Continuous Probability Functions
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].
17. Age is a measurement, regardless of the accuracy used.
21. It means that the value of x is just as likely to be any number between 1.5 and 4.5.
23. 1.5 ≤ x ≤ 4.5
25. 0.3333
27. 0
29. 0.6
31. b is 12, and it represents the highest value of x.
33. 6
35.
37. 4.8
39. X = The age (in years) of cars in the staff parking lot
41. 0.5 to 9.5
43. f(x) = [latex]\frac{{1}}{{9}}[/latex] where x is between 0.5 and 9.5, inclusive.
45. μ = 5
47.
- Check student’s solution.
- [latex]\frac{{3.5}}{{7}}[/latex]
49.
- Check student’s solution.
- k = 7.25
- 7.25
51.
- 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
53.
- X represents the length of time a commuter must wait for a train to arrive on the Red Line.
- X ~ U(0, 8)
- [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
55. 4
57. 0.25
59.
- 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]
63. 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.
65. 5
67. [latex]f(x)={0.2}{e}^{-0.2x}[/latex]
69. 0.5350
71. 6.02
73. [latex]f(x)={0.75}{e}^{-0.75x}[/latex]
75.
77. 0.4756
79. The mean is larger. The mean is [latex]\frac{{1}}{{m}}=\frac{{1}}{{0.75}}=1.33[/latex]
81. continuous
83. m = 0.000121
85.
- Check student’s solution
- P(x < 5,730) = 0.5001
87.
- Check student’s solution.
- k = 2947.73
89.
- 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
91.
- 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
93. 0.3333
95. 2.0794
97.
Let T = the life time of a light bulb.
1.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.
2.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
101.
- [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.).