STATISTICS - LESSON no 1 - MIDTERM-2.pdf

Midterms for Statistics

1. LESSON 5:
CALCULATING MEAN AND
VARIANCE OF A DISCRETE
RANDOM VARIABLE

2. THE MEAN
a.Theaverageofallpossibleoutcomes.Itis
otherwisereferredtoasthe“expectedvalue”
ofaprobabilitydistribution.
b.Meansthatifwerepeatanygivenexperiment
infinitetimes,thetheoreticalmeanwouldbethe
“expectedvalue”.
c.Anydiscreteprobabilitydistributionhasa
mean.

3. THE VARIANCE AND STANDARD
DEVIATION
TheVARIANCE ofarandom
variabledisplaysthevariabilityorthe
dispersionsoftherandomvariables.
Itshowsthedistanceofarandom
variablefromitsmean.

4. THE VARIANCE AND STANDARD
DEVIATION
a.Ifthevaluesofthevarianceandstandarddeviation
areHIGH,thatmeansthattheindividualoutcomes
oftheexperimentarefarrelativetoeachother.In
otherwords,thevaluesdiffergreatly.
b.Alargevalueofstandarddeviation(orvariance)
meansthatthedistributionisspreadout,with
somepossibilityofobservingvaluesatsome
distancefromthemean.

5. THE VARIANCE AND STANDARD
DEVIATION
c.IfthevarianceandstandarddeviationareLOW,that
meansthattheindividualoutcomesofthe
experimentarecloselyspacedwitheachother.In
otherwords,thevaluesarealmostthesamevalues
oriftheydodiffer,thedifferenceissmall.
d.Asmallvalueofstandarddeviation(orvariance)
meansthatthedispersionoftherandomvariableis
narrowlyconcentratedaroundthemean.

6. Mean,variance,and
standarddeviationcan
beillustratedbylooking
patternandanalyzinggiven
illustrationsanddiagrams.

7. DuringTownFiesta,peopleusedtogotoCarnival
thatmostfolkscallit“Perya”.MangBenusedtoplay
“Beto–beto”hopingthathewouldwin.Whileheis
thinkingaboutwhatpossibleoutcomesineveryroll
wouldbe,heisalwayshopingthathisbetisright.
X 1 2 3 4 5 6
P(X) �
�
�
�
�
�
�
�
�
�
�
�
EXAMPLE

8. 1.IstheprobabilityofXliesbetween0and1?
2.WhatisthesumofallprobabilitiesofX?
3.Isthereanegativeprobability?Isitpossible
tohaveanegativeprobability?
4.Howwillyouillustratetheaverageormean
oftheprobabilitiesofdiscreterandom
variable?
QUESTIONS:

9. 1.Istheprobabilityofxliesbetween0
and1?
Yes, the probability of X lies between 0
and 1.
X 1 2 3 4 5 6
P(X) �
�
�
�
�
�
�
�
�
�
�
�
SOLUTION:

10. 2.WhatisthesumofallprobabilitiesofX?
෍????????????=??????�+??????�+??????�+??????�+??????�+??????�
=
�
�
+
�
�
+
�
�
+
�
�
+
�
�
+
�
�
=
�+�+�+�+�+�
�
=
�
�
????????????�
The sum of all probabilities of X is exactly 1.
X 1 2 3 4 5 6
P(X) �
�
�
�
�
�
�
�
�
�
�
�
SOLUTION:

11. 3.Isthereanegativeprobability?Isitpossible
tohaveanegativeprobability?
No negative probabilities because it is
impossible to have it based on the
characteristic of the probability of
discrete random variables.
SOLUTION:

12. Inaseafoodrestaurant,the
managerwantstoknowiftheir
customersliketheirnewrawlarge
oysters.Accordingtotheirsales
representative,inthepast4months,
thenumberofoystersconsumedby
acustomer,alongwithits
correspondingprobabilities,is
showninthesucceedingtable.
Computethemean,varianceand
standarddeviation.
Number of
oysters
consumed
X
Probabilit
y
P(X)
0 �
��
1 �
��
2 �
��
3 �
��
4 �
��
EXAMPLE 1

13. (X)P(X)X ∗P(X) ??????−??????. (??????−??????)² (??????−??????)² ∙ ??????(??????)
0�.���−�.�=
−�.�
−�.�
�
=�.��
�.��∗�.�=�.���
1�.�0.2�−�.�=−�.�−�.�
�
=�.��
�.��∗�.�=�.���
2�.�0.6 �−�.�=�.��.�
�
=�.���.��∗�.�=�.���
3�.�0.6 �−�.�=�.��.�
�
=�.���.��∗�.�=�.���
4�.�0.4 4−�.�=�.��.�
�
=�.���.��∗�.�=�.���

14. 1.What is the mean?
??????=෍[??????∗??????(??????)]
= �+�.�+�.�+�.�+�.�
??????= �.�
SOLUTION:

15. 2.What is the standard deviation?
??????
2
=෍(??????−??????)²∙??????(??????)
=0.448+0.128+0.012+0.288+0.484
??????
�
�.��
SOLUTION:

16. 3.Whatisthestandarddeviation?
??????=σ(??????−??????)²∙??????(??????)??????????????????=??????
2
??????=1.56
??????=�.��
SOLUTION:

17. Mr.Umali,aMathematics
teacher,regularlygivesa
formativeassessmentcomposed
of5multiple-choiceitems.After
theassessment,heusedto
check the probability
distributionofthecorrect
responses,andthedatais
presentedbelow:
TESTITEM
(X)
Probability
??????(??????)
0 0.03
1 0.05
2 0.12
3 0.30
4 0.28
5 0.22
EXAMPLE 2

18. 1.Whatistheaverageormeanofthegiven
probabilitydistribution?
2.Whatarethevaluesofthevarianceofthe
probabilitydistribution?
3.Whatarethevaluesofstandarddeviation
oftheprobabilitydistribution?
QUESTIONS:

19. XP(X)X ∗P(X) X -?????? (X -??????)²(X -??????)² ∗P(X)
00.03
0∗0.03=00−3.41=−3.41(−3.41)²=11.6311.63∗�.��=�.��
10.05
1∗0.05
=0.05
1−3.41=−2.41(−2.41)²=5.815.��∗�.��=�.��
20.12
2∗0.12
=0.24
2−3.41=−1.41(−1.41)²=1.991.��∗�.��=�.��
30.30
3∗0.30
=0.90
3−3.41=−0.41(−0.41)²=0.170.��∗�.��=�.��
40.28
4∗0.28
=1.12
4−3.41=0.59(0.59)²=0.35 0.35∗�.��=�.��
50.22
5∗0.22
=1.10
5−3.41=1.59(1.59)²=2.53 2.53∗�.��=�.��
??????=�.�� ??????
�
=�.��

20. 1.What is the mean?
??????=෍[??????∗??????(??????)]
= �+�.��+�.��+�.��+�.��+�.��
??????= �.��
SOLUTION:

21. 2.What is the standard deviation?
??????
2
=෍(??????−??????)²∙??????(??????)
= �.��+�.��+�.��+�.��+�.��+��
??????
�
=�.��
SOLUTION:

22. 3.Whatisthestandarddeviation?
??????=σ(??????−??????)²∙??????(??????)??????????????????=??????
2
??????=1.59
??????=�.��
SOLUTION:

23. 1.Thenumberofshoessoldperdayataretailstoreisshowninthe
tablebelow.Illustratethemean,variance,andstandarddeviation
ofthisdistribution.
Writeallthenecessaryformulaandshowthecompletesolution.
Formulatobeused:(a)Mean,(b)Variance,(c)Standard
Deviation
X 19 20 21 22 23
P(X) 0.4 0.2 0.2 0.1 0.1
ACTIVITY 1

24. 1.TheLandBankofthePhilippinesManagerclaimed
thateachsavingaccountcustomerhasseveral
creditcards.Thefollowingdistributionshowingthe
numberofcreditscardspeopleown.
a.Calculatethemean,variance,andstandarddeviation
ofadiscreterandomvariable.
X 0 1 2 3 4
P(X) 0.18 0.44 0.27 0.08 0.03
ASSIGNMENT

25. 2.SupposeanunfairdieisrolledandletXbethe
randomvariablerepresentingthenumberofdots
thatwouldappearwithaprobabilitydistribution
below.
a.Calculatethemean,variance,andstandarddeviation
ofadiscreterandomvariable.
OUTCOME (X)1 2 3 4 5 6
Probability ??????(X) 0.1 0.1 0.1 0.5 0.1 0.1

26. 1.Thenumberofcellularphonessoldperdayat
theE-CellRetailStorewiththecorresponding
probabilitiesisshowninthetablebelow.
Computethemean,variance,andstandard
deviationandinterprettheresult.
(??????)
15 18 19 20 22
Probability ??????(??????)
0.30 0.20 0.20 0.15 0.15