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.
- Check student’s solution.
- [latex]\frac{{3.5}}{{7}}[/latex]
45.
- Check student’s solution.
- k = 7.25
- 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.
- Check student’s solution
- P(x < 5,730) = 0.5001
71.
- Check student’s solution.
- k = 2947.73
Continuous Probability Functions – Homework
73. Age is a measurement, regardless of the accuracy used.
The Uniform Distribution – Homework
75.
- X ~ U(1, 9)
- Check student’s solution.
- f(x) = [latex]\frac{{1}}{{8}}[/latex] where 1≤x≤9
- 5
- 2.3
- [latex]\frac{{15}}{{32}}[/latex]
- [latex]\frac{{333}}{{800}}[/latex]
- [latex]\frac{{2}}{{3}}[/latex]
- 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.
- [latex]f\left(x\right)=\frac{1}{8}[/latex] where ≤ x ≤ 8
- 4
- 2.31
- [latex]\frac{{1}}{{8}}[/latex]
- [latex]\frac{{1}}{{8}}[/latex]
- 3.2
79. d
81. b
83.
- 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.
- [latex]P(X>650) = \frac{700-650}{700-300} = \frac{50}{400} = \frac{1}{8} = 0.125[/latex].
- [latex]P(400<X<650) = \frac{650-400}{700-300} = \frac{250}{400} = 0.625[/latex].
- [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.
- 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[latex]\left(\frac{{1}}{{5}}\right)[/latex]
- five
- five
- Check student’s solution.
- 0.1353
- before
- 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
99.
- [latex]\frac{{100}}{{9}}[/latex] = 11.11
- P(X > 10) = 1 – P(X ≤ 10) = 1 – Poissoncdf(11.11, 10) ≈ 0.5532.
- 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].
- [latex]P(T<2) = 1 - e^{- \frac{2}{7}} \approx 0.2485[/latex].
- [latex]P(T>15) = 1 - P(T<15) = 1 - (1- e^{- \frac{15}{7}}) \approx e^{- \frac{15}{7}} \approx 0.1173[/latex].
- [latex]P(T>15 | T>10) = P(T>5) = 1 - (1-e^{- \frac{5}{7}}) = e^{- \frac{5}{7}} \approx 0.4895[/latex].
- 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.