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Name &"1 Worksheet/note sheet! N - (71,5) 1. On the line below, draw J'[ ~5& to I 7/ (etc 10 81 \Sio -(X-I')' P(X)- 1 e 2a' /2no2 cJ.... 16 40- 52 64 76- 88 2. Consider the normal curves above. a) Find the line of symmetry b) Identify the mean(s) '),;l.. c) What are the points ofinflection(s)? :.~ ~ ~) 70 d) Which normal curve has the greatest mean? Sa- v>0e) Which normal curve has the greatest standard deviation? 0-... GO I 3. Given that N - (100,15), estimate the probability that a randomly chosen adult has an IQ between 85 and 145. Draw at the normal curve. 4. The uniform Density Function: / 1 b-a " 5. Which of the following \5a function? a)f(x) b) f(x) c) r.5 O<x<l lO otherwise =j -'"9> 1 a) Let X - Unijorm(5,12). Then the P(X < 8)is: 'if legitimate uniform probabilitf density I;). .5 3 < x < 5 = {0 f(X)=je3 otherwise l<X<l~ o otherwise .25 d) f(x) = {0 -2< x < 2 otherwise . f) All of these are uniform probability density functions 6. A uniform distribution is defmed on the interval [-1,1}. a) Sketch a graph of the probability density function. b) Defme the rule of the probability function; that is, complete f(x) -I.(X:<' l .l ={ ~ +to-v , () 'b uJ I sec) Verify that your rule in part b satisfi,es the conditions for a continuous probability density function. ~;;2. ') -: t d) Calculate the probability P(O < X < 1). Show a sketch to support your -1 calculation. \ ' J- - (i '\ )--~ ----> II. 'd 5.1 Density Functions '1. rex) Consider the function in the sketch below defined on the interval [-1,1]. Note that the height of the triangle is 1. Which of the following statements is true regarding this function? --- --- a) j{x) could not be a continuous probability density function. 2 b) P(-l:Sx:S.5)=) c) P( -1 :S x :S .5) @) P(-l:Sx S.5) = e) P(-lSx:S.5)=- a. a) = -3 4 i 11 12 Which of the following graphs could not be a density function? b) 1 o c) 2 2 Q 1 ----- ~-----_ .._- -------------- -------- e) All could be density functions Warm-Up 11/19 Statistics I. Assume the mean annual consumption of peanuts is normally distributed, with f.l = 5.9. pounds per person and (7 = 1.8 pounds. Estimate the probability that a person consumes between 2.3 pounds and 9.5 pound in a year. rp( ~_ 3 <:. 1<-«:. q,,) ;;2.),5','1 Cf.)--'J!j. 1.'6":::--;< I.%' ~ - P ( -d-~ = c<:::..:2.) (),Q5'4S- 2. The monthly utility bills in a certain city are normally distributed, with a mean of $100 and a standard deviation of$12. A utility bill is randomly selected. a) Find the probability that the utility bill is less than $80. b) Find the probability that the utility bill is between $80 and $115. c) Find the probability that the utility bill is more than $115. d,a..) p( X c <g6) ~ -I. ~~ I -/<sD Ii)- - I ('> - JOO ::- }.3-) :: . rLf 7 I d- n (J{VY)a.(cd /-( ---- II:)' J~ I ?- _ -/.(, 7, I.~-T) YLf- 7 e>;7" I,;)..)r YL(r{W\.d. C C ci t-- ( I. ;;LS',()O ) IDt.o I o. fb CS-;o [ (}JCV tt1- <-<-f a...flu- fj _;;2 I. Given the-'following uniform density distribution: 2 18 «) a) Find the height Ir-;l-/CIJ - 1,,_ V\- ..L I C. b) Write the piece-wise function that describes the picture b) c) Find the P(4 < x < 9) . file) ~ ) '1-4;5 c ~n'5.., 1/ ( -e ..:( G-iO't..V\.. b) ~) c 02. 13) SLcvL: P (~ ) -,. P( JY1) -I, ~6; g- L Z:- L.. t ).<Ktt- It) 11# ~/rr~Vfl.)Ls.e '5(+,.):. ~ 2. Find the indicated area under the stand~rld normal curve: a) To the left ofz = 2.13 -:. . b) Totherightofz=-1.341-:. .'1/00 c) Between z = -1.268 and z = 1.496 -.:. ~~O~ d) P(z < -1.12 or z > 1.48) -:. • ~cm'$ e) P(z> 1.98) :. O. 6~'3q c0 ~..L I. '-19&) . 5.2 Practice Name Kwr 1 - 4 find the indicated probability using the standard normal distribution. Sketch the graph!! ! 2. P(z > -0.25) 1. P(z < 0.45) 3. P(0.45 < z < 1.29) Q-, -;;<-;) 7-'6-;_\ca.ic . 5. An automobile tire brand has a life expectancy that is normally distributed, with a' mean life of 30,000 miles and a standard deviation of 2500 miles. The life spans of four tires selected at random are 35,150 miles 24,750 mile's, 30,000 miles. a) Find the z-scores that correspond to each value ?'S",i~- ~(),~3'O __ .;2C0 ,?4, lS'D-?O,~ -,:), I 3~:._30,C1lC>- 0 ~)&S. ~ ; b) Determine whether any of the values are unusual. t')(J ~,,-c.l ?I..f,,<)(j u...u.. 35' , kJ,,-H\ ;;>,oD Ll. V\ 1.-' \l... 6.-l 6. The level of cholesterol in the blood is important because high cholesterol levels may increase the risk of heart disease. The distribution of blood cholesterol levels in a large population of people of the same age and sex is roughly normal. For l4-year-old boys, the mean is /d = 170 milligrams of cholesterol per deciliter of blood (mg/dl) and the standard deviation is (T = 30 mg/d. Levels above 240 mgldl may require medical attention. D a) What percent of l4-year-old boys have more than 240mg/dl Of Y ChOI~~~I:'n~ ( ..J ) ;;'L(IJ lo[b I 30 (70, = ;<, 3)3 .k( X, '1;Jlf:O) ~) (~ __ . : .~~.q ~_) ,0 l.:::-! 30) I 0/0 ~._. __.__:1. J b) What percent of l4-year-old boys have blood cholesterol between 170 and 240 mgldl? ) - J.:1J)_t .~O 70 __ 0 - L, r :J ( l-W.L.. >< <:....;;J-{() ~ 11'10 L-\- .2) ' 3) Name 5.3 Practice 1. Using the normal distribution of SAT math scores which have j.1 = 514 _ - I J;;/ Lj AI\ II"")" G' = 1l~.1find: P(XLljuD ') ':"J - L/ 5;,) ~ a) What percent of the SAT math scores arelessJ:b.an 500? 56lJ - 5(I-f and "---~ .::;::. 4 r:;"I.:;, -1""-' -I")-~ b) If 1500 SAT math scores are randomly selected, about how many would you expect to be ;;;J~"I greater than 600? I.o?<fD) lOGO ~ 5/4 : \7 (p . (c / , 71,;,) '::.' /1"'3 ,'J;). 't (J <)@.)) ;: 7'("'" P p: ~ '3 ~5i:"_~I_"L.- , . ~0~~~ 2. Given the normal distribution of women's total cholesterol levels with mean of 228 and standard'" deviation of 4;?&. milligrams per deciliter, find: a) What percent of the women have a total cholesterol level less than 239 milligrams per deciliter of blood? cd.. f (~J>d3,:),)-0') V\.~ ~Jq-JO>!- -:. 0,25"'1 ---- Ll3.~ - [0 ,l)qti I \ b) If 200 U.s. women in the 50-59 age group are randomly selected, about how many would yoLieXjlect to have a total cholesterol level greater than 200 milligrams per deciliter of T' (X ." ~) blood? ') -P("2."";>:-;(q~,clp:. ,7~r4> ?42l- O),;(~_:-,103'1 . /..13.8 -:> ~ (?~_):.147:7g [El t~~J 3. Given the normal distribution of monkey weights in which the mean is 15 pounds and the standard deviation is 3 pounds, find: ' a) What percent of the monkeys have a weight that is greater than 20 pounds? L~.75-~ 'Plx';>;;>D) 1- , 'i D.S,-', ?'.,q'):;(.) - ,?:9-t.C;-:: ~ /JI-{7:;? . 1.'0(,,1 ;, - , b) If 50 rhesus monkeys were randomly selected, about how many would you expect to weigh .,...-\ .._~s\s I? - l'l_;. ~?- than 12 pounds? D 4. l- :3. - I =t-:. ' ,I \'67 Ig;) ~7. '1) ~ Using the normal distribution of utility bills for which the mean is $100 and the standard deviation is $12, find: a) ~.ercent ----1, ~(/)ol \ "____.. \. of the utility biUs are more than $125? . -p(x. '> I~~) . I&. 5" - iOJ ;. ,0-- b) If ~tility ..!;2,? "'j (;, \ J <----- :1..0 g bills were randomly selected, about how many would you expect to be less than (X t.qo") I qa - -~l--f--':'::- 100, -, ::10:< (~):;. loa Co ¥,~ Iu-- / ~t 5.4 Extra Practice I Y\Vl'lqJl'Ittc. ,,0 ) 1) Find the z-score that corresponds to each of the following percentiles a) P.o b. Second quartile 50"l. c. P99 -l,~'l d. 0.2090 ;X. 30?~ 0 - -,g091 2. The monthly utility bills in a city have a mean ~ and a standard deviation of $8. Find the x-values that correspond to z-scores of-o.75, 4.29, -1.82. What can you .Pl:. 70 "G:-~O)-I 'Ci':~~,8 ~olooo?-(£70 be Co Lj I t-:)). L I ~ lN t-k.e-'t'\ '-eGL--.-.l D q .) ?- L.o it. \::J (11.J{ +t'\..L fVU.Cu,\' LL Y\-k S L-LA-1 I 3. The braking distances of a sample ofFord F-150s are normally distriJmted. On a dry surface, the mean braking distance was lS,KJeet and the standard deviation was 6.51 feet. What is the longest braking distance on a dry surface one of these Ford F-150s could have and still be in the best 1%? ,...M::. l 5 ~ ~ -:.. v S I 17 ( X I. I ~o ) ( . . /rIV(\()Y"MCO I'): -~,,;I - ;;to '3 ;;t (, ;:. X.- (p 1st . 1.>. I:"':.--_-T ~('43..f+) A': l(j;;l.¥/o 4. The length oftime employees have worked at a corporation is normally distributed, with a mean of 11.2 years and a standard deviation of 2.1 years. In a company cutback, the lowest 10% in seniority are laid off. What is the maximum length of time an employee could have worked and still be laid off? '2; -,,,U1\.a'l~ (,10)'" _I-;~_X-I\,~ ,0'- -- X:: -\. 01,\ rSI a ~8 'P( X~ 10) 5. In a survey of men in the U.S. (ages 20-29), the me~.2 ~ standard deviation of2.9 inches. a) What height represents the 90th percentile? inches with a II ~y;l~ y- (.,q. .2 £:.. IJ'( \ I v\ II I'\O{..(. 'IOJ d- , "\ [X~1~.qIJ b) What height represents the first quartile? \' rl V Y\(Jy'\¥I(,,)5)~ -D. 10 74)~ - 0 <a 745": X. (,,'lJ . , ~.9 6. A college requires applicants to have an ACT~e The ACT scores are normally distributed, with ~ ~ top 12% of all test scores. and a standard deviation of ~I:-O'). !" . a) Find the lo:vest test score that a student could get and still meet)l;le college's requirement. ) (\ 11 r\.IJ, l'\ ( J ~ , I J..) " r1V (\(JY iVl (. r~);:. q X -;f I 1,175:.. I, II 5' "';.(d~.s.;l. Y'I b) If 1500 students were randomly selected, how many would be expected to have a test score that would meet the college's requirement? 'c) How does the answer to part (a) change if the colh;,ge decides to agcept the top 18% of all test scores? LVI v f\(fY'1I1 ( 1- I /'() - (}CjIS' <t D,g/54:. '1- y:.dS.30 ~I 11v- s ~ ~, l (,(J1'1( I Cf'..tM" fo 7. The lengths of individual shellfish in a population of 10,000 shellfish are approximately normally distributed with mean I 0 centimet~rs and standard deviation 0.2 centimeter. What is the shortest interval that contains approximately 3,QQ.Q shellfish lengths? i i}- ,L\O (fll?o -'tOrflc:(o ). y I fli_: , . 0 . L V1 tr Y) &r m(. "30);:' ' -,5 ;).L/l/ -:, 5.;;J.4L/ -:..)<- 10_ o,~ x:..CJ;%q.s o"b ;;;. ()A :.. 30'/0 {O CJ: O. ~ :;;{~ 36. Name \-...- Pg_ Candy Bar Weights Suppose that the wrapper of a candy bar lists its weight as 8 ounces. The actual weights of individual candy bars naturally vary to some extent, however. Suppose that these actual weights vary according to a normal distribution with mean fJ = 8.3 ounces and standard deviation (}"= 0.125 ounces. a) What proportion of the candy bars weigh less than the advertised 8 ounces? « -?.3 p( XL- g), - O,,~ ;; -.;>.'-1~ ,~--- YLdVrYLlL{uH ( --exIJ, -;;J. -- o03~ f:;; -.v{ ) ' '-l - ~~:2"7o~ What proportion of the candy bars weigh more than 8.5 ounces? "•.5 - ~.3 .=. Yl ()r ). Co . YI1.a..-lcd f O.l~::'- {). Co I 00) ().O ') ( 5, '-i g 5~oj c) What is the weight such that only 1 candy bar in 1000 weighs less than that amount? p-_.J-;; 60 I ?.(. •aJ I l<SlID • - 3- 0 q: X - c).I;>'S 'if, 3 ~---- (X -- 7. 9 <.------ d) If the manufacture)' wants to adjust the production process so that only 1 candy bar in 1000 weighs less than the advertised weight, what should the mean of the actual weights be (assuming that the standard deviation of the weights remain .125 ounces)? 3 ~ .d'1~ D. ,1<) e) If the manufacturer wants to adjust the production process so that the mean remains at 8.3 ounces but only 1 candy bar in 1000 weighs less than the advertised weight, how small does the standard deviation of the weights need to be? - '3.0'1 . ~.3 U ~ 'J Name Practice 5,6 Normal Approximations to Binomial Distributions, '''-.-/ I, Match the binomial probability P(x < 23) with the correct statement @P(there are fewer than 23 successes) B) P(there are at most 23 successes) C) P(there are more than 23 successes) D) P(there at least 23 successes 2, Ten percent of the population is left-handed, A class of 100 students is selected, Convert the binomial probability P(x > 12) to a normal probability by using the correction for continuity, ,'-:::;1, ~ A) P(x < 11.5) B) P(x ,.,; 125) ~p (x ~ 11.5) (t~ \j9!(X ;12.5)Y + 3, Only 8% of people in the U,S, feel that the nation is more patriotic today than it was decades ago, You randomly select 70 people in the US, and ask each ifhe or she feels the nation is ,more patriotic today than itwas decades ago, What is the probability that more than 10 respond yes? ' a. Determine whether you can use the normal distribution to approximate the binomial variable rOCCR) -:.S.0 70 C. y~) ::-04 L( b. Find the mean p. and the standard deviation a for the distribution. J\:()f (5-:. {npi : c. Apply the appropriate continuity c~ion lOS - 5; (, I :::~,I\p :1,0),7 J. d. Find the corresponding z-score. VlOfrVlq,.Lcd-f (:;U fa, and sketch a graph. (oG.') _ /\ /\til':. ;>.t(o ",,,</tIt e. Find the probability to the.left of z. <' () (') L2f The failure rate in a statistics class is 20%. In a class of 30 students, find the probability that exactly We students will fail. Use the normal distribution to approximate the binomial distribution. p(4S '- )(L..\.~) 4.'5- .[;-:--,&Ztq d. ---r!1.~ I (\ P(-- <0 ~,.:.C;-S-f.o :C-;:.IiO J., (Cf ,,-:4 _(=1 )(io) v-t ":. () ::-e,., (\ q o ~ I & ~ V\P2 ~ f/}O) (;e}i'l ,:jy () 1 ~').,11 Name_~_-_______ _ IB/AP Statistics Quiz B (5 pts each question) When solving probability questions, be sure to list: a) probability statement, b) the graph, c) z-score or equation, d) conclusion. ALL problems should have a graph except for #6 and #9) 1. The times for completing one circuit of a bicycle course are normally distributed with a mean of 72.5 minutes and a standard deviation of 6.5 minutes. An association wants to sponsor a race but only wants the top 25% of riders included. In a trial run, what should be the cutoff time? In a trial run, what should be the cutoff time? (Be careful- you want the best times for the top 25%!) lV\Vy)6YI"1(.~S-)= - ,/":'745 2. A college class uses a normal curve to distribute grades. On the last exam, the mean was 85'Yowith a standard deviation of 6%. If you want to be in the top 10% (because this is an A), what score must you get? tAn LJ t'\(JY1"Y\ (, '10) ~ ),0'-'6 _ I J-~: )( - ~5 G X -- CJ~."7 3. r>eople use their computer~ an average of 2.4 years before upgrading to a new machine. The standard deviation is .5 owner is selected at random, find the probability that he or she will ~ year. If a computer . use it for less than 2 years before upgrading. Assume the variable x is normally distributed. p('>(<:..~) p:.. ,;;111'1 -,6 4. Compare the scores: a score of220 on a test with a mean of200 and a standard deviation of21, and a score of 90 on a test with a mean of 80 and a standar eviation-{)f-S :which_score is better? Why? I, J S-- qo - )0 "6 R )C-L;Ye 5. Find the z score that corresp0l)ds to the 20th percentile. L Vl iJ nUy rl, (, ~O) .l.L) " -i lr-) cz D "1 6. Let X - Uniform(3, 10), a. Sketch the uniform density function ~J~I U'I,",L ;, '0 b. Find the height /0 - '3-; 7 c. Write the piece-wise function that describes the picture '1 3<-)CC ~o ,0 D ~yl.<.J l s--t.. d. Find the P(4 < x < 7) W.-LJ-J '0 7. IQ test scores are normally distributed with a mean of 100 and a standard deviation of 15. An individual's IQ score is found to ~. Find the z-score corresponding to this value. .. ~ jO-/::::.;. IS- r ,CaVa! ] c.M.. 8. The average sales price of an existing single-family house in the United States is $125,700. You randomly select 12 single-family houses. What is the probability that the mean sales price is more that $100,000? Assume that the sales prices are normally distributed with a standard deviation of PCX ., $26,000. IOO,ooD) I()O I crm:> - . cdf nO{~) I(l s" ,,<17) "(/Q _ ;2 (p,<J6l) , 0 0 f.Cj~g /~) =- . u. v0 3~iLT ~ r-3.N <?•• JI'"\ 9. Find the sum of the areas under the standard normal curve to the left of z = -1.25 and to the right of z = 1.25. ... ~ - p.),. -t.~'j . ~. 0 r (p +- .,J,. u)(.. 011. t ~ Of•• 5.5 Central Limit Theorem I. The number of eggs a female housefly lays during her lifetime is normally distributed, with a mean of 800 eggs and a standard deviation of 100 eggs. Random samples of size 15 are drawn from this population and the mean of each sample is determined. a) Can the Central Limit be used to find the mean and standard error of the mean? Explain YES, BECAUSE IT IS NORMALLY DISTRIBUTED b) Find the mean and standard error. MEAN= 800; STANDARD ERROR= 1001.tiS = 25.82 2. During a certain week the mean price of bread in California was J1 = $1.722 per loaf. A random sample of 38 stores is drawn from this population. What is the probability that x, the mean price for the sample, was between $1.727 and $1.7317 Assume 0" = $0.049. a) Can the Central Limit be used to find the mean and standard error of the mean? Explain YES, THE SAMPLE SIZE IS GREATER THAN 30. MEAN = $1.722; STANDARD ERROR=0.049 I -J38 = .0079 b) Find the z-scores Z = 1.727 -1.722 - .6329 Z = 1.737 -1.722 1.899 .0079 .0079 c) Sketch the problem d) Find the probability Normalcdf(.6329, 1.899) = .235 or 23.5% 3. The mean height of men in the u.S. (ages 20-29) is J1 = 69.2 inches. Ifa random sample of60 men in this age group is selected, what is the probability that x, the mean height for the sample, is greater than 70 inches? Assume 0" = 2.9 inches. 70-69.2 Z = .J6O = 2.14 P(x > 70) = normalcdf(2.14, 1000) = .0162 or 1.62% 2.91 60 4. A manufacturer claims that the mean weight of its ice cream cartons is 10 ounces with a standard deviation of 0.5 ounces. Assume the weights are normally distributed. You test 25 cartons and find their mean weight is 10.15 ounces. a) Assuming the manufacturer's claim is correct, what is the probability the mean of the sample is 10.I 5 ounces or more? P(x> 10.15) z= 10.15-10 0.5/J25 1.5 Probability: normalcdf(1.5, 1000)=.0668 or6.68% b) Using your answer from part (a), what do you think of the manufacture's claim? The claim is accurate. c) Would it be unusual to have an individual carton with a weight of 10.15 ounces? Why or why not? Not unusual, assuming the manufacturer's claim is true, because 10.15 is less than 20" away from the mean for an individual ice cream carton: Name FRAPPY FRIDAY! 1. A local radio station plays 40 rock-and-roll songs during each 4-hour show. The program director at the station needs to know the total amount of airtime for the 40 songs so that time can be programmed during the show for news and advertisements. The distribution of the lengths of rock-and-roll songs, in minutes, is roughly symmetric with a mean length of 3.9 minutes and a standard deviation of 1.1 minutes. ~:. Des' sampling distribution of the sample mean song lengths to random - s mples of 40 ock-and-roll songs. ov'"iC'e +he ~lcC i; la..-r'-qe (n-;Jjo) +hl c..-eV\~o...l a..p(.)lr~J~I\A~.e5CJ-W\pte dl's-IYrhi.\...J.,lM ts Sa...m.l"le IIM,..t +V1-lQ(tWl {A PP (HI Mo.-ok I~- Y\orrY\.'t/l~ d i ~fyi !ot.<.kd. i'h~ <...M:)( ::? q IN\' A ~-k5 a...r0 '+-h~ S+a..r.d a..rd de W\..lA..v1 1/I a.+, ~ . t5 . b. If the program manager schedules 80 minutes of news and advertisements for the 4-hour (240 minute) show, only 160 minutes are available for music. Approximately what is the probability that the total amount of time needed to play 40 randomly selected rock-and-roll songs exceeds the available airtime? j"t'-t t?rzrbl(t-j I(~ +h.a.t ~he frJfo.! ~1I'+(M.e ()+ 40 ro..",d~11 sel.ebkd. S~J ek(-e~J5 -t-ke I&OMLI\.~4eJ rJ e~urva.i4l1\.t -+0 ~~ "pvo!oa.I.:>III~ e-f. +l1~ sa...~ple (V)ea.-"", !-e.rt41V1 (Sf +he.. 40 SOV\.Oj~ t 5 a v~-kr ~f(.Yl J fJO ~ 1oV\,'I\J .=: Lf . , v L{ 0 e 1:/:. of ~""4) "P (x '7 L./.o) ~ p( ~ '7 4.00.174 - ,.q \ ) :: P (-i- ( 0.5" 7 ) - - -,' ~-- \ I _ _ rvt.Vry 0, ;l8'••. ....., , Su»>