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Unformatted text preview: Problem 1. You purchase a stock for $1000. The stock gains 20% or loses
10% each day, each with a probability 0.5. Its returns on consecutive days
are independent of each other. You plan to sell the stock after two days. a. Find the distribution of the values of the stock after two days (in other
words, what are the possible values of the stock after two days, and what is
the probability of each value?) Give it in the form of a table. 100/ “#ka b. What is the probability that the stock is worth more after two days than
the $1000 you paid for it? FL x 7 \oaaj : Faniosokﬂxﬂmuo") :: cots Ar (9.25) :KOTS c. What is the mean value of the stock after two days? HX ,~.~ (gm) {0.95) + gage) 0,5; 1‘. QWO) (0125)
: 2015‘ «\— 5—40 ragga ..’) a ..._
ﬁg. 2,16; Problem 2. Suppose that 20% of the pepulation has Type A blood. If a
random sample of 81 people is selected, what is the approximate probability
that 15 or fewer people in the sample have Type A blood? a. What is explicitly the random varable X having a. Binomial distribution? X 2 1d; oé} \Deapiﬂ Lotx 335M drape, A \ohoodn W0
1960 FM .
$2 : ID . .2 ) n f; gl
c. Use the Normal approximation with continuity correction to ﬁnd the
desired probability, use X~ N(np,\/np(1—p)).
X m Nkugﬂgj \ 31,20") tbs 5.1153: H X: «553 Problem 3. A researcher makes a measurement in a biology laboratory and
records the result in his laboratory report. The standard deviation of his lab
measurements is cr = 10 miligrams. The researcher repeats the measurement
3 times and records X of his 3 measurements. a. What is the standard deviation of of his result? CST—:L: FL) %ijq%€
Y on (g 55:. b. How many times must the researcher repeat the measurement to reduce
the standard deviation of X to 2? a l0
.— ((1, __s
3 «2—23 SSH.7..— 26— L o3_8’\ (C’onﬁ’rrnmﬁ erecdqbrx‘)
:51=>(z s «aswe) remiss N 1.. V7 Problem 4. You have three instruments with which to measure the height
of a tower. If the true height is 50 meters, measurements with the ﬁrst instru—
ment vary with mean 50 and standard deviation 1.2 meters. Measurements
with the second instrument vary with mean 50 and standard deviation 0.80
meters. Measurements with the third instrument vary with mean 50 and
standard deviation 1.3 meters. You make one measurement with each in—
strument. Your results are X; for the ﬁrst, X 2 for the second and X3 for the
third7 and are independent of each other. a. To combine the three measurements you might average them, (X1 + X2 + X3)
———3 . What are the mean and standard deviation of Y? y: :[VLXL :HXB :: 50 '3 63L\ '1" ‘3‘2‘, gig‘29:?) 5‘32 2“?) \i 1: HA): witx3 a§@x\+HX1* vxgﬁteﬁﬂ )7; ea
2 3 :2 U'x\+m.+Mg _ ( “‘xdxﬁxﬁ cth “7‘3” 63‘1" x33] 3 ”"i _ ::: oJ—t \ 5‘ '
b. Deﬁne another averaging for the measurement: J
1 1 3
W = —X + —X + —X
8 1 8 2 4 3 what are the mean and standard deviation of W?
W = JgWWJgHth 3mm) —
0‘31}:— eraa semie scif" rig,
(mu) mg”) (06“) e (ff?) UM)
2: oozzg IK— (904 f 00V30t13 4 1: 0,0175 30—3
W deems r» M , _ ‘0‘th 9.) A—Loﬁs) k (LL23 Problem 5. A random sample of 400 lightening ﬂashes in a certain region
resulted in a sample average radar echo duration of 0.64 sec and the standard
deviation is assumed to be a = 0.20. a. Calculate a 92% conﬁdence intervai for the true average echo duration [1. 0.0L} got: {ITS 5; o. 7. “social ’55:. 0‘5“
. ._ ‘1
Leg CI . x 4— "17‘3“ r“ ,1 0.64 1 gjslb )
7 ,— “ i405)
7124.75 2 was + 0.0175 c [5.6.2.252 0.6615]
‘22: . b. What dees' the conﬁdence interval mean for p? Describe in your own
words. 1\\ We euxa a234, Qvﬂg’kJLQ/l—‘m WOLFE" dim. We mean
_\u'{\\ \sc «(‘n C e 61x5 0 b§1€§1 ” '
or \\ u; we do saw; q 104: OS.— EmirMS 0% 9:22. 1400
sagas. m m [3.5125, 0.69753wa @an TL U
c. If the conﬁdence interval is [0.625, 0.655], what is the level of conﬁdence? meg“05‘” m : o.6‘35—0.b1§ __ {9.0175 .3: ,3; (0.2.) ' €(ﬂof '21 ‘lqoa
62:17} (“r0‘5) : 4.5 4— QQOJDbE’gS 85:3) 19.366” E .s 5. 6i A. c L l
at sample size we need to ensure that we get[ .59, 0.69] as a conﬁdence interval with 95% conﬁdence level?
0 5 cl’ 0 c3 6‘ ~— EXS— ’\~CH: r margiﬂ Oér C(fof rm: _____________._._. .
J M? 1W 'Mm L—w‘g "a" ”‘5
' ' 22* Problem 6. What value of z* in a large sample conﬁdence interval for ,u.
should be used to obtain each of the following conﬁdence levels?
a. 76% b. 94%
z“: 4.86 Problem 7. Determine the conﬁdence level for each of the. following conﬁ—
dence intervals for p. u a
X :l: 0.97% ml»; c.1660
' . (LL 2 4,. 19am: 11(10Jt50) WT! : o. (368’ b Xi3.03% 86.87; C L— emote—es) Problem 8. A Gallup poll ﬁnds that 30% of adults visited a restaurant in
the past 12 months, and that 6% ordered the chef ’s menu. These results
come ﬁom a random sample of 1000 adults. For an SRS of size n = 1000: (hint: use 33 ~ N(p,1/E:—m))
a. What is the probability that the sample proportion 13, (of adults visited a restaurant in the past 12 months), is between 0.29 and 0.31 if the p0pulation
roportion is 0.30? (0.5){ot'rz “9'0qu E000 fCD‘Q’Ol$%6 will): f3( (9.291.433 {is o.31»®.%) PLF—o.é°\ 5% g 0.601.)
—: 4‘» Qfo.aq5\):o.socxsg b. Suppose that the true population proportion is p = 0.30. If we take bigger
and bigger sample sizes, what happens to P[0.2 < 13 < 0.4]? (calculate this
probability for sample sizes 1:. 2 100, 1000, 10000, and try to predict what
happens for very very big numbers... ) the? ~S “Us!" (0.9...0.3<%< viii— ~03 HM !( .0, a “(.0 '1) ‘l (0 3)(.°~1l Wee \<% some) ' W. l’C—osse s z 3 Meg): ns’soo
PClLtli’ffézsltmg) hatooo
PL—ltsgﬁ 7'— 6 ”5875391 n: tojaoo en o1_‘~%‘°‘H) a; «t‘ iew “2&5 "w  —.—————~———m—v—m——e—————————._—.—————_._~_W.m.m..m__——ﬂ1u~u—————————Ewm.mmun (c) Test whether the true mean of age of men is something ther than 24 at
signiﬁcance level a = 0.04. 2% Problem 9. Ages for three randomly chosen men are 19, 21, 23. Ages of
men is a Normal random variable with 0' = 2.1. (a) Compute the sample mean 51': and sample standard deviation 3 for the
sample. :32: lﬁlzlﬁ‘zg .__.__ 2‘ 3
52:4.[23‘4r tie£13: it :> $5.2
9'2, &—.— (b) Assuming that the population distribution of “age of men” has a Normal
distribution, compute a. 96% conﬁdence interval for p, the mean age of men “ 2‘ + 2.1}8‘3 = Lﬂgsxs /., 1:q%5:l {s—D (\o'l‘ \I\ m ":CQ'STQ/Um\ (,Cl) .HMUL M W” m tall: en: e224 amok <1th (ﬁ’l HT! 1 ‘Adglq, (20points)?roblem 10. (BONUS Problem) A p—coin is a. coin such that
P[heads] = p. A p—coin is tossed untill it turns “heads" , where p = 0.6. Toss
this pcoin untill it comes heads for the ﬁrst time. a. What is the sample space for this experiment?(VVhat are possible out~
comes of the experiment?) e:\ A) T U.) TTU,TTTH,TTTTHJ_ ,_ 1; b. Let X =number of tosses needed until the first heads. What are the
possible values of X ? (in other words What are the possible times you can
have the ﬁrst heads...) y:— ‘l, 2,) 8, Ll.) S, ,. _._. “21. EAT“ (1000ka *mle i (S c.Wha.t are the events {X = 1},{X = 2}, ~  {X = k} in terms of heads and
tails? , seem, (es—Lima, _ g x, 14,, éTW ”f H é
TERA) T s
d. Find the mass function of X i. e ﬁnd P[X = k] for all k in past a.(Find
P[X = 1],P[X = 2],~  and so on... ) slid) e, H H7: 0 E 4? 1% ,fm HFH) Q" LOU6)
—~\
(0%)“ . (“It”). \(u—lj T‘s ﬂ 9 Formulae 11—1 Conditional probability deﬁnition Hate) = W. Mean and variance of a diSCIete random variable: tux : ka(k)a 0% = Z“: " ﬂx)2P(k) Standard deviation of the sample mean 32’ Gir=a/\/7_a Large sample conﬁdence interval is given by a=Zenszcerta—a2=ei(zaL :5”)  or
Xi*—~—
a“5 Binomial distribution: X is a Binomiel(n,p) 7?. W) = PM 2 k) = (9m — gark Central Limit theorem: Sampling from a population which is distributed ,
with random variable X. Mean of X is a}; and standard deviation of X is ox. If sample size n is large, then a X approximatein (p, (TX/ﬂ). Ifthe random variable X is normal then the above approxiamtion is exact.
Mean of sum random variables is the sum of the means. Variance of the sum
of independent random variables is the sum of the variances. i i i 10 l
i l  l ' ' ! ...
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 Summer '08
 NANE

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