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

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

1. X represents the length of time a commuter must wait for a train to arrive on the Red Line.
2. Graph the probability distribution.
3. $f(x)=\frac{1}{8}$ where 0 ≤ x ≤ 8
4. four
5. 2.31
6. $\frac{1}{8}$
7. $\frac{1}{8}$

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 5.40
2. P(19 < X < 22) = (22 – 19) $\left(\frac{1}{9}\right)$ = $\frac{3}{9}$ = $\frac{1}{3}$.

   

   Figure 5.41
3. 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) = $\frac{P(x>21\cap x>18)}{P(x>18)}$
     = $\frac{P(x>21)}{P(x>18)}$ = $\frac{(25-21)}{(25-18)}$ = $\frac{4}{7}$.

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

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

1. X = the time (in years) after reaching age 60 that it takes an individual to retire
2. X is continuous.
3. five
4. five
5. Check student’s solution.
6. 0.1353
7. before
8. 18.3

a

c

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.