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Unformatted text preview: AElvl Eli) Name (inﬂame % Spring 230’? II).
Section
Prelim [l
Please lIEirele the Correct Answer. Each multiple choice question is worth 2 points. I.) The following table presents the probability distribution function for the number of
claims processed per hour at an insurance agency. Which of' the following is true? m T
ms] at? ens ﬂlﬂ P{X : 4) = ens
.) P{}{ 5 4} = oar
c.) P(X v 2) = Loo
a.) POI 3 e) = 0.1 2.} Which of the following statements about the binomial distribution is not correct? a.) Each trial results in a success or failure. 1}.) Trials are independent of each other. c. The probability of success remains constant from trial to trial.
The random variabie of interest is continuous. e.) The experiment consists of n identical trials. 3.} The random variable x is known to be uniformlyr distributed between "iii and 9D.
The probability of it having a value between EU to 95 is: a.) are
{15 o. 3.135 d.) l e.) None of the above trustvars is correct. 4.} The relationship between the standard normal random variable 2 and normal
random variable X is that: {int}r the normal random variable X is continuous. Gulls.r the standard normal variable 2 is continuous. The standard normal variable Z counts the number of standard deviations
that the value of the normal random variable X is aura};r from its mean.
d.) The values of the standard normal randan variable Z cannot be negative.
e.) The values of the normal random variable X cannot be negative. '3'?”
Lav 5.] The binomial distribution can be approximated by a normal distribution.
Consider the following statements: {i} The normal approximation improves as sample size increases. {ii} Certain corrections are necessary as the normal approximation involves
fitting a curve to a discrete distribution. (iii) For a given sample size, n, the goodness of approximation depends on p,
the probability of success on one trial. Which of these statements best describes the normal approximation to the
binomial: a.) (1) and (iii)
a.) (i) and {ii} c. (ii) and {iii}
QT]? {i}, (a) and (iii)
e. (i) 6.) The standard error of the mean is a name for the standard deviation of the .} sample
.) parent population
.) sampling distribution of the variance sampling distribution of the mean
e. raw scores nuns: T.) Which of the following statements is NUT true aooording to the Central Limit
Theorem? a.) An increase in sample size from n : 16 to n = 25 will produce a sampling distribution with a smaller standard deviation
@ The mean of a sampling distribution of sample means is equal to the population mean divided by the square root of the sample size. c.) The mean of the sampling distribution of sample means for samples of size
n = 15 will be the same as the mean of the sampling distribution for samples
of size n = ill]. d.) The larger the sample size, the more the sampling distribution of sample
means will resemble a normal distribution. 8.} A 95% conﬁdence interval indicates that: 95% of the intervals constructed using this process based on samples from
this population will include the population mean
b.) 95% of the time the interval will include the sample mean
c.) 95% of the possible population means will be included by the interval
d.) 95% of the possible sample means will be included by the interval 9.) 1e.) 1].) 12.) A conﬁdence interval increases in width as: a.) The level of conﬁdence increases
b.) It decreases
s increases c. £39 All ofthe above
The USA Today AD Track (BEING) examined the effectiveness of the new ads
involving the Pets.com Seek Puppet (which is new extinct). In particular, they
conducted a nationwide poll of 423 adults who had seen the Pets.com ads and
asked for their opinions. They found that 33% of the resmndents said they liked
the ads. If you wanted to cut the margin of error {sampling error) in half the next
time you took a poll like this, what sample size would you need?I lDne that is twice the size of what it was before.
l[Zine that is half the size of what it was before. lDne that is a quarter the size of what it was before.
None of the above. a.)
b.) The reason that the t score cannot be used for constructing conﬁdence intervals for
the population proportion with small samples is: the t score cannot be used when the population distribution is not normal. b } The 1 score cannot be used when the sample size is under 3i}.
c.) The t score cannot be used unless up and n(l—p} are greater than or equal to
5.
d.) the t score cannot be used unless n f: and n[l  ,5) are greater than or equal to 5 1When determining the sample size for a proportion for a given level of conﬁdence
and sampling error, the closer p is estimated to be to .Sﬂ the _the
sample size required. a} smaller larger Sample size is not affected. The effect cannot be determined from the information given. Please show all your work and eirele your ﬁnal answer. 13.) Suppose that the times required for a cable eiznn‘ipartyf to ﬁx eahle problems in its
customers” homes are uniformlyr distributed bBtween 4t] minutes and 65 minutes. a,] Mat is the probability that a randomly selected eahle repair visit will take
at least El} minutes? “5 a“) ‘0‘” ‘ so 5:: e5 )‘5 In.) 'What is the [Jl‘ﬂbﬂhiliif that a randornlg,r selected eahle repair visit falls f as.
within 1 standard deviation of the mean? 1 Q!
(a s we”) a: 4M wage w H : HE is
{a} CD 3* Q
Popsiggrﬁx '5 59,?50 r: (msvﬂev) = l4.) Aeeording to Fortune [April 1994), proﬁts as a percentage of sales in the oil
industry during 1993 ranged from —5% to 7’93. If we wish to estimate average
proﬁts of sales with a 99% eonﬁdeaee, what size sample should be drawn?
Assume a margin oferror of 2%. (6} 4 1 z 1,3. (52,5?) (3)  MW» 0 15.) {5) 16.} {5) (2i (5} A contractor must pay a $49,999 penalty if construction of an expensive home requires more than 16 weeks. He will receive a bonus of $19,999 if the home is
completed within 8 weeks. Based on experience with this type of project, the
contractor feels there is a 9.2 chance the heme will require more than 115 weeks
for completion, and there is a 93 chance it will he ﬁnished within 8 weeks. if the
price of the heme is $359,999 before any penalty or bonus adjustment, how much can the buyer expect to pay for her new MW (D Time it (a) pa) 1:39 rdﬁ ‘ .31 15:53—92
we » Itasca 7f ‘ was $41,000 69005
ruse: o . a 0 “agrees: nae 6"” ‘9 “Jam *5 ﬂit; sewed iii£153“ F5000 “—‘ ﬁreman +0551“); $345000 V According to the Internal Revenue Service (IRS), the chances of your tax return
being audited are about *5 in l,999 if your income is less than $59,999; 19 in 1,999
if'your income is between $59,999 and $99,999; and 49 in 1,999 if your income is
$199,999 or more (Statistical Abstract oftite United States; 1995}. a.) lfﬁve taxpayers with incomes under $59,999 are randomly selected, what
is the probability that more than one will be audited? “RT5 é) @ Part) 1 F" (99“ "’h “03%) E b.) What assumptions did you have to make in order to answer part a? 0 ?rabab'i:+y oi: derﬂj anottied remains amine1d" 6’ (it’d: Q :0 Hheiiur‘ ﬂxpayars acre mica! er nef
independeni' dyad£5 c.} If two taxpayers with incomes under $59,999 are randomly selected and two with incomes more than attention are randomly seiected, what is the
probability that the IRS will audit none of these taxpayers? Fe ,ccg, 75(0): ,9??? Q]
f):,0‘t"? 70(5):.90‘7/ .999 (.904) = Q as: QJ 71:2, d.) If ltlﬂﬂ taxpayers with incomes more than $lllﬂ,l]llﬂ are randomly
selected. what is the probability that more than a of them will be audited? 7.25m“ lg: 0‘1“? fed awvégsss = am
{00,55 boo) rs Pact: ‘?ed ‘5) 2:69—47“? T, /,g g
é.t§9 Q
P (27t.é3')=.5n‘t535 . The production manager for the Breitman manufacturing companyr is concerned
that the customer orders are being shipped late. He asked one of his planners to
check the timeliness of shipments for 2000 orders. The planner randoij;r selected
2WD orders from a population of SIIIU and found dtat 43f} orders were shipped
late. Construct the 96% conﬁdence interval for the proportion of orders shipped late. ﬂ Q Q
m) EGGS = , 2‘7"
\/ aoeo 5000*3000
ﬂood—r
(. .9356. ass) as on. up The average starting salaryr for this year's graduates at a large universityr (LU) is
$4ﬂ,ﬂﬂﬂ with a standard deviation of $843313. Furthermore, it is known that the
starting salaries are normally distributed. 4 r” {'5} 1f... .._ aroﬁ
(13;) .Jl‘t'I ,ﬂfﬁ , Q4 . 18.) a.) Individuals with starting salaries of less than $35,621!) receive a lovi.r income tax break. What percentage of the graduates receives the tax break?
1 5 5 @D 55; £05; — deﬁed
Fooo—F# ’0(3a‘55‘> = ,§~.3033":99f2.
CD CD What are the minimum and the maximum starting salaries of the middle
95% of the LU graduates? E {4} s.) (6) 32969 = X“ — Hojooo Q) sooo
: L959 XML : 55:; Xt—tfoJooO XL. #31520 @ ____...——+ 3000 e.) If 189 of the recent graduates have salaries of at least $52,24t}, hoytr many
students graduated this year from this university? (a z 1 5:." use J 451.000 8000 GD
szv*t53)s fie.?3? = . #606?) a x99 [/l/r 30001 GD 19.} The length of time it takes a stock analyst to eomplete an evaluation of a
company‘s earnings forecast is normallyr distributed with a mean of 6.? hours.
From looking at the time spent by an analyst evaluating 35 randomly selected
stocks you ﬁnd that the sample mean was 6.4 hours. If you are told tl'tat the
probability of getting a sample mean this small or smaller is tilt}, what must the
sample standard deyiation have been? <6? K~ﬂ(s.¥, 55 7:: 3s 3364 ﬁKfs (9.4) = “"0 C1) : 3&9? : "IO—Li 2U.) [inline customer service is a key element of online retailing. According to the
W5} Market Data Group, 315% of Princeiinecotn customers take advantage of
this service. If random samples of 2'31) Priceline.com customers are selected, what
is the probability that the sample proportion of customers who use on—line customer service is between 35% and 4t}%? 5 J”; .. $5
“ 5.3;
(a (.354: eye) mew F g M5
:7 I 0 “‘2 I: r
} Li— ae Q vmji .3?5(.sas) $.95
Joe @ '5 v90 69 Pfﬂﬂaaeeﬂﬁ) = 1( does) an (9 EXTRA CREDIT: You do NUT have to do this problem. 21.] A large state university is funded depending on the mean number of credit units a
student takes. A preliminary survey involving T2 students yielded the mean
number of f = 13.4 units with a standard deviation s = 2.1 units. Suppose there are
EELEIJU students enrolled in the university and the institution is funded at the rate of
$3M per student unit. At the 99 percent conﬁdence level, estimate the interval that
covers the total funding of the university for the year. (4) my: §.5?5 (MA/$3) £5.49? c. any
a; 2
(more: 5 £4.09) a»: 300 e 510300
(ii timed. we deionised) 952352 75M @369 ...
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This note was uploaded on 10/18/2008 for the course AEM 2100 taught by Professor Vanes,c. during the Spring '08 term at Cornell.
 Spring '08
 VANES,C.

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