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Unformatted text preview: 1iFtp‘hat is the signiﬁcance of the mean of a probability‘ dismbution? Choose the correct answer below. It gives information about how the outcomes H" [t is the expected value of a discrete random variable. Determine whether the statement is due or false. [fit is false. rewrite it as a true statement. The mean of a random variable represents the "theoretical average" of a probabilitv experiment
and sometimes is not a possible outcome. Choose the correct answer below. V“ True. Decide whether the random variable x is discrete or continuous.
X represents the length of time it takes to get to work. Is the random variable x discrete or continuous? Choose the correct answer below. '1'" Continuous Discrete Decide whether the random variable x is discrete or conu'nuous.
1: represents the total number of die rolls reqtu'red for an individual to roll a ﬁve. Is the random variable x discrete or conu'nuous? Choose the correct answer below. V Discrete C ontinu ous Determine whether the dismbution is a discrete probability disuibution. x Pﬂx} ﬂ  0.01 l — [LEE
2 2.66
3 — [LEE
4 — ELUl Is the probabilitv dism'bution a discrete disn‘ibution’.’ Wth Choose the correct answer below. H" No. because some of the probabilities have values greater than 1 or less than D. Use the ﬁequencv disuibution to conscht a probabilitv distribution. The number of dogs per household in a small town Dogs El 1 2 3 4 S
Households 1485 429 loll 42 29 13
1485
PEG) = F =5 [USES
21:8 The completed table is shown below. P‘Ex) 0.638
0.199
0.094
U. [ll 9
9.913
0.006 U‘IPU‘IINJI—‘gﬁ Eh} Find the mean of the prohahﬂitv dism'hution.
The mean of a discrete random variable is given hv the following formula.
it = Z KP Ex)
Each value of x is multiplied hf; its corresponding prohahilitv and the products are added. The completed prohahilitv disnihulion table is shown below. rounded to the nearest
hundredth. x Pﬂx} xPﬂx}
El III.583 [l 1 0.199 [LE 2 [I'ﬂ—H 0.15 3 0.019 [LUIS
4 0.013 [LBS 5 0.006 CLUE The mean of the prohahﬂitv dismhution is shown below. .L= = 0— [LE — [L15 — ELUIfr— [LEI'S — CLUE = [L49
Thus the mean is approximater I15.
(c) Find the variance of the prohahﬂitv disnihution. The variance of a discrete random variable is given hv the following formula. '52: EEK—H}2PEX} x Pix) 1—11 (11—1111 rx—olatx)
[1 [1.683 — [1.451 [1.24 [1.165 1 [1.1 E151 [1.51 [1.215 [1. [152 2 [1.[1'1'4 1.51 2.28 [1.1651 5 [1.[1151 2.51 15.5 [1.12 4 [1.015 3.51 12.52 [1.1I5 5 [1.0116 4.51 2[1.3=1 [1.122 Finally. to ﬁnd the yariance of the probability disuihutiort sum the (x  p.) 2I" [11) products
found in the previous step. 52: Z (x—1931191):odes—oosa—o_1119—o_12—o_1s—o_1aa=o_1aa Thus the variance of the probability disuihulion is approximately [1.8. 1
[n the game of roulette. a player can plate a $15 bet on the number 55 and have a E probability of winning. If the metal hall lands on 33. the player wins $525. Otherwise. the casino
takes the player's $15. 1iWhat is the expected value of the game to the player? If you played the
game 1[1[1[1 times how much would you expect to lose? 1: Pﬂx) x ' Pﬂx}
$525ﬁwim1ing) [1.0263115 $118159
— $15 (losing) [1.9131584 — 14.:1osao I‘Cow compute E xPﬂx}.
13.3159—[— Isl[10526013 — [1.29 Thus the player's expected yalue is approximately — $0.29. A binomial experiment is a probability experiment that satisﬁes the follots'ing conditions. 1. The experiment is repeated for a ﬁxed number oftriala where each dial is independent of the
other dials. 2. There are only two possible outcomes of interest for each dial. The outc ornes can be
classiﬁed as a success (3) or as a failure (F). 3. The probability of a success HE) is the same for each dial. 4. The random variable x counts the number of successful dials. Find the mean. variance and standard deviation of the binomial dismbution with the given values
ofn and p. n = 120. p = 0.4
The mean of a binomial dismbution is given bf; 
ll 2 11? where n is the number of times a dial is repeated and p is the probabilitf; of success in a
single dial. Find the mean bf; substituling n = 120 and p = [L4 into the formula for the mean.
u=123  [Lil=43 The variance of a binomial dismbution is given bf; .l
4 D" =ﬂpq where q is the probabilitf; of failure in a single dial (q: 1 p). Find the variance bf; subintuIing n = 120. p = [t4 and q = or» into the formula for the variant e.
53=1ao  o4  o.o=2as
The standard deviation of a binomial dismbution is given bf; the following formula. 0': ﬂpq Find the standard deviation bf; subintuIing n = 120. p = [L4 and q = on into the formula for the
standard deviation. 5: U 120  [L4  on £55.4 29% of adults sav cashews are their favorite kind of nut. You randomlv select 12 adults and a!
each to name his or her favorite nut. Find the probabilitv that the number who sav cashews arr their favorite nut is (a) exactlv three. (b) at least four. and (c) at most two. If convenient use
technologv to ﬁnd the probabilities. Ln a binomial experiment the probabilitv of exacdv x successes in n trials is given bv the
following formula. n! Pﬁxl= C p“o“"‘=—p"q“"‘
“ x [nx)!x! (a) Using 11 = 12. p = I129. q = EL? 1. and x = 3. ﬁnd the probabilitv that exactlv three sav
cashews are their favorite nut. PG}: negroaoﬂovug
:5 oars (b) The probabilitv that at least four sav that cashews are their favorite nut is the sum of
PG). H5). H6}. . P02). However. one can also solve this problem bv using the
complement rule. P(xa4)=1—P(x<4) To ﬁnd the probability that x is less than four__ ﬁnd the sum of Pﬂﬂl Pﬂll PG} and PG). 10(0): 12:::3(0_20}'3(0_?1)12
:5 0.010410 10(1): 12e1(0_20)1(0_?1)11
:5 0.000401 10(2): 12:::2(0_20)3(0_?1)1'3
s: 0.100000 10(3): 12:3(0_20)3(0_?1)9
:5 0.240004 Now ﬁnd the probability that less than four sag; eashews are their favorite nut. P111614): P(U)—P(1)—P(2)—P(3)
= 0.010410—0000431—0.100000—0.240004
= 0.523531 Finalh: ﬁnd the probability that at least four sag: that eashews are their favorite nut. P(x24j = 1Pﬁxi4)
= 1—0.523531
=5 [Lil7'6 (e) To ﬁnd the probability that x is at most two: ﬁnd the sum ofP[ﬂ]__ PU} and PG}. P(x£2 = P[0)—P(1)—P(2)
= 0.010410—0000431—0100000
s: 0200 Ihe graph below shows the results of a
su1vev of drivers who were asked to name
the most annoying habit of other drivers. You
randomlv select six people who participated in
the su1vev and ask each one of them to name
the most annoying habit of other drivers. Let
x represent the number who named talking
on cell phones as the most annoying habit.
Complete parts (a) through (c). :1;
Talking on 131111 phone 35°F:
Slaw drim in fast lane 22%
Pull}: drivers who tailgat a 13%
1Veasrethru‘ugh trai'ie 13°F: 1 1‘?'«'n Dn—ro ad pet peeve 5 Use the binomial probabilitv formula given
below to ﬁnd the probabilities for the
possible values of x. Pei) : ﬂcxpan—x n!
_ x nx
— [II—x)!le q Find the probabilitv for x = El. :5! [oasﬂomﬁ'ﬂ
(6D)!ﬂ!
:5 ones PUD) = Consuuct a binomial dismbulzion for the
possible values of 3c rounding to the
nearest thousandth. P (x)
El'ﬂlﬁE"
[LE 32
0.325
0.245
U. 103
0.323
0.302 DinlhUJIUI—iclibd h) Find the prphahilr'ts' that exactly two
people wﬂl narue "talking on cell phones." When the driver narues talking on cell
phones as the must arumsirlg hahit it is
classiﬁed as a success. Ihus ﬁnd P[2}__ the
probability pr successes in 6 uials. Use the part (a) binomial disuihuticrn to ﬁnd
the probability. P(2)=t+_326 c) Find the prphahﬂity that at least ﬁve
people will name "talking on cell phones." The prphahilr'ts' of 5 or more successes is the sum of the probability of 5 successes
and the probability of IS successes. Plﬁxas) = Fiﬁ—P03)
= [LUBE[LUKE Thus P(x 2 s] = ates. The following histograms each represent
binomial distributions. Each dismbulion has
the same number ofuials n but different probabilicl' 'es of success p. [EJPH 0.? E} 0.? E} 0. 0. 0.5 0.5 0. 0. 0. 0. 0.2 0.2 0.1 _ 0.1 _
D 01234 i‘ D 01234 3‘
'3ng I” l} 0. 0.5 0. 0. 0.2 0.1 _ D 01234 }‘ 1234 Using n=4. p= [L5. and q: [*5 obtain the
following probability disuibulion. x [l 1 2 3 4
P 0.003 [L250 [L37'5 [LE'U [L063 Ihe corresponding histogram is shown
below. Match p = [+5. p = [+5. p = [Ll with the
correct graph. To consuuct the binomial dismbulion. ﬁnd
the probability for each yalue of :1. Recall that the probability of exactly 3:
successes in n uials is given by the formula
below. P (x) = anp‘qn'” L'sing n =4. p= [P.El. and q = 0.7". obtain the
following probability dismbu‘cion. x [l 1 2 3 4
P [l.24[l [L412 [[2155 0.030 [L008 Now. graph the probability dismbulion
using a histogram as shown below. Finally using n =4__ p= [Ll and :1: [Li
obtain the following prohahﬂity distribution. x U 1 2 3 4
P [LUBE [LU"Jo 0.265 [L412 0.240 Graph the probability distribution.
PEX}
? I: III.
El.
El.
III.
III.
III.
El. 01234 Thu; conclude that (a) graph corrosponds to p = [Li Eh} graph corrosponds to p = [Li and (c) graph corresponds to p = [Ll ...
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 Spring '09
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 Statistics, Probability, Probability theory, El III.583

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