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Unformatted text preview: AEM 211] Name:
Spring 2MB Seetian: "
Section 2
l.) The vice president of marketing at a large corporation is wandering haw man}r of the company's employees arrive at work before he does. He decide: tn eelleet
same data by Donating the number of cars that are in the parking let when he arrives at 6:30 AM.
23 24 24 25 25
25 25 26 26 3?
Find the sample mean at the data. :3 3’” WI (F33) a) Findthemedian efthe data.
mad/Wt =' .35 mar {19.33) Find the range of the data. (5551—35 5”) QB? C9
13} 69 made '= Wear; (1,4,;
or. [9 Find the standard deviation at" the data.
5: 561»? am ff" a) e.) If the vice president wanted to eatimate the typical manber efernpleyees
who arrive atworkbeferehedeea, wnnidthenieanarthemedianhea
@ better statistic to use? Why ‘? Warhecatauiambw I'i't/Ll'f' afﬁcofql bUﬁw
mume I...) 1‘.) Is the standard deviatian or the range more reliable for these data? Wh
Wat cit/Lu Mme.»  We it“ if terr aﬁeﬁdcf @ ham MW 0f 3’? M5 2+} g.) Find the s—seore of the data value of3? ears. Does it conﬁrm yotn‘ ﬁdonﬂtatﬂie data value is mum; 15 377k!" ml.” “milk
t": e A“? w are an r war/U 1
5.5;} [IE—l beer [:3 Mmﬁﬁcf, 33’ 0“” LIE: W166 N 9% Mother cumsma game Wd (FIB) Refer to the data from Section 1, # 2 (the dollar value of the change in your
pocket). :1.) Provide relevant measures of spread for this data Write a semenee describing your ﬁndings. . . .
 ﬂusoﬁfhﬂhi writ vat
am e 
magmas: > W“ ‘9’ (Mi) 8
aidarm b.) If the data is mound shaped (refer to your histogram in Section 1), the
empirical rule states that approximately 95% of the data lies widrin 2
standarddeviations ofthemean. Thisirnpliesthattheringeofadata
set 1will be about 4 standard deviations. Check this approximation, is.
divide the range by 4, is this value close to the standard deﬂation? @ Why do you think the approximation did or did not mrk wel for this @ @ dataset'? 9...: m; (m will cart) tug] mire/1 (Fiﬁ) mu' m‘fm/ did aﬁwavlc‘vdm mi“ be!
”ﬂit” mﬂgm‘baﬁdﬁmpfﬂmi rule {7% m m
M c.) Find the 90th and 25th percentile for this data. Describe in words what (F 3;?)
these values mean. in (ctr/W)“ .: [anthem F0/ Wm Tau/MIMI
£1.qu! if m rim u'rx rel—W bylaw ____. if a. (stamps Pi‘ihm 1W 95*“ Wt" It ‘7 3.) Millions of American work ﬂ'orn ofﬁoes in their homes. Following is a sample of
age data for individuals who work at home. E_——
E5 3} Cmnputethemeau, modiauaud modefortlﬂsdﬂta I Egg/5253939“? 3
d‘ i '3' J J
w ML” all??? a a,“
{1% : :5; EH ‘1‘ apt1:55.25)?
m ﬁfty55) 5 5,1? b.) The median age of the popiﬂation ofa]! adults is 36 years {The World
Almanac, 2096}. Use the median age of the preceding data to eonuuent on
wdtether the athome workers tend to be younger or older than the population of all adults. .
@ #m WW: WA 10 lx/sLtﬂWt’la ”W5“
than 1m populous?” aFAU ndHHj e.) Compute the ﬁrst and third quartiles How would you interpret these (f 3:?)
values? @ a}: (Mﬁﬂﬂyﬁﬁ): 5 $15.65*? 3 2515(01‘1 Q9 ¢3={Waﬂ)(201= 15 to: am; assign:
s ' lac/laud 9'55 5.
gusts Him m or; M“ M
:: lift" 3.5.“ a? .mg, (1411., or: as?“ of berm? LIB53:13. 4.) The results of a national survey of 1154 adults showed that on average, adults
sleep 6.9 hours per day during the workweek {2WD Omnibus Sleep in America
Poll). Suppose that the standard deviation is 1.2 hours. T3, @ What percentage of individuals sleep between 4.5 and 9.3 hours per day’imlfg m I'lflz’.:l'5 W g use“ are a—Fmdr‘vl'duatr stat? W W W ”+5 ﬂﬂﬁl ”*3 “Dug/Mg'
as} 27"?“
b.) What percentage of individuals sleep more than 1'15 hours per day?
I fish .94 Wt least 93% of the individuals sleep between how many hours? (PH?) .43 = 1.. fig: (as seams), as Hearst”) @ Jana? (amtﬁehm]
I :11 5'9 as eat"
d.) Aﬁume that the number of hours of sleep follows a mound shaped
distribution. Calculate the percentage of individuals who sleep between
I 4.5 and 9.3 hours per day, and more than 1&5 hours. Hovr do these ﬁsults mag to the values that you obtained in parts {a} and (b)? C1045, . . . . . lm/
A? ramnatal as f. 0F jMEUfﬁLuﬂlf ELI/0F til5' ”l5 “it?
Lorri segmtlmatst 0% at? indt'wlfiuiti W ”’5'" l"rt/it's}?
Me 5.5 #5 ‘5? be 6] ﬁts of @ ”19ng mt” W Mali [5 F ﬁl 591%
W Ltddf H5 19 cadsladﬂ/ mm; radtqtdualf s #157. W
l l IE .1 45“ 01,15 LIN/Ad and L5“ _M‘ , a mate HIP5' it"s/d“ 
I.” {l mow state {S Lust(«YEW “WM ﬂl'ﬁ’ WM. 5.) A student took two national aptitude tests. The national average and standard
deviation were 415 and It“), respectively, for the ﬁrst test and Bit and 8,
respectively, for the second test. The student scored 625 on the ﬁrst test and 45
on the second test. Determine on which exam the student performed better C FLTB)
relative to t other test takers. '27." : {Kltr” Z7}? (KE‘QU)
a” T @m semtforﬁmd
aw i #54” retaﬁvcllﬂ W m ma —'“”’ {mm a.
[email protected] lﬂfﬂ@ E.) For each of the following arguments, explain why the conclusion dream is M
valid. a.) A real estate agent notea that the mean housing price for an area is (1:19 $125381] and concludes that half of the houses in the area cost more than (P 33)
that amt: describe}; WWﬂMm mmn waisted
usi tndt'sttc haw” mam/OI slata pomt5 an} absi’t/ or he! am! it. b.) A businesswoman calculates that the median cost of the ﬁve business 'ps
that she took in a month is Hill] and concludes that the total cost must 1133:) havebee $ , t
 whims army! the Middle value on an M
$054135 tilF Jimmyaﬁ Wm“ can“ M “attempting?”
11; ﬁnd Wt “5+: (appPsywaerpmslsat c.) A company executive concludes that an accountant must have made a mistake because she prepared a report saying that seat of the companies
(I? employees earn less than the mean salary. 5: P35)
1 my: ls isn't a wotfake W ma waste! irist M
indl'ﬁﬁrw #1st ataxia ppmhr arc mm N ha
I1t and Martians 34;. My” man meld be greatlg mammal ”ti sum“ d.) A restaurant owner decides that more than half of her customers prefer chocolate ice cream because chocolate is the mode when customers are
@ ofﬂoad their choice of chocolate, vanilla and strawberry. {P 33 3
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that” silica“ mast ﬁWfﬁM plat/{Hit Wsaniéj
55L} as (stews sf auto stwo: meow, burr
W m‘ rs mtg sit/W : 35%
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 Spring '08
 VANES,C.

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