Continuous Probability Functions – Practice

1. Uniform Distribution

3. Normal Distribution

5. P(6 < x < 7)

7. 1

9. 0

11. 1

13. 0.625

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

The Uniform Distribution – Practice

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.

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) = [latex]\frac{{1}}{{9}}[/latex] where x is between 0.5 and 9.5, inclusive.

41. μ = 5

43.

1. Check student’s solution.
2. [latex]\frac{{3.5}}{{7}}[/latex]

45.

1. Check student’s solution.
2. k = 7.25
3. 7.25

The Exponential Distribution – Practice

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. [latex]f(x)={0.2}{e}^{-0.2x}[/latex]

53. 0.5350

55. 6.02

57. [latex]f(x)={0.75}{e}^{-0.75x}[/latex]

59.

61. 0.4756

63. The mean is larger. The mean is [latex]\frac{{1}}{{m}}=\frac{{1}}{{0.75}}=1.33[/latex]

65. continuous

67. m = 0.000121

69.

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

71.

1. Check student’s solution.
2. k = 2947.73

The Uniform Distribution – Homework

75.

1. X ~ U(1, 9)
2. Check student’s solution.
3. f(x) = [latex]\frac{{1}}{{8}}[/latex] where 1≤x≤9
4. 5
5. 2.3
6. [latex]\frac{{15}}{{32}}[/latex]
7. [latex]\frac{{333}}{{800}}[/latex]
8. [latex]\frac{{2}}{{3}}[/latex]
9. 8.2

77.

1. X represents the length of time a commuter must wait for a train to arrive on the Red Line.
2. X ~ U(0, 8)
3. Graph the probability distribution.
4. [latex]f\left(x\right)=\frac{1}{8}[/latex] where ≤ x ≤ 8
5. 4
6. 2.31
7. [latex]\frac{{1}}{{8}}[/latex]
8. [latex]\frac{{1}}{{8}}[/latex]
9. 3.2

79. d

81. b

83.

1. The probability density function of X is [latex]\frac{{1}}{{25-16}}=\frac{{1}}{{9}}[/latex] P(X > 19) = (25 – 19) [latex]\left(\frac{{1}}{{9}}\right)=\frac{{6}}{{9}}=\frac{{2}}{{3}}[/latex]

2.P(19 < X < 22) = (22 – 19) [latex]\left(\frac{{1}}{{9}}\right)=\frac{{3}}{{9}}=\frac{{1}}{{3}}[/latex]

3.The area must be 0.25, and 0.25 = (width)[latex]\left(\frac{{1}}{{9}}\right)[/latex] so width = (0.25)(9) = 2.25. Thus, the value is 25 – 2.25 = 22.75.

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

85.

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

The Exponential Distribution – Homework

87.

1. X = the useful life of a particular car battery, measured in months.
2. X is continuous.
3. X ~ Exp(0.025)
4. 40 months
5. 360 months
6. 0.4066
7. 14.27

89.

1. X = the time (in years) after reaching age 60 that it takes an individual to retire
2. X is continuous.
3. X ~ Exp[latex]\left(\frac{{1}}{{5}}\right)[/latex]
4. five
5. five
6. Check student’s solution.
7. 0.1353
8. before
9. 18.3

91. a

93. c

95. Let T = the life time of a light bulb.

a.The decay parameter is m = 1/8, and T ∼ Exp(1/8). The cumulative distribution function is P(T<t) = 1-[latex]{e}^{-\frac{{t}}{{8}}}[/latex]≈ 0.1175.

b.We want to find P(6 < t < 10),To do this, P(6 < t < 10) – P(t < 6)= [latex]\left(1-{e}^{-\frac{{t}}{{8}}*10}\right)-\left(1-{e}^{-\frac{{t}}{{8}}*6}\right)[/latex]≈ 0.7135 – 0.5276 = 0.1859

c. We want to find 0.70 =P(T>t)=1-[latex]1-\left(1-{e}^{-\frac{{t}}{{8}}}\right)={e}^{\frac{{-t}}{{8}}}[/latex]. [latex]{e}^{\frac{{-t}}{{8}}}=0.70[/latex], so [latex]\frac{-t}{8}[/latex]=ln(0.70)≈ 2.85 years.

d. We want to find [latex]0.02 = P(T<t) = 1-e^{- frac{1}{8}}[/latex].
Solving for t, [latex]e^{- \frac{t}{8}} = 0.98[/latex], so [latex]- \frac{t}{8}[/latex] = 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 [latex]\frac{ln(\text{area to the right})}{(-m)} = \frac{ln(1-0.2)}{- \frac{1}{8}} = 0.1616[/latex].e. 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) = [latex]1-(1-e^{- \frac{1}{8}}) = e^{- \frac{1}{8}} \approx 0.8825[/latex]
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) =[latex]\frac{{{3}^{0}{e}^{-3}}}{{0!}}={e}^{-3}[/latex]≈ 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 13 season. For the exponential, µ = 13.
Therefore, m = [latex]\frac{{1}}{{\mu}}[/latex] = 3 and T ∼ Exp(3).

a. The desired probability is P(T > 1) = 1 – P(T < 1) = 1 – (1 – [latex]{e}^{-3}[/latex]) = [latex]{e}^{-3}[/latex] ≈ 0.0498.
b. 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.
c. 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.

1. [latex]\frac{{100}}{{9}}[/latex] = 11.11
2. P(X > 10) = 1 – P(X ≤ 10) = 1 – Poissoncdf(11.11, 10) ≈ 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 μ = 9 and m =[latex]\frac{{1}}{{9}}[/latex]. The cumulative distribution function of X is P(X<x)=1 – [latex]{e}^{\frac{-x}{9}}[/latex]), thus P(X > 20) = 1 – P(X ≤ 20) = 1 – ([latex]{e}^{\frac{-20}{9}}[/latex])≈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 [latex]{\left(\frac{8}{9}\right)}^{20}[/latex].
(The geometric distribution is more appropriate than the exponential because the number of people between Type B 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 [latex]m = \frac{1}{7}[/latex].  The cdf is [latex].

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