is a season. Assume that X is Poisson with mean λ = 3. Then P(X > 3) = 1 – P(X ≤ 3) = 0.3528.

99.

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

NOTE

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

101.

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

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

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