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midterm2Key

# midterm2Key - Statistics 20 Midterm 2 Spring 2007 Name\L Q...

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Unformatted text preview: Statistics 20: Midterm 2. Spring 2007. Name: \L Q/ Student ID: Instructions You have untii 3:30pm to ﬁnish this test. You may use one sheet of notes and a calculator. Please show your work and circle your answers to each question. 1. Suppose X and Y are independent random variables such that XN ~Bin(5 1/2) and Y~ Bin(10, 1/3). Find sd(X Y). 25~<i()( ~27) lwi X-WafY We; 2. Suppose I randomly pick 5 cards with replacement from a standard deck of cards. What is the probability that three of the cards are Kings? ,ﬂ;% I403; NBE(§,%‘L (P( ii 56 14230:) f <2 2%? :3” 0.003? {11:31 -‘ 3. Mr. Smith’s gardener is not depe2ndableg'the probability that he will forget to water the rosebush during Smith’s absence is g. The rosebush' 1s in questionable condition anyhow; if watered the probability of its withering IS -2-, but if it is not watered, the probability of its withering is %. Upon returning, Smith ﬁnds that the rosebush has withered. What 1s the probability that the gardener did not water the rosebush? 49170111 wary-172), , cpwaev) V3 ‘ ?( N0 Wale-r (willed; w ;C3/\>(2 3 W 17(wll’L-A I No WRAMM waleA +lem-Ml 111M)? Width“ 34. ‘93 ,1 4. Suppose that a random variable has the following density 1/3 for —-1 < a: < 2 otherwise. f(w)= {0 Find the probability that this random variable will be less than 0. Find the variance of this random variable. 5. 7 The probability that a man will. hit a target is g. If he shoots at the target until he hits it for the ﬁrst time, ﬁnd the probability that it will take him 5 shots to hit the target. 6"" Sinis- t-e are :1 4mg 2,. m - HM w (W ) 6. The lengths of full—grown scorpions of a certain variety have a mean of 1.96 inches and a standard deviation of 0.08 inches. Assuming that the distribution of these lengths is normal, ﬁnd the probability that a full-grown scorpion has a length of 2.20 inches or more. Also find the probability that a full—grown scorpion has a length of at least 1.8 inches. L FIN/mt, @0713”) 7 a 2221mm : _ g" P(L’Z'Z): P( 050% . 000’? > ZNA/[é‘i‘] ‘Pimﬂt W22” W 45-3-3} (3617729 7. Let X1,. .,X20 be independent copies of a random variable that is normally dis tributed with expected value 55 and standard deviation 10 Let Xm Jifoﬂﬁm. Find P(.X > 56). X AMBER '0 > E32: 51;? §9Q(§)3 ma zip/Mos) 8. T03 11110tme3 .eLtYr cordithenu nur‘nber 0efhds nersosthﬁth and letXr ordth:0 numerb o'fhe as1d tnhe 13311615513363.0111th P1(Y= 2)B.11-dP(X:a1). Are X311in 11de ned nt?Sh0wy0 easo ning? WBM '4) x0334 1/3) I P(Y'gst 261.4! L2>é_ ' a :- 0131(‘1 VD“? iglG P '" ‘0705f3f 7 I —o - -0 ‘ ‘ I I ' / ﬂénggm :fé X3: are. View" ‘WM (102) dY~ Bi n(--1,0.4) A130 suppose P(X= 0&Y= 0)= 048 HX=oeygn=om " P(X=1&Y=0)=0.12 P(X=1&Y=1)m0.08 Coueemptth O_fX+2Y MIX UP?) (03%“) 01132 WK [03PM 13- (~:(03>00 032 KM JL '7? . MAN 3 (021(06') on) I -3 J PX,.[W )L;\r £02.) 01‘):— 003 ‘ \$W ...
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