# 5.11: Chapter 5 Solution (Practice + Homework)

- Page ID
- 5570

**1**.

Uniform Distribution

**3**.

Normal Distribution

**5**.

\(P(6 < x < 7)\)

**7**.

one

**9**.

zero

**11**.

one

**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**.

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

**19**.

\(1.5 \leq *x* \leq 4.5\)

**21**.

0.3333

**23**.

zero

**24**.

0.6

**26**.

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

**28**.

six

**30**.

**33**.

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

**35**.

0.5 to 9.5

**37**.

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

**39**.

\(\mu = 5\)

**41**.

- Check student’s solution.
- \(\frac{3.5}{7}\)

**43**.

- Check student's solution.
- \(k = 7.25\)
- 7.25

**45**.

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.

**47**.

five

**49**.

\(f(x)=0.2 e^{-0.2 x}\)

**51**.

0.5350

**53**.

6.02

**55**.

\(f(x) = 0.75e^{-0.75x}\)

**57**.

**59**.

0.4756

**61**.

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

**63**.

continuous

**65**.

\(m = 0.000121\)

**67**.

- Check student's solution
- \(P(x < 5,730) = 0.5001\)

**69**.

- Check student's solution.
- \(k = 2947.73\)

**71**.

Age is a measurement, regardless of the accuracy used.

**73**.

- Check student’s solution.
- \(f(x)=\frac{1}{8} \text { where } 1 \leq x \leq 9\)
- five
- 2.3
- \(\frac{15}{32}\)
- \(\frac{333}{800}\)
- \(\frac{2}{3}\)

**75**.

- \(X\) represents the length of time a commuter must wait for a train to arrive on the Red Line.
- Graph the probability distribution.
- \(f(x)=\frac{1}{8} \text { where } 0 \leq x \leq 8\)
- four
- 2.31
- \(\frac{1}{8}\)
- \(\frac{1}{8}\)

**77**.

d

**78**.

b

**80**.

- 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
<22)> - 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: \(P(x>21 | x>18)=\frac{P(x>21 \cap x>18)}{P(x>18)}\)

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

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

**82**.

- \(P(X>650)=\frac{700-650}{700-300}=\frac{50}{400}=\frac{1}{8}=0.125\).
- \(P(400
<650)>

**84**.

- \(X =\) the useful life of a particular car battery, measured in months.
- \(X\) is continuous.
- 40 months
- 360 months
- 0.4066
- 14.27

**86**.

- \(X =\) the time (in years) after reaching age 60 that it takes an individual to retire
- \(X\) is continuous.
- five
- five
- Check student’s solution.
- 0.1353
- before
- 18.3

**88**.

a

**90**.

c

**92**.

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 \(\lambda = 3\).

Therefore, \((X=0)=\frac{3^{0} e^{-3}}{0 !}=e^{-3} \approx 0.0498\)

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 \(\frac{1}{3}\) season. For the exponential, \(\mu = \frac{1}{3}\).

Therefore, \(m = \frac{1}{\mu} = 3\) and \(T \sim E x p(3)\)

The desired probability is \(P(T>1)=1-P(T<1)=1-\left(1-\mathrm{e}^{-3}\right)=\mathrm{e}^{-3} \approx 0.0498\).

- 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 \(\lambda = 3\). Then \(P(X > 3) = 1 – P(X \leq 3) = 0.3528\).

**94**.

- \(\frac{100}{9}=11.11\)
- \(P(X>10)=1-P(X \leq 10)=1-\text { Poissoncdf }(11.11,10) \approx 0.5532\).
- 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 \(\mu = 9\) and \(m = \frac{1}{9}\). The cumulative distribution function of \(X\) is \(P(X<x)>20)=1-P(X \leq 20)=1-\left(1-e^{-\frac{20}{9}}\right) \approx 0.1084\).

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 \(\left(\frac{8}{9}\right)^{20} \approx 0.0948\). (The geometric distribution is more appropriate than the exponential because the number of people between Type B people is discrete instead of continuous.)

**96**.

Let \(T =\) duration (in minutes) between successive visits. Since patients arrive at a rate of one patient every seven minutes, \(\mu = 7\) and the decay constant is \(m = \frac{1}{7}\). The cdf is \(P(T<t)>

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