in this chapter, you will learn about simple model statistics. enjoy it

1. Answer Task 1
[1] first,make a table toeasyunderstandingfromamanufacturerof jeanshasplants.
No. Data Amountpairof jeans
1. CA (California) 9
2. AZ (Arizona) 8
3. TX (Texas) 8
Total 25
a. Constructeddatato a pie chart
*For Californiawe get32 % because
8
25
𝑥100% = 32% …. That equationforArizonaandtexastoo
Bar Chart
b. Proportionof the jeansare made in texas is32% or 0.32 basedon pie chart.
c. State produce the most jeansinthe groupis CA ( California) because inbarchart the highestis
California.
d. From pie chart we knowif a manufacturerof jeansbigsectoris Californiaandfrombarchart we
knowCaliforniahave bigamountamanufacturerof jeans.Soconclusionthese dataisbiggeror
roughly.
36%
32%
32%
A Manufacturerof Jeans
CA ( California ) AZ ( Arizona ) TX ( Texas )
7.5 8 8.5 9 9.5
CA ( California )
AZ ( Arizona )
TX ( Texas )
A Manufacturerof Jeans
Amount

2. [2]. The lengthof time (inmonths) betweenthe onsetof aparticularillnessanditsresurrence was
recordedfor50 patients.
No. Length of time Amount
1. 0.00-5.00 22
2. 5.00-10.00 14
3. 10.00-15.00 5
4. 15.00-20.00 5
5. 20.00-25.00 2
6. 26.00-30.00 1
7. 30.00-35.00 1
Total 50
a. Constructa relative frequencyhistogramforthe data
b. Wouldyoudescribe the shape asroughlysymmetric,skewedrightorskewedleft.
From the table as we knowshape isskewedrightbecause the table isstart withlargervalue and
smalleruntil the lengthof time 30.00-35.00.
c. Give the fractionof recurrence timeslessthanorequal to10 months.
As we knowfromthe table : the lengthof time is0.00 to 10.00
(22+14)/50 = 36/50 = 0.72
So the fractionof recurrence time lessthanor equal to10 monthsis 0.72
[3].An experimentalpsychologistmeasuredthe lengthof time ittookfora rat tosuccessfullynavigatea
maze on eachof five days.
Day 1 2 3 4 5
Time (s) 45 43 46 32 25
a. Create a line chart to describe the data
22
14
5 5
2
1 1
0
5
10
15
20
25
0.00-5.00 5.00-10.00 10.00-15.00 15.00-20.00 20.00-25.00 25.00-30.00 30.00-35.00
AxisTitle
The length of time
A Relative Frequency Histogram

3. From line chartwe got inday 3 ismode the lengthof time ittook fora rat to successfullynavigate a
maze on eachof five days. Meanis 7.64 withmedianis3rd
days withthe lengthof time is46.
[4].An estimate of these numbers(inmillions) isshowninthe table.
Religion Members(inmillions)
Buddhism 375
Christianity 2100
Hinduism 851
Islam 1300
Judaism 15
Sikhism 25
Other 21
a. Contructa pie chart to describe the total membershipinthe world’sorganizedreligion
Basedon pie chart we knowtotal membershipinthe world’sorganizedreligionthereare :
Christianity:45% Islam: 28% Hinduism:18% Buddhism:8%
Sikhism : 1 % Judaism: 0% Other : 0%
Total forall : 45+28+18+8+1+0+0 = 100%
45
43
46
32
25
1 2 3 4 5Day
Line Chart
Time
Buddhism
8%
Christianity
45%Hinduism
18%
Islam
28%
Judaism
0%
Sikhism
1% Other
0%
Statistics of the world religion

4. b. Constructa bar chart to describe the total membershipinthe world’sorganizedreligion.
Total membershipinthe world’sorganizedreligionfromabar chart is 4687 members.The
largestmembershipisChristianityandthe smallestmembershipisJudaism.
c. Orderthe religiousgroups fromthe smallesttothe largestnumberof members.Consructa
paretochart to describe the data.Whichof the three displaysismosteffective ?
375
2100
851
1300
15
25
21
Buddhism
Christianity
Hinduism
Islam
Judaism
Skhism
Other
Statistics of the world's religion
Members (In Milions)
Religion Members (In Millions Cumulative Amount Percentage
judaism 15 4687 100%
other 21 4672 100%
sikhism 25 4651 99%
buddhism 375 4626 99%
hinduism 851 4251 91%
islam 1300 3400 73%
christianity 2100 2100 45%
15 21 25
375
851
1300
2100100% 100% 99% 99%
91%
73%
45%
0%
20%
40%
60%
80%
100%
120%
0
500
1000
1500
2000
2500
Members (In Millions
Percentage

5. Most effective todescribethe total membershipinthe world’sorganizedreligionisparetochart
because thischart completelyhave datatoeasierunderstandingaboutstatisticsof the world’s
religion.Thatisusing2 indicator,eachindicatorshowinganestimate of these numbesbasedon
table.
[5].10 measurements:3, 5, 4, 6, 10, 5, 6, 9, 2, 8.
Firstwe can arrangement10 measurementsfromsmallesttolargest,so:
2, 3, 4, 5, 5, 6, 6, 8, 9, 10.
a. Calculate mean
X=
Σ𝑥1
𝑛
=
+3+4+5+5+6+6+8+9+10
10
= 5.8
b. Findmedian
n = 10 the positionof medianis
𝑛+1
2
so
11
2
= 5.5 and the medianisthe average the 5th
and6th
measurements OrMedian=(5+6)/2 = 5.5
c. Findmode
Mode inmeasurementsis5and 6 whichboth occur twice indata
[6] a sample of 25 householdsproducedthe followingmeasurementsonthe numberof DVDsinthe
household.
a. Calculate the mean,medianandmode forthese measurements
Arrangementof measurementsis
0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3.
The mean isX=
Σ𝑥1
𝑛
=
27
25
= 1.08
Medianis
𝑛+1
2
so
26
2
= 13 so the medianisin13th
inmeasurementsis1
Mode forthese measurementsis1whichthe highestfrequencyindata.
b. Is the distributionof the numberof DVDsinthhouseholdsymmetric orskewed?Explain
6
13
4
2
0 1 2 3
Number of DVDs
Amount

6. There are 2 opinionfirst, Fromchart we know data skewedintorightwithmostof the data are
on the leftside of the chart but secondisbecause mean,mode andmedianisalmostsame so
chart is symmetric
c. Draw a relative frequencyhistogramforthe dataset.Locate the mean,medianandmode along
the horizontal axis.Are youanswerforpart (b) correct?
There 2 opinioninhere first, Medianandmode isin1 on relative frequencyhistogramwith
meanis 1.08 Because mean> medianandmode so chart skewedintoright.The secondissame
all of locatedaboutmean,mode andmedianis1 so chart is symmetric
[7]. Arrangementof measures:1, 3, 3, 4, 4, 5, 5, 6.
a. Calculate the range
The smalleris1 andthe highestis6 so the range is 6-1 = 5. So the range is5
b. Calculate the sample mean
The sample meanis
1+3+3+4+4+5+5+6
8
=
31
8
= 3.875
c. Calculate the sample variance
S2=
1
7
[(1-3.875)2
+(3-3.875)2
+(3-3.875)2
+(4-3.875)2
+(4-3.875)2
+(5-3.875)2
+(5-3.875)2
+(6-3.875)2
]
=
1
7
[16.875] = 2.41071429
Sample standarddeviationisS= √ 𝑆2 =√2.41071429 = 1.55264751
[8] the monthlyutilitybillsforahouseholdinriverside,California,were recordedfor12 consecutive
monthsstartinginJanuary2006
Month Amount Month Amount
September $ 343.50 October $ 226.80
August $ 335.48 March $ 219.41
July $ 306.55 November $ 208.99
June $ 289.17 May $ 187.16
January $ 266.63 February $ 163.41
December $ 230.46 April $ 162.64
6
13
4
2
0
2
4
6
8
10
12
14
0 1 2 3
Mean 1.08
Mode andMedian

7. a. Calculate the range of the utilitybillsforthe year2006
The highestamountinmonthSeptemberandthe smallestamountinmonthapril
So the range is $ 343.50 - $ 162.64 = $ 180.86
b. Calculate the average monthlyutilitybillsforthe year2006
AmountfromSeptemberintoapril fromthe table is2940.2 dividedby 12 so the average
monthlyutilitybillsforthe year2006 is $ 245.0167
c. Calculate the standarddeviationforthe 2006 utilitybills.
S2 =
1
11
[(343.50-245.0167)2
+(335.48-245.0167)2
+(306.55-245.0167)2
+(289.17-245.0167)2
+(266.63-245.0167)2
+(230.46-245.0167)2
+(226.80-245.0167)2
+(219.41-245.0167)2
+(208.99-245.0167)2
+(187.16-245.0167)2
+(163.41-245.0167)2
+(162.64-245.0167)2
]
S2 =
1
11
[ 43375.908] = $ 3943.264
So the standarddeviationisisS= √ 𝑆2 =√3943.264= $ 62.795
[9] the table belowshowsthe namesof the 42 presidentsof the unitedstatesalongwiththe numberof
theirchildren.
Children Amount
0 6
1 2
2 9
3 6
4 7
5 3
6 5
7 1
8 1
9 0
10 1
11 0
12 0
13 0
14 0
15 1
a. Calculate arelative frequencyhistogramtodescribe the data.How wouldyoudescribe the
shape of the distribution.
23
15
2 1 1
0
5
10
15
20
25
0 to 3 4 to 6 7 to 9 10 to 12 13 to 15
Amount
Children
A relative frequency Histogram

8. The shape of the distributionisskewedintorightwith mostof the dataare on the leftside of
the chart.
b. Calculate the meanandthe standard deviationforthe dataset.
Mean is total amountof childrendividedby16 so meanis42/16 = 2.625
the standard deviation is
S2 =
1
15
[(0-2.625)2
+(0-2.625)2
+(0-2.625)2
+(0-2.625)2
+(0-2.625)2
+(0-2.625)2
+(1-2.625)2
+(1-2.625)2
+(2-2.625)2
+(2-2.625)2
+(2-2.625)2
+(2-2.625)2
+(2-2.625)2
+(2-2.625)2
+(2-2.625)2
+(2-2.625)2
+(2-2.625)2
+(3-2.625)2
+(3-2.625)2
+(3-2.625)2
+(3-2.625)2
+(3-2.625)2
+(3-2.625)2
+(4-2.625)2
+(4-2.625)2
+(4-2.625)2
+(4-2.625)2
+(4-2.625)2
+(4-2.625)2
+(4-2.625)2
+(5-2.625)2
+(5-2.625)2
+(5-2.625)2
+(6-2.625)2
+(6-2.625)2
+(6-2.625)2
+(6-2.625)2
+(6-2.625)2
+(7-2.625)2
+(8-2.625)2
+(10-2.625)2
+(15-2.625)2
]
S2 =
1
15
[393.65625] = 26.24375
So the standarddeviationisS= √ 𝑆2 =√26.24375= 5.122865409
c. Constructthe intervalsx±s, x±2s, x±3s. find the percentage of measurementsfallingintothese
three intervalsandcompare withthe correspondingpercentagegivenbytchebyseff’stheorem
and the empirical rule.
X is2.625 and S is5.123
X + S = 2.625 + 5.123 = 7.748 X - S = 2.625 – 5.123 = -2.498
X +2S = 2.625 + 2(5.123) = 12.871 X - 2S = 2.625 – 10.246 = -7.621
X +3S = 2.625 + 3(5.123) = 17.994 X - 3S = 2.625 – 15.369 = -12.744
Form the interval,
(X-S,X+S) = (-2.498, 7.748) the numberrepresents40/42,or 95%
(X-2S,X+2S) = (-7.621, 12.871) the numberrepresents41/42, or 98%
(X-3S,X+3S) = (-12.744, 17.994) the numberrepresent42/42,or 100%
*For empirical rule,standarddeviationpercentagesagree fairlywellwiththe approximationof 68%,
95% and100%.
*for Chebyshev’sRule that nomatterwhatthe shape of the distribution.In2standarddeviationof Xis
75% andin 3 standarddeviationof Xis89%.
-15
-10
-5
0
5
10
15
20
X-3S X-2S X-S X X+S X+2S X+3S
Intervals
Value

9. I thinkif interval compare withusingempirical rule isalmostsame between(X-2S,X+2S) and(X-
3S, X+3S) but notsame withchebyshev’srule.Withempirical rule we know formoundshaped
distributionbutif we lookatthe intervalschartthe distributionisnotreallymoundshaped,norisit
extremelyskewedintoright.Of course,thatnomatterwhat the shape of the distributionusing
chebyshev’srule.
Aria Wibisono
1410911059
Mechanical Engineering
Andalas University