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 = $\frac{{3}}{{2}}$ to x = 4 above the x-axis and up to f(x) = $\frac{{1}}{{3}}$.

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

45. μ = 5

47.

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

49.

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

51.

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

53.

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. $f\left(x\right)=\frac{1}{8}$ where ≤ x ≤ 8
4. 4
5. 2.31
6. $\frac{{1}}{{8}}$
7. $\frac{{1}}{{8}}$
8. 3.2

55. 4

57. 0.25

59.

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 < X < 22) = (22 – 19) $\left(\frac{{1}}{{9}}\right)=\frac{{3}}{{9}}=\frac{{1}}{{3}}$

3.The area must be 0.25, and 0.25 = (width)$\left(\frac{{1}}{{9}}\right)$ 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 $\frac{{1}}{{(25-18)}}=\frac{{1}}{{7}}$ So, P(x > 21|x > 18) = (25 – 21)$\left(\frac{{1}}{{7}}\right)=\frac{{4}}{{7}}$
• Use the formula: P(x > 21|x > 18) =$\frac{{P(x>21 and x>18)}}{{P(x>18)}}=\frac{{P(x>21)}}{{P(x>18)}}=\frac{{4}}{{7}}$

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

69. 0.5350

71. 6.02

73. $f(x)={0.75}{e}^{-0.75x}$

75.

77. 0.4756

79. The mean is larger. The mean is $\frac{{1}}{{m}}=\frac{{1}}{{0.75}}=1.33$

81. continuous

83. m = 0.000121

85.

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

87.

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

89.

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

91.

1. X = the time (in years) after reaching age 60 that it takes an individual to retire
2. X is continuous.
3. X ~ Exp$\left(\frac{{1}}{{5}}\right)$
4. five
5. five
6. Check student’s solution.
7. 0.1353
8. before
9. 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-${e}^{-\frac{{t}}{{8}}}$≈ 0.1175.

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

3. We want to find 0.70 =P(T>t)=1-$1-\left(1-{e}^{-\frac{{t}}{{8}}}\right)={e}^{\frac{{-t}}{{8}}}$. ${e}^{\frac{{-t}}{{8}}}=0.70$, so $\frac{-t}{8}$=ln(0.70)≈ 2.85 years.99.

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) =$\frac{{{3}^{0}{e}^{-3}}}{{0!}}={e}^{-3}$≈ 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 13 season. For the exponential, µ = 13.

Therefore, m = $\frac{{1}}{{\mu}}$ = 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.

101.

1. $\frac{{100}}{{9}}$ = 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 =$\frac{{1}}{{9}}$
4. The cumulative distribution function of X is P(X<x)=1 – ${e}^{\frac{-x}{9}}$), thus P(X > 20) = 1 – P(X ≤ 20) = 1 – (${e}^{\frac{-20}{9}}$)≈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 ${\left(\frac{8}{9}\right)}^{20}$.
   (The geometric distribution is more appropriate than the exponential because the number of people between Type B people is discrete instead of continuous.).