9.11: Solutions

Last updated
Save as PDF

Page ID: 261828

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

$ \newcommand{\dsum}{\displaystyle\sum\limits} $

$ \newcommand{\dint}{\displaystyle\int\limits} $

$ \newcommand{\dlim}{\displaystyle\lim\limits} $

$ \newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$

( \newcommand{\kernel}{\mathrm{null}\,}\) $ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\id}{\mathrm{id}}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\kernel}{\mathrm{null}\,}$

$ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$

$ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$

$ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\AA}{\unicode[.8,0]{x212B}}$

$ \newcommand{\vectorA}[1]{\vec{#1}} % arrow$

$ \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$

$ \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vectorC}[1]{\textbf{#1}} $

$ \newcommand{\vectorD}[1]{\overrightarrow{#1}} $

$ \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} $

$ \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} $

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

$\newcommand{\avec}{\mathbf a}$ $\newcommand{\bvec}{\mathbf b}$ $\newcommand{\cvec}{\mathbf c}$ $\newcommand{\dvec}{\mathbf d}$ $\newcommand{\dtil}{\widetilde{\mathbf d}}$ $\newcommand{\evec}{\mathbf e}$ $\newcommand{\fvec}{\mathbf f}$ $\newcommand{\nvec}{\mathbf n}$ $\newcommand{\pvec}{\mathbf p}$ $\newcommand{\qvec}{\mathbf q}$ $\newcommand{\svec}{\mathbf s}$ $\newcommand{\tvec}{\mathbf t}$ $\newcommand{\uvec}{\mathbf u}$ $\newcommand{\vvec}{\mathbf v}$ $\newcommand{\wvec}{\mathbf w}$ $\newcommand{\xvec}{\mathbf x}$ $\newcommand{\yvec}{\mathbf y}$ $\newcommand{\zvec}{\mathbf z}$ $\newcommand{\rvec}{\mathbf r}$ $\newcommand{\mvec}{\mathbf m}$ $\newcommand{\zerovec}{\mathbf 0}$ $\newcommand{\onevec}{\mathbf 1}$ $\newcommand{\real}{\mathbb R}$ $\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$ $\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$ $\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$ $\newcommand{\laspan}[1]{\text{Span}\{#1\}}$ $\newcommand{\bcal}{\cal B}$ $\newcommand{\ccal}{\cal C}$ $\newcommand{\scal}{\cal S}$ $\newcommand{\wcal}{\cal W}$ $\newcommand{\ecal}{\cal E}$ $\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$ $\newcommand{\gray}[1]{\color{gray}{#1}}$ $\newcommand{\lgray}[1]{\color{lightgray}{#1}}$ $\newcommand{\rank}{\operatorname{rank}}$ $\newcommand{\row}{\text{Row}}$ $\newcommand{\col}{\text{Col}}$ $\renewcommand{\row}{\text{Row}}$ $\newcommand{\nul}{\text{Nul}}$ $\newcommand{\var}{\text{Var}}$ $\newcommand{\corr}{\text{corr}}$ $\newcommand{\len}[1]{\left|#1\right|}$ $\newcommand{\bbar}{\overline{\bvec}}$ $\newcommand{\bhat}{\widehat{\bvec}}$ $\newcommand{\bperp}{\bvec^\perp}$ $\newcommand{\xhat}{\widehat{\xvec}}$ $\newcommand{\vhat}{\widehat{\vvec}}$ $\newcommand{\uhat}{\widehat{\uvec}}$ $\newcommand{\what}{\widehat{\wvec}}$ $\newcommand{\Sighat}{\widehat{\Sigma}}$ $\newcommand{\lt}{<}$ $\newcommand{\gt}{>}$ $\newcommand{\amp}{&}$ $\definecolor{fillinmathshade}{gray}{0.9}$

Uniform Distribution

Normal Distribution

P(6 < x < 7)

one

zero

11.

one

13.

0.625

15.

The probability is equal to the area from x = 3232" role="presentation" style="position: relative;">32 $/*<![CDATA[*/ 3 2 3 2 /*]]>*/$ to x = 4 above the x-axis and up to f(x) = 1313" role="presentation" style="position: relative;">13 $/*<![CDATA[*/ 1 3 1 3 /*]]>*/$ .

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 ≤ x ≤ 4.5

21.

0.3333

23.

zero

25.

0.6

27.

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

29.

six

31.

This graph shows a uniform distribution. The horizontal axis ranges from 0 to 12. The distribution is modeled by a rectangle extending from x = 0 to x = 12. A region from x = 9 to x = 12 is shaded inside the rectangle.

Figure 5.52

33.

4.8

35.

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

37.

0.5 to 9.5

39.

f(x) = 1919" role="presentation" style="position: relative;">19 $/*<![CDATA[*/ 1 9 1 9 /*]]>*/$ where x is between 0.5 and 9.5, inclusive.

41.

μ = 5

43.

Answers may vary.
3.573.57" role="presentation" style="position: relative;">3.57 $/*<![CDATA[*/ 3.5 7 3.5 7 /*]]>*/$

45.

Answers may vary.
k = 7.25
7.25

47.

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.

49.

five

51.

f(x) = 0.2e^-0.2x

53.

0.5350

55.

6.02

57.

f(x) = 0.75e^-0.75x

59.

This graph shows an exponential distribution. The graph slopes downward. It begins at the point (0, 0.75) on the y-axis and approaches the x-axis at the right edge of the graph. The decay parameter, m, equals 0.75.

Figure 5.53

61.

0.4756

63.

The mean is larger. The mean is 1m=10.75≈1.331m=10.75≈1.33" role="presentation" style="position: relative;">1m=10.75≈1.33 $/*<![CDATA[*/ 1 m = 1 0.75 ≈1.33 1 m = 1 0.75 ≈1.33 /*]]>*/$ , which is greater than 0.9242.

65.

continuous

67.

m = 0.000121

69.

Answers may vary.
P(x < 5,730) = 0.5001

71.

Answers may vary.
k = 2947.73

73.

Age is a measurement, regardless of the accuracy used.

75.

X ~ U(1, 9)
Answers may vary.
f(x)=18f(x)=18" role="presentation" style="position: relative;">f(x)=18 $/*<![CDATA[*/ f(x)= 1 8 f(x)= 1 8 /*]]>*/$ where 1≤x≤91≤x≤9" role="presentation" style="position: relative;">1≤x≤9 $/*<![CDATA[*/ 1≤x≤9 1≤x≤9 /*]]>*/$
five
2.3
15321532" role="presentation" style="position: relative;">1532 $/*<![CDATA[*/ 15 32 15 32 /*]]>*/$
333800333800" role="presentation" style="position: relative;">333800 $/*<![CDATA[*/ 333 800 333 800 /*]]>*/$
2323" role="presentation" style="position: relative;">23 $/*<![CDATA[*/ 2 3 2 3 /*]]>*/$
8.2

77.

X represents the length of time a commuter must wait for a train to arrive on the Red Line.
X ~ U(0, 8)
Graph the probability distribution.
f(x)=18 f ( x ) = 1 8 " role="presentation" style="position: relative;">f(x)=18 $/*<![CDATA[*/ f ( x ) = 1 8 f ( x ) = 1 8 /*]]>*/$ where 0≤x≤80 ≤ x ≤ 8" role="presentation" style="position: relative;">0≤x≤8 $/*<![CDATA[*/ 0 ≤ x ≤ 8 0 ≤ x ≤ 8/*]]>*/$
four
2.31
18 1 8 " role="presentation" style="position: relative;">18 $/*<![CDATA[*/ 1 8 1 8 /*]]>*/$
18 1 8 " role="presentation" style="position: relative;">18 $/*<![CDATA[*/ 1 8 1 8 /*]]>*/$
3.2

79.

81.

83.

The probability density function of X is 125−16=19125−16=19" role="presentation" style="position: relative;">125−16=19 $/*<![CDATA[*/ 1 25−16 = 1 9 1 25−16 = 1 9 /*]]>*/$ .
P(X > 19) = (25 – 19) (19)(19)" role="presentation" style="position: relative;">(19) $/*<![CDATA[*/ ( 1 9 ) ( 1 9 ) /*]]>*/$ = 6969" role="presentation" style="position: relative;">69 $/*<![CDATA[*/ 6 9 6 9 /*]]>*/$ = 2323" role="presentation" style="position: relative;">23 $/*<![CDATA[*/ 2 3 2 3 /*]]>*/$ .
19) = 2/3."> 19) = 2/3." width="968" height="456" data-lazy-src="/apps/archive/20240130.204924/resources/8e6879719f6f0b2c01f4a67c4c3e282fbdcfbc27">
Figure 5.54
P(19 < X < 22) = (22 – 19) (19)(19)" role="presentation" style="position: relative;">(19) $/*<![CDATA[*/ ( 1 9 ) ( 1 9 ) /*]]>*/$ = 3939" role="presentation" style="position: relative;">39 $/*<![CDATA[*/ 3 9 3 9 /*]]>*/$ = 1313" role="presentation" style="position: relative;">13 $/*<![CDATA[*/ 1 3 1 3 /*]]>*/$ .

Figure 5.55
The area must be 0.25, and 0.25 = (width)(19)(19)" role="presentation" style="position: relative;">(19) $/*<![CDATA[*/ ( 1 9 ) ( 1 9 ) /*]]>*/$ , 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 1(25−18)1(25−18)" role="presentation" style="position: relative;">1(25−18) $/*<![CDATA[*/ 1 (25−18) 1 (25−18) /*]]>*/$ = 1717" role="presentation" style="position: relative;">17 $/*<![CDATA[*/ 1 7 1 7 /*]]>*/$
  So, P(x > 21|x > 18) = (25 – 21)(17)(17)" role="presentation" style="position: relative;">(17) $/*<![CDATA[*/ ( 1 7 ) ( 1 7 ) /*]]>*/$ = 4/7.
- Use the formula: P(x > 21|x > 18) = P(x>21 AND x>18)P(x>18)P(x>21 AND x>18)P(x>18)" role="presentation" style="position: relative;">P(x>21 AND x>18)P(x>18) $/*<![CDATA[*/ P(x>21 AND x>18) P(x>18) P(x>21 AND x>18) P(x>18) /*]]>*/$
  = P(x>21)P(x>18)P(x>21)P(x>18)" role="presentation" style="position: relative;">P(x>21)P(x>18) $/*<![CDATA[*/ P(x>21) P(x>18) P(x>21) P(x>18) /*]]>*/$ = (25−21)(25−18)(25−21)(25−18)" role="presentation" style="position: relative;">(25−21)(25−18) $/*<![CDATA[*/ (25−21) (25−18) (25−21) (25−18) /*]]>*/$ = 4747" role="presentation" style="position: relative;">47 $/*<![CDATA[*/ 4 7 4 7 /*]]>*/$ .

85.

P(X > 650) = 700−650700−300=50400=18700−650700−300=50400=18" role="presentation" style="position: relative;">700−650700−300=50400=18 $/*<![CDATA[*/ 700−650 700−300 = 50 400 = 1 8 700−650 700−300 = 50 400 = 1 8 /*]]>*/$ = 0.125
P(400 < X < 650) = 650−400700−300=250400650−400700−300=250400" role="presentation" style="position: relative;">650−400700−300=250400 $/*<![CDATA[*/ 650−400 700−300 = 250 400 650−400 700−300 = 250 400 /*]]>*/$ = 0.625
0.10 = width700−300width700−300" role="presentation" style="position: relative;">width700−300 $/*<![CDATA[*/ width 700−300 width 700−300 /*]]>*/$ , so width = 400(0.10) = 40. Since 700 – 40 = 660, the drivers travel at least 660 miles on the furthest 10% of days.

87.

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

89.

X = the time (in years) after reaching age 60 that it takes an individual to retire
X is continuous.
X ~ Exp(15)(15)" role="presentation" style="position: relative;">(15) $/*<![CDATA[*/ ( 1 5 ) ( 1 5 ) /*]]>*/$
five
five
Answers may vary.
0.1353
before
18.3

91.

93.

95.

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−t8P(T<t)=1−e−t8" role="presentation" style="position: relative;">P(T<t)=1−e−t8 $/*<![CDATA[*/ P(T*/$

Therefore, P(T < 1) = 1 – e^{–18–18" role="presentation" style="position: relative;">–18 $/*<![CDATA[*/ – 1 8 – 1 8 /*]]>*/$} ≈ 0.1175.
We want to find P(6 < t < 10).
To do this, P(6 < t < 10) – P(t < 6)
=(1–e–18*10)–(1–e–18*6)=(1–e–18*10)–(1–e–18*6)" role="presentation" style="position: relative;">=(1–e–18*10)–(1–e–18*6) $/*<![CDATA[*/ =( 1– e – 1 8 *10 )–( 1– e – 1 8 *6 ) =( 1– e – 1 8 *10 )–( 1– e – 1 8 *6 ) /*]]>*/$ ≈ 0.7135 – 0.5276 = 0.1859

Figure 5.56
We want to find 0.70 =P(T>t)=1–(1–e−t8)=e−t8.=P(T>t)=1–(1–e−t8)=e−t8." role="presentation" style="position: relative;">=P(T>t)=1–(1–e−t8)=e−t8. $/*<![CDATA[*/ =P(T>t)=1–( 1– e − t 8 )= e − t 8 . =P(T>t)=1–( 1– e − t 8 )= e − t 8 . /*]]>*/$
Solving for t, e^{–t8–t8" role="presentation" style="position: relative;">–t8 $/*<![CDATA[*/ – t 8 – t 8 /*]]>*/$} = 0.70, so –t8–t8" role="presentation" style="position: relative;">–t8 $/*<![CDATA[*/ – t 8 – t 8 /*]]>*/$ = ln(0.70), and t = –8ln(0.70) ≈ 2.85 years.
Or use t = ln(area_to_the_right)(–m)=ln(0.70)–18≈2.85 yearsln(area_to_the_right)(–m)=ln(0.70)–18≈2.85 years" role="presentation" style="position: relative;">ln(area_to_the_right)(–m)=ln(0.70)–18≈2.85 years $/*<![CDATA[*/ ln(area_to_the_right) (–m) = ln(0.70) – 1 8 ≈2.85 years ln(area_to_the_right) (–m) = ln(0.70) – 1 8 ≈2.85 years /*]]>*/$ .
2.85) = 0.70."> 2.85) = 0.70." width="981" height="564" src="/apps/archive/20240130.204924/resources/443d2d74b21ce16f1b3033b0eaee2d8aa8fe7a3d">
Figure 5.57
We want to find 0.02 = P(T < t) = 1 – e^{–t8–t8" role="presentation" style="position: relative;">–t8 $/*<![CDATA[*/ – t 8 – t 8 /*]]>*/$}.
Solving for t, e^{–t8–t8" role="presentation" style="position: relative;">–t8 $/*<![CDATA[*/ – t 8 – t 8 /*]]>*/$} = 0.98, so –t8–t8" role="presentation" style="position: relative;">–t8 $/*<![CDATA[*/ – t 8 – t 8 /*]]>*/$ = 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 ln(area_to_the_right)(–m)=ln(1–0.2)–18ln(area_to_the_right)(–m)=ln(1–0.2)–18" role="presentation" style="position: relative;">ln(area_to_the_right)(–m)=ln(1–0.2)–18 $/*<![CDATA[*/ ln(area_to_the_right) (–m) = ln(1–0.2) – 1 8 ln(area_to_the_right) (–m) = ln(1–0.2) – 1 8 /*]]>*/$ = 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–(1–e–18)=e–18≈0.88251–(1–e–18)=e–18≈0.8825" role="presentation" style="position: relative;">1–(1–e–18)=e–18≈0.8825 $/*<![CDATA[*/ 1–( 1– e – 1 8 )= e – 1 8 ≈0.8825 1–( 1– e – 1 8 )= e – 1 8 ≈0.8825 /*]]>*/$
Therefore, P(T < 8|T > 7) = 1 – 0.8825 = 0.1175.

97.

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) = 30e–30!30e–30!" role="presentation" style="position: relative;">30e–30! $/*<![CDATA[*/ 3 0 e –3 0! 3 0 e –3 0! /*]]>*/$ = 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 1313" role="presentation" style="position: relative;">13 $/*<![CDATA[*/ 1 3 1 3 /*]]>*/$ season. For the exponential, µ = 1313" role="presentation" style="position: relative;">13 $/*<![CDATA[*/ 1 3 1 3 /*]]>*/$ .
Therefore, m = 1μ1μ" role="presentation" style="position: relative;">1μ $/*<![CDATA[*/ 1 μ 1 μ /*]]>*/$ = 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.

99.

10091009" role="presentation" style="position: relative;">1009 $/*<![CDATA[*/ 100 9 100 9 /*]]>*/$ = 11.11
P(X > 10) = 1 – P(X ≤ 10) = 1 – Poissoncdf(11.11, 10) ≈ 0.5532.
The number of people with Type B positive blood encountered roughly follows the Poisson distribution, so the number of people X who arrive between successive Type B positive arrivals is roughly exponential with mean μ = 9 and m = 1919" role="presentation" style="position: relative;">19 $/*<![CDATA[*/ 1 9 1 9 /*]]>*/$ . The cumulative distribution function of X is P(X<x)=1−e−x9P(X<x)=1−e−x9" role="presentation" style="position: relative;">P(X<x)=1−e−x9 $/*<![CDATA[*/ P( X*/$ . Thus, P(X > 20) = 1 - P(X ≤ 20) = 1−(1−e−209)≈0.1084.1−(1−e−209)≈0.1084." role="presentation" style="position: relative;">1−(1−e−209)≈0.1084. $/*<![CDATA[*/ 1−( 1− e − 20 9 )≈0.1084. 1−( 1− e − 20 9 )≈0.1084. /*]]>*/$

NOTE

We could also deduce that each person arriving has a 8/9 chance of not having Type B positive blood. So the probability that none of the first 20 people arrive have Type B positive blood is (89)20≈0.0948(89)20≈0.0948" role="presentation" style="position: relative;">(89)20≈0.0948 $/*<![CDATA[*/ ( 8 9 ) 20 ≈0.0948 ( 8 9 ) 20 ≈0.0948 /*]]>*/$ . (The geometric distribution is more appropriate than the exponential because the number of people between Type B positive people is discrete instead of continuous.)

101.

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 = 1717" role="presentation" style="position: relative;">17 $/*<![CDATA[*/ 1 7 1 7 /*]]>*/$ . The cdf is P(T < t) = 1−et71−et7" role="presentation" style="position: relative;">1−et7 $/*<![CDATA[*/ 1− e t 7 1− e t 7 /*]]>*/$

P(T < 2) = 1 - 1−e−271−e−27" role="presentation" style="position: relative;">1−e−27 $/*<![CDATA[*/ 1− e − 2 7 1− e − 2 7 /*]]>*/$ ≈ 0.2485.
P(T > 15) = 1−P(T<15)=1−(1−e−157)≈e−157≈0.11731−P(T<15)=1−(1−e−157)≈e−157≈0.1173" role="presentation" style="position: relative;">1−P(T<15)=1−(1−e−157)≈e−157≈0.1173 $/*<![CDATA[*/ 1−P( T<15 )=1−( 1− e − 15 7 )≈ e − 15 7 ≈0.1173 1−P( T<15 )=1−( 1− e − 15 7 )≈ e − 15 7 ≈0.1173 /*]]>*/$ .
P(T > 15|T > 10) = P(T > 5) = 1−(1−e−57)=e−57≈0.48951−(1−e−57)=e−57≈0.4895" role="presentation" style="position: relative;">1−(1−e−57)=e−57≈0.4895 $/*<![CDATA[*/ 1−( 1− e − 5 7 )= e − 5 7 ≈0.4895 1−( 1− e − 5 7 )= e − 5 7 ≈0.4895 /*]]>*/$ .
Let X = # of patients arriving during a half-hour period. Then X has the Poisson distribution with a mean of 307307" role="presentation" style="position: relative;">307 $/*<![CDATA[*/ 30 7 30 7 /*]]>*/$ , X ∼ Poisson(307)(307)" role="presentation" style="position: relative;">(307) $/*<![CDATA[*/ ( 30 7 ) ( 30 7 ) /*]]>*/$ . Find P(X > 8) = 1 – P(X ≤ 8) ≈ 0.0311.

Search

Text Color

Text Size

Margin Size

Font Type

NOTE

NOTE