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

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.

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

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

*X*represents the length of time a commuter must wait for a train to arrive on the Red Line.*X*~*U*(0, 8)- $f(x)=\frac{1}{8}$ where ≤
*x*≤ 8 - four
- 2.31
- $\frac{1}{8}$
- $\frac{1}{8}$
- 3.2

- The probability density function of
*X*is $\frac{1}{25-16}=\frac{1}{9}$.*P*(*X*> 19) = (25 – 19) $\left(\frac{1}{9}\right)$ = $\frac{6}{9}$ = $\frac{2}{3}$. *P*(19 <*X*< 22) = (22 – 19) $\left(\frac{1}{9}\right)$ = $\frac{3}{9}$ = $\frac{1}{3}$.- 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.
- 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) = $\frac{P(x>21\text{AND}x18)}{P(x18)}$

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

- Draw the graph where a is now 18 and b is still 25. The height is $\frac{1}{(25-18)}$ = $\frac{1}{7}$

*P*(*X*> 650) = $\frac{700-650}{700-300}=\frac{50}{400}=\frac{1}{8}$ = 0.125.*P*(400 <*X*< 650) = $\frac{650-400}{700-300}=\frac{250}{400}$ = 0.625- 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 farthest 10 percent of days.

*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

*X*= the time (in years) after reaching age 60 that it takes an individual to retire*X*is continuous.*X*~*Exp*$\left(\frac{1}{5}\right)$- five
- five
- Check student’s solution.
- 0.1353
- before
- 18.3

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

- Therefore,
*P*(*T*< 1) = 1 – e^{$\u2013\frac{1}{8}$}≈ 0.1175. - We want to find
*P*(6 <*t*< 10).

To do this,*P*(6 <*t*< 10) –*P*(*t*< 6)

= $=\left(1\u2013{e}^{\u2013\frac{1}{8}*10}\right)\u2013\left(1\u2013{e}^{\u2013\frac{1}{8}*6}\right)$ ≈ 0.7135 – 0.5276 = 0.1859 - We want to find 0.70 $=P(T>t)=1\u2013\left(1\u2013{e}^{-\frac{t}{8}}\right)={e}^{-\frac{t}{8}}.$

Solving for*t*, e^{$\u2013\frac{t}{8}$}= 0.70, so $\u2013\frac{t}{8}$ =*ln*(0.70), and*t*= –8*ln*(0.70) ≈ 2.85 years

Or use*t*= $\frac{ln\text{(area\_to\_the\_right)}}{(\u2013m)}=\frac{ln\text{(0}\text{.70)}}{\u2013\frac{1}{8}}\approx 2.\text{85years}$. - We want to find 0.02 =
*P*(*T*<*t*) = 1 –*e*^{$\u2013\frac{t}{8}$}.

Solving for*t*,*e*^{$\u2013\frac{t}{8}$}= 0.98, so $\u2013\frac{t}{8}$ =*ln*(0.98), and t = –8*ln*(0.98) ≈ 0.1616 years, or roughly two months.

The warranty should cover light bulbs that last less than 2 months.

Or use $\frac{\mathrm{ln}\text{(area\_to\_the\_right)}}{\text{(}\u2013\text{m)}}=\frac{\mathrm{ln}\text{(1}\u20130.2)}{\u2013\frac{1}{8}}$ = 0.1616. - 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) = $1\u2013\left(1\u2013{e}^{\u2013\frac{1}{8}}\right)={e}^{\u2013\frac{1}{8}}\approx 0.8825$

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) = $\frac{{3}^{0}{e}^{\u20133}}{0\text{!}}$ = e^{–3} ≈ 0.0498

You could let *T* = duration of time between no-hitters. Since the time is exponential and there are three no-hitters per season, then the time between no-hitters is $\frac{1}{3}$ season. For the exponential, *µ* = $\frac{1}{3}$.

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.

- $\frac{100}{9}$ = 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*= $\frac{1}{9}$. The cumulative distribution function of*X*is $P\left(X<x\right)=1-{e}^{-\frac{x}{9}}$. Thus hus,*P*(*X*> 20) = 1 -*P*(*X*≤ 20) = $1-\left(1-{e}^{-\frac{20}{9}}\right)\approx \mathrm{0.1084.}$

### Note

We could also deduce that each person arriving has a $\frac{8}{9}$ chance of not having type B blood. So the probability that none of the first 20 people arrive have type B blood is ${\left(\frac{8}{9}\right)}^{20}\approx 0.0948$. (The geometric distribution is more appropriate than the exponential because the number of people between type B people is discrete instead of continuous.)

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*) = $1-{e}^{\frac{t}{7}}$

*P*(*T*< 2) = 1 - $1-{e}^{-\frac{2}{7}}$ ≈ 0.2485.*P*(*T*> 15) = $1-P\left(T<15\right)=1-\left(1-{e}^{-\frac{15}{7}}\right)\approx {e}^{-\frac{15}{7}}\approx 0.1173$.*P*(*T*> 15|*T*> 10) =*P*(*T*> 5) = $1-\left(1-{e}^{-\frac{5}{7}}\right)={e}^{-\frac{5}{7}}\approx 0.4895$.- 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.