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Nursing Assessment
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Nursing assessment











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Nursing assessment
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Nursing assessment

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Uploaded on
January 31, 2026
Number of pages
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Chapter 7 :
SHORT ANSWER. Write the word or phrase that best completes each statement or
answers the question.
136)
The!!California !!Department !!of!!Education!!wants!!to!!gauge!!the!!difficulty!!of!!a !!new!!exam!!b
y!!having!!a!!sample!!of!!students!!at!!a !!particular!!school !!take!!the!!exam.!!The!!quality!!of!!the!!students
!!at !!the!!chosen!!school!!varies!!widely !!and !!the!!school !!administrators!!are!!allowed!!to!!choose !!who!!g

ets!!to!!take!!the!!exam.!!The!!administrators!!have!!a!!strong!!incentive!!for !!the!!school!!to!!do!!well!!on!!t
he!!exam.!!Do!!you !!think!!the!!results!!will!!represent!!the!!true!!ability!!of!!the!!students!!at!!school?!!Wh
at!!kind!!of!!bias,!!if!!any,!!do!!you!!think!!will!!be!!present? !!Explain.


Answer: !!!!Results!!will!!be!!biased!!because!!of!!selection!!bias.
Explanation: !!!!The!!administrators!!will!!probably!!systematically!!choose!!the!!good!!students!!to!!tak
e!!the!!test.!!This!!is!!a!!systematic!!exclusion!!of!!the!!bad!!students!!from!!the!!sample!!and!!will!!lead!!to!!
a!!biased!!score.!!This!!is!!called!!selection!!bias.

Results!!will!!be!!biased!!as!!Its!!selection!!bias!!,!!the!!administrator!!are!!systematically!!choosing!!the!!goo
d!!students!!and!!!!they!!excluded!!the!!bad!!students!!..!!


137)
The!!campaign!!manager !!for !!a !!candidate!!for!!governor !!in!!Arizona!!wants!!to!!conduct!!a!!pol
l!!to!!better !!understand!!his!!candidate’s!!chances!!for !!the!!upcoming!!election.
!!!!!!a.What!!is!!the!!population !!of!!interest?

!!!!!!!!b.Why !!may!!the!!poll !!be!!biased !!if!!a!!simple!!random!!sample!!of!!voters!!in!!the!!last!!gubernatorial

!!election!!(four!!years!!prior)!!is!!taken?


Answer: !!!!
a.!!Voters!!in!!the!!upcoming!!election!!
b.!!Selection!!bias
Explanation: !!!!The!!campaign!!manager!!wants!!to!!know!!how!!voters!!in!!the!!upcoming!!election!!vie
w!!his!!candidate.!!That!!is!!the!!population!!of!!interest.!!Voters!!from!!the!!last!!election!!may!!not!!vote!!
in!!the!!coming!!election.!!Voters!!who!!were!!too!!young!!to!!vote!!four !!years!!ago!!are!!not!!included!!us
ing!!this!!sampling!!method.!!Additionally, !!migration!!into!!and!!out!!of!!Arizona !!will!!change!!the!!vot
er!!over !!a !!four-year !!period.

,138)
It!!is!!known!!that!!college!!students!!at!!a !!local!!community!!college!!study!!12!!hours!!per !!week
!!with !!a !!standard !!deviation !!of!!five!!hours.!!What!!are!!the!!expected !!value!!and!!variance!!for!!a!!samp

le!!of!!nine!!students?


Expected!!value!!is!!12!!hrs

!! variance!!=!!!!(standard!!deviation!!)2 /!!n!! !!




= ( 5)*2!!/!!9!!=!!2.7779!!
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !! !!

!!


Answer: !!!!Expected!!value!!equals!!12,!!and!!variance!!equals!!2.7779.
Explanation: !!!!The!!expected!!value!!of!!
is!!the!!same!!as!!the!!expected!!value!!of!!individual!!observation, !!that!!is,!!E(
!!

)!!= !!E(X) !!=!!μ.!!The!!variance!!of!!the!!sample!!mean!!is!!computed!!as!!
Var( )!!= !! !!!!=5^2/9 !!= !!2.7779.




139) A!!fast-
food!!restaurant!!uses!!an!!average!!of!!110!!grams!!of!!meat!!per !!burger !!patty.!!Suppose!!the!!amount!!o
f!!meat!!in!!a!!burger !!patty!!is!!normally!!distributed!!with!!a!!standard!!deviation!!of!!20!!grams.!!What!!i
s!!the!!probability!!that!!the!!average!!amount!!of!!meat!!in!!four !!randomly!!selected!!burgers!!is!!less!!tha
n!!105!!grams?


Answer: !!!!0.3085
Explanation: !!!!If!! !!is!!normal,!!we!!can!!transform!!it!!into!!a!!standard!!normal!!random!!variable!!as!!

Z!!=!! ,!!and!!any!!value!!of!! !!on!! !!has!!a!!corresponding!!value!!z!!on!!Z!!given!!by!!


Z!!=!! .!!Compute!!P( !!<!!105).!!Use!!z!!table.

The!!appropriate!!Excel!!function!!is!!=NORM.DIST(105,110,20/SQRT(4),TRUE) !!=!!0.3085

,140) Suppose!!residents!!in!!a!!well-to-
do!!neighborhood!!pay!!an!!average!!overall!!tax!!rate!!of!!25% !!with!!a!!standard!!deviation!!of!!8%.!!
Assume!!tax!!rates!!are!!normally!!distributed.!!What!!is!!the!!probability!!that!!the!!mean!!tax!!rate!!of!!16
!!randomly!!selected !!residents!!is!!between !!20% !!and!!30%.




=NORM.DIST(30%,25%,8%/SQRT(16),TRUE)-NORM.DIST(20%,25%,8%/SQRT(16),TRUE)

Answer: !!!!0.9876
Explanation: !!!!If!! !!is!!normal,!!we!!can!!transform!!it!!into!!a!!standard!!normal!!random!!variable!!as!!

Z!!=!! ,!!and!!any!!value!!of!! !!on!! !!has!!a!!corresponding!!value!!z!!on!!Z!!given!!by!!


Z!!=!! .!!Compute!!P(20%!!≤!! !!≤!!30%).!!

Note!!that!!P(z1 !!≤!!Z!!≤!!z2)!!=!!P(Z!!≤!!z2) !!–!!P(Z!!≤!!z1).!!Use!!z!!table.

The!!appropriate!!Excel!!function!!is!!=NORM.DIST(0.3,0.25,0.08/SQRT(16),TRUE)-
NORM.DIST(0.2,0.25,0.08/SQRT(16),TRUE) !!=!!0.9876



141)
Suppose!!the!!average!!casino!!patron!!in!!Las!!Vegas!!loses!!$110!!per!!day,!!with!!a!!standard!!d
eviation!!of!!$700.!!Assume!!winnings/losses!!are!!normally!!distributed.
!!!!!!a.What!!is!!the!!probability !!that !!a !!random !!group!!of!!nine!!people!!averages!!more!!than!!$500 !!in!!

winnings!!on!!their !!one-day!!trip!!to!!Las!!Vegas?
!!!!!!!!b.What!!is!!the!!probability !!that !!a !!random!!group!!of!!nine!!people!!averages!!more!!than !!$500!!in !

!losses!!on!!their!!one-day !!trip!!to!!Las!!Vegas?
!!!!!!


Answer:!!!!
a.!!0.0045!!
b.!!0.0473
Explanation: !!!!
a.!!The!!standard!!deviation!!of!!
!!is!!calculated !!as!!the!!positive!!square!!root!!of!!the!!variance.!!We!!call !!it!!the!!standard!!error!!of!!the!!s


ample!!mean,!!and!!it!!is!!computed!!as!!se( )!!=!! .!!If!!


is!!normal,!!we!!can!!transform!!it!!into!!a!!standard!!normal!!random!!variable!!as!!Z!!= !!
!!

, ,!!and!!any!!value!!of!! !!on!! !!has!!a!!corresponding!!value!!z !!on!!Z!!given!!by!!Z!!= !! .!!

Compute!!P( !!>!!500).!!Note!!that!!P(Z!!> !!z) !!= !!1!!–!!P(Z!!≤!!z).!!Use!!z!!table.!!

The!!appropriate!!Excel!!function!!is!!=1-NORM.DIST(500,-110,700/SQRT(9),TRUE) !!= !!0.0045!!

b.!!Compute!!P( !!!!<!!−500).!!Use!!z !!table.!!The!!appropriate!!Excel!!function!!is!!=NORM.DIST(-
500,-110,700/SQRT(9),TRUE)!!=!!0.0473




142)
A!!ski!!resort!!gets!!an!!average!!of!!2000!!customers!!per!!weekday!!with!!a!!standard!!deviation!!
of!!800!!customers.!!Assume!!the!!underlying!!distribution!!is!!normal.!!What!!is!!the!!probability!!a!!ski!!
resort!!averages!!between!!1500!!customers!!and!!3000!!customers!!per!!weekday!!over!!the!!course!!of!!f
our!!weekdays?


=NORM.DIST(3000,2000,800/SQRT(4),TRUE)-NORM.DIST(1500,2000,800/SQRT(4),TRUE)!!!!=!!0.8881



Answer:!!!!0.8881
Explanation: !!!!If!! !!is!!normal,!!we!!can!!transform!!it!!into!!a!!standard!!normal!!random!!variable!!as!!

Z!!=!! ,!!and!!any!!value!!of!! !!on!! !!has!!a!!corresponding!!value!!z!!on!!Z!!given!!by!!


Z!!=!! .!!Compute!!P(1,500!!≤!! !!≤!!3,000).!!

Note!!that!!P(z1 !!≤!!Z!!≤!!z2)!!=!!P(Z!!≤!!z2) !!–!!P(Z!!≤!!z1).!!Use!!z!!table.
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