Solutions

Uniform Distribution

Normal Distribution

P(6 < x < 7)

one

zero

one

0.625

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

It means that the value of x is just as likely to be any number between 1.5 and 4.5.

1.5 ≤ x ≤ 4.5

0.3333

zero

0.6

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

six

4.8

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

0.5 to 9.5

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

μ = 5

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

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

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.

five

f(x) = 0.2e^-0.2x

0.5350

6.02

f(x) = 0.75e^-0.75x

0.4756

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

continuous

m = 0.000121

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

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

Age is a measurement, regardless of the accuracy used.

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

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. $f(x)=\frac{1}{8}$ where $0\le x\le 8$
5. four
6. 2.31
7. $\frac{1}{8}$
8. $\frac{1}{8}$
9. 3.2

d

b

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

   Figure 5.54
2. P(19 < X < 22) = (22 – 19) $\left(\frac{1}{9}\right)$ = $\frac{3}{9}$ = $\frac{1}{3}$.

   Figure 5.55
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)$\left(\frac{1}{7}\right)$ = 4/7.
   • Use the formula: P(x > 21|x > 18) = $\frac{P(x>21\text{ AND }x>18)}{P(x>18)}$
     = $\frac{P(x>21)}{P(x>18)}$ = $\frac{(25-21)}{(25-18)}$ = $\frac{4}{7}$.

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

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

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

a

c

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 $P(T<t)=1-e^{-\frac{t}{8}}$

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

   Figure 5.56
3. We want to find 0.70 $=P(T>t)=1–\left(1–e^{-\frac{t}{8}}\right)=e^{-\frac{t}{8}}.$
   Solving for t, e^{$–\frac{t}{8}$} = 0.70, so $–\frac{t}{8}$ = ln(0.70), and t = –8ln(0.70) ≈ 2.85 years.
   Or use t = $\frac{ln\text{(area\_to\_the\_right)}}{(–m)}=\frac{ln\text{(0}\text{.70)}}{–\frac{1}{8}}\approx 2.\text{85 years}$.

   Figure 5.57
4. We want to find 0.02 = P(T < t) = 1 – e^{$–\frac{t}{8}$}.
   Solving for t, e^{$–\frac{t}{8}$} = 0.98, so $–\frac{t}{8}$ = 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 $\frac{\ln \text{(area\_to\_the\_right)}}{\text{(}–\text{m)}}=\frac{\ln \text{(1}–0.2)}{–\frac{1}{8}}$ = 0.1616.
5. 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) = $1–\left(1–e^{–\frac{1}{8}}\right)=e^{–\frac{1}{8}}\approx 0.8825$
   Therefore, P(T < 8|T > 7) = 1 – 0.8825 = 0.1175.

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) = $\frac{3^0{e}^{–3}}{0\text{!}}$ = e^–3 ≈ 0.0498

1. The desired probability is P(T > 1) = 1 – P(T < 1) = 1 – (1 – e^–3) = e^–3 ≈ 0.0498.
2. 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.
3. 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.

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 = $\frac{1}{9}$. The cumulative distribution function of X is $P\left(X<x\right)=1-e^{-\frac{x}{9}}$. Thus, P(X > 20) = 1 - P(X ≤ 20) = $1-\left(1-e^{-\frac{20}{9}}\right)\approx \mathrm{0.1084.}$

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 = $\frac{1}{7}$. The cdf is P(T < t) = $1-e^{\frac{t}{7}}$

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