Exam 1 (lab 4)
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Exam 1 (lab 4)
Page 1 of 18
file:///Users/mathabib/Documents/R%20files/Exam1/Exam1.html
Exam 1 (lab 4)
Mathew H
2023-10-18
library
(magrittr)
library
(formattable)
library
(LearningStats)
library
(knitr)
Question 1: CONFRIRMATION NEEDED
#Manual Rankings:
Pond ID: Ranking: _________ ________ 1 7 (tie) 2 10 3 6 4 7 (tie) 5 3 (tie) 6 5 7 1 (tie) 8 11 9 1 (tie) 10 3 (tie) 11 7
(tie)
#R rankings
bullfrogs=c(34,65,23,34,18,20,15,70,15,18,34)
rank(bullfrogs)
## [1] 8.0 10.0 6.0 8.0 3.5 5.0 1.5 11.0 1.5 3.5 8.0
##The ranking is the same, although R delineates ties by outputting a .5 value.
Question 2:
student_data=read.csv("/Users/mathabib/Documents/R files/Exam1/Student Data.csv")
malepulse=student_data$Pulse[1:77]
hist(malepulse,breaks=15,main="male pulses",col="blue",xlim=c(40,150),ylim=c(0,20),xl
ab="pulse")
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Exam 1 (lab 4)
Page 2 of 18
file:///Users/mathabib/Documents/R%20files/Exam1/Exam1.html
femalepulse=student_data$Pulse[78:165]
hist(femalepulse,breaks=15,main="female pulses",col="pink",xlim=c(40,150),ylim=c(0,2
0),xlab="pulse")
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Exam 1 (lab 4)
Page 3 of 18
file:///Users/mathabib/Documents/R%20files/Exam1/Exam1.html
## The distributions are not similar. While both seem to be centralized around a pulse of 80, the female
distribution is much more broad, with both lower minimums and higher maximums than are observed in the
male pulse graph.
Question 3:
bass=read.csv("/Users/mathabib/Documents/R files/Exam1/bass.csv")
bassL=bass$length.mm.
bins=seq(60,460,by=10)
hist(bassL,breaks=bins,main="frequency of bass lengths",col="green",xlim=c(50,550),xl
ab="lengths in mm")
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Exam 1 (lab 4)
Page 4 of 18
file:///Users/mathabib/Documents/R%20files/Exam1/Exam1.html
#table
bins=seq(60,460,by=10)
cutbassL=cut(bassL,bins)
basstable=transform(table(cutbassL))
names(basstable)[1]="length in mm |"
names(basstable)[2]="Frequency"
formattable(basstable)%>%
kable("html",escape=F,caption="lengths of largemouth")
lengths of largemouth
length in mm |Frequency
(60,70]
2
(70,80]
0
(80,90]
1
(90,100]
1
(100,110]
0
(110,120]
3
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Exam 1 (lab 4)
Page 5 of 18
file:///Users/mathabib/Documents/R%20files/Exam1/Exam1.html
(120,130]
1
(130,140]
1
(140,150]
0
(150,160]
1
(160,170]
2
(170,180]
6
(180,190]
3
(190,200]
9
(200,210]
5
(210,220]
1
(220,230]
0
(230,240]
2
(240,250]
3
(250,260]
2
(260,270]
2
(270,280]
0
(280,290]
4
(290,300]
3
(300,310]
4
(310,320]
8
(320,330]
6
(330,340]
4
(340,350]
4
(350,360]
2
(360,370]
5
(370,380]
2
(380,390]
2
(390,400]
3
(400,410]
2
(410,420]
3
(420,430]
0
(430,440]
2
(440,450]
0
(450,460]
1
Question 4:
10/26/23, 3
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Exam 1 (lab 4)
Page 6 of 18
file:///Users/mathabib/Documents/R%20files/Exam1/Exam1.html
#table
basscumulative=cumsum(table(cutbassL))
cumulativecolumn=transform(basscumulative)$X_data
cumulativebasstable=data.frame(basstable,cumulativecolumn)
names(cumulativebasstable)[1]="lengths in mm |"
names(cumulativebasstable)[2]="frequency |"
names(cumulativebasstable)[3]="cumulative frequency"
formattable(cumulativebasstable)%>%
kable("html",escape=F,caption="lengths of largemouth")
lengths of largemouth
lengths in mm |frequency |cumulative frequency
(60,70]
2
2
(70,80]
0
2
(80,90]
1
3
(90,100]
1
4
(100,110]
0
4
(110,120]
3
7
(120,130]
1
8
(130,140]
1
9
(140,150]
0
9
(150,160]
1
10
(160,170]
2
12
(170,180]
6
18
(180,190]
3
21
(190,200]
9
30
(200,210]
5
35
(210,220]
1
36
(220,230]
0
36
(230,240]
2
38
(240,250]
3
41
(250,260]
2
43
(260,270]
2
45
(270,280]
0
45
(280,290]
4
49
(290,300]
3
52
(300,310]
4
56
(310,320]
8
64
(320,330]
6
70
(330,340]
4
74
(340,350]
4
78
(350,360]
2
80
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10/26/23, 3
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Exam 1 (lab 4)
Page 7 of 18
file:///Users/mathabib/Documents/R%20files/Exam1/Exam1.html
(360,370]
5
85
(370,380]
2
87
(380,390]
2
89
(390,400]
3
92
(400,410]
2
94
(410,420]
3
97
(420,430]
0
97
(430,440]
2
99
(440,450]
0
99
(450,460]
1
100
#histogram
hist(cumsum(table(cutbassL)),breaks=25,main="cumulative bass length frequency histogr
am",col="darkolivegreen",xlim=c(0,100),xlab="cumulative lengths")
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Exam 1 (lab 4)
Page 8 of 18
file:///Users/mathabib/Documents/R%20files/Exam1/Exam1.html
#polygon
plot(cumsum(table(cutbassL)),xlab="Cumulative lengths",ylab="cumulative frequencies",
main="cumulative bass length frequency polygon")
polygon(c(0,cumsum(table(cutbassL)),0),col="darkolivegreen")
Question 5:
kingfisher=read.csv("/Users/mathabib/Documents/R files/Exam1/kingfisher.csv")
kingL=kingfisher$Bill_length
mean(kingL)
## [1] 55.69762
median(kingL)
## [1] 56.5
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Exam 1 (lab 4)
Page 9 of 18
file:///Users/mathabib/Documents/R%20files/Exam1/Exam1.html
#mode
table(kingL)
## kingL
## 47 47.8 48 48.1 48.8 50.1 50.4 50.8 51 51.8 52.3 52.5 53.4 55.2 55.6 55.8 ## 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 ## 56 56.2 56.3 56.5 56.8 57.1 57.5 57.7 57.8 58 59.2 59.3 59.8 59.9 60.2 61.1 ## 1 1 1 2 3 1 1 1 1 1 2 2 1 1 1 1 ## 61.2 61.5 61.8 62.4 ## 1 1 1 1
## mode is 56.8 with 3 appearances
#range
range(kingL)[2]-range(kingL)[1]
## [1] 15.4
IQR(kingL)
## [1] 6.85
var(kingL)
## [1] 18.43097
sd(kingL)
## [1] 4.293131
#coeff var.
(sd(kingL))/(mean(kingL))
## [1] 0.07707925
boxplot(kingL,col="red",main="Kingfisher bill length",ylab="bill length")
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Exam 1 (lab 4)
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Mean: 55.69762 Median: 56.5 Mode: 56.8 (3 occurrences) Range: 15.4 IQR: 6.85 Variance: 18.43097 SD:
4.293131 Coe
ffi
cient of Variation: 0.07707925
Question 6:
shell_L=c(95,105,115,125,135,145,155,165)
shellwt=c(4,10,9,6,9,3,5,1)
weighted.mean(shell_L,shellwt)
## [1] 123.5106
#The weighted mean comes out to 123.5106, which is very close to the actual computed mean of 123.6. The
wieghted mean is less than a tenth of a mm o
ff
of the computed mean.
Question 7:
10/26/23, 3
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Exam 1 (lab 4)
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file:///Users/mathabib/Documents/R%20files/Exam1/Exam1.html
mosquitofish=read.csv("/Users/mathabib/Documents/R files/Exam1/Mosquitofish(1).csv")
mmfish=mosquitofish$FishLength[1:854]
tenfish1=sample(mmfish,10,replace=F)
tenfish2=sample(mmfish,10,replace=F)
tenfish3=sample(mmfish,10,replace=F)
tenfish4=sample(mmfish,10,replace=F)
tenfish5=sample(mmfish,10,replace=F)
twentyfish1=sample(mmfish,20,replace=F)
twentyfish2=sample(mmfish,20,replace=F)
twentyfish3=sample(mmfish,20,replace=F)
twentyfish4=sample(mmfish,20,replace=F)
twentyfish5=sample(mmfish,20,replace=F)
thirtyfish1=sample(mmfish,30,replace=F)
thirtyfish2=sample(mmfish,30,replace=F)
thirtyfish3=sample(mmfish,30,replace=F)
thirtyfish4=sample(mmfish,30,replace=F)
thirtyfish5=sample(mmfish,30,replace=F)
#Tenfish
mean(tenfish1)
## [1] 24
sd(tenfish1)
## [1] 1.699673
mean(tenfish2)
## [1] 23.4
sd(tenfish2)
## [1] 2.412928
mean(tenfish3)
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Exam 1 (lab 4)
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## [1] 22.4
sd(tenfish3)
## [1] 1.505545
mean(tenfish4)
## [1] 22.8
sd(tenfish4)
## [1] 2.347576
mean(tenfish5)
## [1] 23.3
sd(tenfish5)
## [1] 2.002776
#Twentyfish
mean(twentyfish1)
## [1] 24
sd(twentyfish1)
## [1] 3.077935
mean(twentyfish2)
## [1] 23.8
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Exam 1 (lab 4)
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sd(twentyfish2)
## [1] 2.764436
mean(twentyfish3)
## [1] 24.55
sd(twentyfish3)
## [1] 3.103055
mean(twentyfish4)
## [1] 24.45
sd(twentyfish4)
## [1] 1.932411
mean(twentyfish5)
## [1] 23.7
sd(twentyfish5)
## [1] 2.792848
#Thirtyfish
mean(thirtyfish1)
## [1] 23.7
sd(thirtyfish1)
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Exam 1 (lab 4)
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file:///Users/mathabib/Documents/R%20files/Exam1/Exam1.html
## [1] 2.718392
mean(thirtyfish2)
## [1] 22.96667
sd(thirtyfish2)
## [1] 2.918234
mean(thirtyfish3)
## [1] 23.6
sd(thirtyfish3)
## [1] 2.372253
mean(thirtyfish4)
## [1] 22.83333
sd(thirtyfish4)
## [1] 2.841604
mean(thirtyfish5)
## [1] 23.56667
sd(thirtyfish5)
## [1] 2.473073
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Exam 1 (lab 4)
Page 15 of 18
file:///Users/mathabib/Documents/R%20files/Exam1/Exam1.html
The means and srandard deviations of the
samples are not the same as the population
mean and standard deviation. This is
because the sample has a much larger size,
and therefore has the possibility of having
significant outliers that can influence the
data. There is a low possibility of including
these outliers given the relatively small
sample size, so the disparities in mean and
sd are observed.
Question 8:
#A:
dbinom(0,25,.10)
## [1] 0.0717898
#B:
numblind=0:25
prob=dbinom(numblind,25,0.1)
blindtable=data.frame(numblind,prob)
names(blindtable)[1]="number of color blind people |"
names(blindtable)[2]="probability"
formattable(blindtable)%>%
kable("html",escape=F,caption="probability of being colorblind")
probability of being colorblind
number of color blind people |probability
0 0.0717898
1 0.1994161
2 0.2658881
3 0.2264973
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Exam 1 (lab 4)
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4 0.1384150
5 0.0645937
6 0.0239236
7 0.0072150
8 0.0018038
9 0.0003786
10 0.0000673
11 0.0000102
12 0.0000013
13 0.0000001
14 0.0000000
15 0.0000000
16 0.0000000
17 0.0000000
18 0.0000000
19 0.0000000
20 0.0000000
21 0.0000000
22 0.0000000
23 0.0000000
24 0.0000000
25 0.0000000
Question 9:
wormspeel=c(0:10)
wormsweight=c(17,8,7,6,4,5,3,5,4,3,4)
wormsmean=weighted.mean(wormspeel,wormsweight)
wormsmean
## [1] 3.530303
#A:
n=17
wormprob=dpois(0:n,wormsmean)
wormprob
## [1] 2.929604e-02 1.034239e-01 1.825588e-01 2.148293e-01 1.896032e-01
## [6] 1.338713e-01 7.876772e-02 3.972485e-02 1.753009e-02 6.876283e-03
## [11] 2.427536e-03 7.790853e-04 2.292006e-04 6.224212e-05 1.569525e-05
## [16] 3.693933e-06 8.150440e-07 1.692560e-07
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Exam 1 (lab 4)
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wormtotal=sum(wormprob)
#B:
wormtable=data.frame(0:n,wormprob)
names(wormtable)[1]="# of worms. |"
names(wormtable)[2]="Probability"
formattable(wormtable)%>%
kable("html",escape=F,caption="probability of being eels containing x number of wor
ms")
probability of being eels
containing x number of
worms
# of worms. |Probability
0 0.0292960
1 0.1034239
2 0.1825588
3 0.2148293
4 0.1896032
5 0.1338713
6 0.0787677
7 0.0397248
8 0.0175301
9 0.0068763
10 0.0024275
11 0.0007791
12 0.0002292
13 0.0000622
14 0.0000157
15 0.0000037
16 0.0000008
17 0.0000002
expectedworms=sum(0:n)*wormprob #expected Frequency values
actualworms=c(wormsweight,rep(0,n+1-length(wormsweight)))
wormtableb=data.frame(0:n,actualworms,expectedworms)
names(wormtableb)[1]="worms per eel"
names(wormtableb)[2]="observed frequency"
names(wormtableb)[3]="expected frequency"
wormtableb
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Exam 1 (lab 4)
Page 18 of 18
file:///Users/mathabib/Documents/R%20files/Exam1/Exam1.html
## worms per eel observed frequency expected frequency
## 1 0 17 4.482294e+00
## 2 1 8 1.582385e+01
## 3 2 7 2.793150e+01
## 4 3 6 3.286889e+01
## 5 4 4 2.900928e+01
## 6 5 5 2.048231e+01
## 7 6 3 1.205146e+01
## 8 7 5 6.077902e+00
## 9 8 4 2.682104e+00
## 10 9 3 1.052071e+00
## 11 10 4 3.714130e-01
## 12 11 0 1.192000e-01
## 13 12 0 3.506769e-02
## 14 13 0 9.523044e-03
## 15 14 0 2.401374e-03
## 16 15 0 5.651718e-04
## 17 16 0 1.247017e-04
## 18 17 0 2.589617e-05
The observed and expected frequencies do not match up very well. This suggests that the data set does not
fit a poisson distribution very well.
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Question 21. Let i denote the square root of -1. Suppose w=2– 3i. Find three cube roots of w in
the exponential form.
Note the arguments should be given as principal arguments [-T, T) in radians up to 3 decimal places.
|
a) wo = V13e0.3:
,-0.328i
b) wo = V13e
,-0.983i
c) wo = V13e
V13e-0.328i
W1 = V13e2.422i
Wi = V13e2.422i
Wi = V13e5.300i
V13e2.422i
w2 = V13e4.516i
,-1.767i
= V13e-
W2 = V13el1.583i
W2 = V13e-1.767i
d) wo
W1
||
e) not in the list
Question 22. Let i denote the square root of -1. Suppose z = 5+2i and…
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Foundation Student Handbook x - Course: Science and Engineeri X
9 SAMPLE-final exam_SEF041_2 X
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Question 18. Consider the function g(x) = sin?x, in the interval [0, 1].
Use the Trapezium Rule, dividing the interval [0, 7] into strips of equal width d = T/2, to find an
estimate for the definite integral
| 8(x)dx.
2
Answer:
Question 19. A circular area of r-0 is swept out as a rod of fixed length r=1 m pivots about a
point through an angle of 0 radians. If area is increasing at a rate of m² per second, find the rate of
increase of the angle 0 in rad /s. Enter your answer correct to 2 decimal places without…
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If the average number of yards per game of all the HS wide receivers catches with 50 attempts in the 2010 season averaged 49 yards per game. A sample of 20 wide receivers from 2010 averaged 46.54 yards per game.
Is 49 yards a parameter or statistic? and provide correct statistical notation
Is 46.54 years a parameter or statistic? and provide correct statistical notation
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book/MTH 215T 57059731/chapter/2/section/2
> MTH/215T: Quantitative Reasoning I home > 2.2: Adding and subtracting
PARTICIPATION
ACTIVITY
Check
2) 110+85=
2.2.7: Mentally adding and estimations: Bake sale.
Drew ran a bake sale that made 108, 86, 101, and 52 dollars for the four nights of a school play. The school principal asks, "About
how much did the snack bar make?" Drew estimates by rounding to the nearest 5 and adding.
1) 108+86 is about 110+
115
Check
Show answer
Check
Show answer
3) 195+101 is about if 101 is rounded down to 100 first.
Show answer
zy Section 2.2-MTH/215T: Quantitative Reasoning I | zyBooks
4) 295+52 is about if 52 is rounded down to the nearest 5 first.
Check Show answer
vailab.com
site helps w/writting
monts. (use this site)
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A
Incorrect
The nearest 5 to 86 is 1 less than 86.
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Keen den in
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Age dependency Ratio:Old-Age dependency ratio:Child dependency Ratio:3. Explain the significance of each ratio by writing them in complete sentences. Why would someone bother to compare the populations of the relevant age categories?
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KS
Hawkes Learning Blackboard Le X
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MAT 71: Foundations of Precalculus (4194_B5Z1)
Course Content
Lesson: 8.4 Point-Slope Form
Save & End Practice
LORIE TRIVETTE
2/10
Question 3 of 10, Step 1 of 1
Correct
Incorrect
-3
Find the equation for the line that passes through the point (-1, 2), and that is perpendicular to the line with the equation x
11
5
х
10°
10
-5
-5
10|
Send to Instructor
Skip
Submit Answer
Try Similar
Tutor
O2019 Hawkes Learning
7:42 PM
84
W
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the link to the data is given below. please help asap i will upvote!!
https://drive.google.com/file/d/18JRXEQEk8c-voKyNTUhUM9aZGhCWEr1N/view?usp=sharing
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Please do question 10 B part. Thanks
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A study is conducted comparing birth weights (in
pounds) of infants born to mothers of various ages.
The following data can also be found in the file
(a6_q1.xlsx) in the Module "Session 8":
Group Information:
1 = "< 20 Years"
2 = "20-29 Years"
3 = "30+ Years"
GROUP
BIRTHWT
1
8.4
1
7.3
1
9.1
1
7.8
1
8.4
7.5
2
6.3
2
6.9
2
5.4
2
7.1
3
6.9
3
7.1
3
5.7
3
6.5
3
6.6
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了
* 00
R
Frequency
cc.edu/webapps/assessment/take/launch.jsp?course_assessment_id=_842675_1&course_id%=_859154_1&content_id%3_19492605_1&step
* Question Completion Status:
QUESTION 1
The following histogram represents the number of body piercings for students in a statistics class.
40
0 2
4.
9
Number of Piercings
8
a. Is this graph symmetric or skewed?
b. For this graph, is mean or median the best choice for the measure of center?
c. For this graph, is standard deviation or IQR the best choice for the measure of variation?
d. For this graph, the mean expected to be
(higher than, lower than, or about the same as?) the median.
QUESTION 2
Click Save and Submit to save and submit. Click Save All Answers to save all answers.
Save
2
étv
MacBook Air
DD
F7
F8
08
F 4
F5
F3
2$
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#3
9.
6.
5.
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