ࡱ> LNK!` bjbj\\ 4>>H,!!!!d8"D,5"j&j&j&j&j&j&j&4444444$6h9^4(j&j&((4j&j&}5---(j&j&4-(4--V24@4j&|" nċ!*r4 4505~4 {9-.{94{94j&&^-L'L'Bj&j&j&44- j&j&j&5((((,,,,,,,,, CHAPTER 9 REVIEW 1) The average adult has completed an average of 11.25 years of education with a standard deviation of 1.75 years. A random sample of 90 adults in obtained. What is the probability that the sample will have a mean (a) greater than 11.5 years? (b) between 11 and 11.5 years? ANS: The sampling distribution of x-bar has  EMBED Equation.3  Because the sample size is large (n=90), the central limit theorem tells us that large sample techniques are appropriate. Accordingly, (a) P( EMBED Equation.3 or normalcdf(11.5, 1E99, 11.25, 0.184) (b) normalcdf(11, 11.5, 11.25, 0.184) = 0.8258 2) Over the years, the scores on the final exam for AP Calculus have been normally distributed with a mean of 82 and a standard deviation of 6. The instructor thought that this years class was quite dull and, in fact, they only averaged 79 on their final. Assuming that this class is a random sample of 32 students from AP Calculus, what is the probability that the average score on the final for this class is no more than 79? Do you think the instructor was right? ANS: P( EMBED Equation.3 or normalcdf(-1E99, 79, 82, 1.06) If this group really were typical, there is less than a 1% chance of getting an average this low by random chance alone. That seems unlikely, so we have some pretty good evidence that the instructor was correct. 3) Harold fails to study for his statistical final. The final has 100 multiple choice questions, each with 5 choices. Harold has no choice but to guess randomly at all 100 questions. What is the probability that Harold will get at least 30% on the test? ANS: Because 100(0.2) and 100(0.8) are both greater than 10, we can use the normal approximation to the sampling distribution of p-hat. Because p = 0.2, the sampling distribution of p-hat has  = 0.2 and EMBED Equation.3  Therefore, P( EMBED Equation.3 or normalcdf(0.3, 1E99, 0.2, 0.04) Harold should have studied. 4) What is the probability that a sample of size 35 drawn from a population with mean 65 and standard deviation 6 will have a mean less than 64? ANS: The sample size is large enough that we can use large-sample procedures. Hence,  EMBED Equation.3 or normalcdf(-1E99, 64, 65, 1.014) 5) Approximately 10% of the population of the United States is known to have blood type B. If this is correct, what is the probability that between 11% and 15% of a random sample of 50 adults will have type B blood? ANS: If p = 0.10,  EMBED Equation.3  Then,  EMBED Equation.3 or normalcdf(0.11, 0.15, 0.10, 0.042) 6) A tire manufacturer claims that his tired will last 40,000 miles with a standard deviation of 5000 miles. (a) Assuming that the claim is true, describe the sampling distribution of the mean lifetime of a random sample of 160 tires. Describe means discuss center, spread, and shape. (b) What is the probability that the mean lifetime of the sample of 160 tires will be less than 39,000 miles? Interpret the probability in terms of the truth of the manufacturers claim. ANS:  EMBED Equation.3 miles (a) With n = 160, the sampling distribution of x-bar will be approximately normally distributed with mean equal to 40,000 miles and standard deviation 395.28 miles. (b)  EMBED Equation.3 or normalcdf(-1E99, 39000, 40000, 395.28) If the manufacturer is correct, there is only about a 0.6% chance of getting an average this low or lower. That makes is unlikely to be just a chance occurrence and we should have some doubts about the manufacturers claim. 7) Crabs off the coast of Northern California have a mean weight of 2 lbs. with a standard deviation of 5 oz. A large trap captures 35 crabs. (a) Describe the sampling distribution for the average weight of a random sample of 35 crabs taken from this population. (b) What would the mean weight of a sample of 35 crabs have to be in order to be in the top 10% of all such samples? ANS:  EMBED Equation.3  (a) With samples of size 35, the central limit theorem tells us that the sampling distribution of x-bar is approximately normal with mean 32 oz. and standard deviation 0.338 oz. (b) In order for x-bar to be in the top 10% of samples, it would have to be at the 90th percentile, which tells us that its z-score is 1.28. Hence,  EMBED Equation.3  Solving, we have x-bar = 32.43 oz. A crab would have to weigh at least 32.43 oz., or about 2 lb. 7 oz., to be in the top 10% of samples of this size. 8) A certain type of light bulb is advertised to have an average life of 1200 hours. If, in fact, light bulbs of this type only average 1185 hours with a standard deviation of 80 hours, what is the probability that a sample of 100 bulbs will have an average life of at least 1200 hours? ANS: EMBED Equation.3   EMBED Equation.3 or normalcdf(1200, 1E99, 1185, 8) 9) Opinion polls in 2002 showed that about 70% of the population had a favorable opinion of President Bush. That same year, a simple random sample of 600 adults living in the San Francisco Bay Area showed found only 65% that had a favorable opinion of President Bush. What is the probability of getting a rating of 65% or less in a random sample of this size if the true proportion in the population was 0.70? ANS: If p = 0.70,  EMBED Equation.3  Then,  EMBED Equation.3 or normalcdf(-1E99, 0.65, 0.70, 0.019) 10) A polling organization asks an SRS of 1500 employees if they took things from their place of work for personal use. It is believed that on a national level, 35% of employees engage in white-collar crimes and take things from their place of employment. What is the probability that a random sample of 1500 will give a result within 2 percentage points of the true population proportion of 35%? ANS: n = 1500 and p = 0.35. We are interested to find EMBED Equation.3  Can we use the normal distribution to approximate the sampling distribution of p-hat? np = 1500(0.35) = 525 and n(1-p) = 1500(0.65) = 975. Both are much larger than 10. Since the conditions are met, the normal approximation is an appropriate model. The sampling distribution of p-hat has  EMBED Equation.3   EMBED Equation.3 or normalcdf(0.33, 0.37, 0.35, 0.0123) 11) In a large high school of 2500 students, 21% of them are seniors. A simple random sample of 150 students is taken and the proportion of seniors calculated. What are the mean and standard deviation of the sample proportion, p-hat? ANS: In this case, the population proportion is p = 0.21 and the sample size is n = 150. The mean of the sample proportion p-hat is  EMBED Equation.3 and the standard deviation is  EMBED Equation.3  12) In a large high school of 2500 students, 21% of them are seniors. A simple random sample of 150 students is taken and the proportion of seniors calculated. What is the probability that the sample will contain less than 15% seniors? ANS: Since the binomial conditions np = 150(0.21) = 31.5 and n(1-p) = 150(0.79) = 118.5 are both greater than 10the sampling distribution of p-hat can be approximated by a normal distribution with mean 0.21 and standard deviation 0.033 (the results from (11)).  EMBED Equation.3 or normalcdf(-1E99, 0.15, 0.21, 0.033) Thus, the chance of obtaining a sample with less than 15% seniors is about 3.6%--not very likely. 13) In a large high school of 2500 students, the mean number of cars owned by students families is 2.35 with a standard deviation of 1.06. A simple random sample of 36 students is taken and the mean number of cars owned is calculated. (a) What are the mean and standard deviation of the sample mean, x-bar? (b) What is the probability that the sample mean is greater than 2.5 cars? (c) Below what value is the lowest 5% of all possible sample means? ANS: (a) The population mean is  = 2.35 cars, the standard deviation is  = 1.06 cars, and the sample size is n = 36. The mean of the sample mean x-bar is  EMBED Equation.3  (b)  EMBED Equation.3 or normalcdf(2.5, 1E99, 2.35, 0.177) (c) The 5th percentile in Table A corresponds to a z-score of -1.65. Therefore,  EMBED Equation.3  Solving for x-bar, we get 2.958. The lowest 5% of all sample means will be below 2.958 cars. 14) The GPAs of graduating students at a large university are normally distributed, with a mean GPA of 2.8 and a standard deviation of 0.5. A random sample of 50 students is taken from all the graduating students. (a) Find the probability that the mean GPA of the sampled students is above 3.0. (b) Find the probability that the mean GPA of sampled students is between 2.7 and 3.0. ANS: Sample size 50 is sufficiently large for us to assume approximate normality for the sampling distribution of x-bar with  EMBED Equation.3  In other words, X ~ N(2.8, 0.071) approximately. (a)  EMBED Equation.3 or normalcdf(3.0, 1E99, 2.8, 0.071) There is less than a 1% chance (0.24%) that the mean GPA of the 50 sampled students will exceed 3.0. (b)  EMBED Equation.3 or normalcdf(2.7, 3.0, 2.8, 0.071) There is almost a 92% chance that the mean GPA of the 50 sampled students will be between 2.7 and 3.0. 15) The mathematics department at a state university notes that the SAT math scores of high school seniors applying for admission into their program are normally distributed with a mean of 610 and standard deviation of 50. (a) What is the probability that a randomly chosen applicant to the department has an SAT math score above 700? (b) What is the shape, mean, and standard deviation of the sampling distribution of the mean of a sample of 40 randomly selected applicants? (c) What is the probability that the mean SAT math score in an SRS of 40 applicants is above 625? (d) Would your answers to (a), (b), or (c) be affected if the original population of SAT math scores were highly skewed instead of normal? Explain. ANS: (a)  EMBED Equation.3 or normalcdf(700, 1E99, 610, 50) (b) It is roughly normal with N(610,  EMBED Equation.3 =N(610, 7.91) (c)  EMBED Equation.3 or normalcdf(625, 1E99, 610, 7.91) (d) The answer to part (a) would be affected because it assumes a normal population. The other answers would not be affected because for large enough n, the central limit theorem gives that the sampling distribution will be roughly normal regardless of the distribution of the original population. 16) A jar of Jiffy Peanut Butter (labeled as a 32 oz. jar) is selected randomly off the end of the assembly line. The weight of the jar is measured. If X measures this weight then X has the N(32.3, 0.40) distribution (the filling machine is calibrated this way). That is, the mean fill is 32.3 oz., the standard deviation of the fills is 0.40 oz. (a) Identify the response variable being measured. Is it categorical or quantitative? (b) Identify the population and parameters. (c) What is the average fill of all Jiffy jars? Is this value a statistic? (d) Whats the probability a randomly selected jar is filled with less than 32.0 fl oz? Such a can would be under volume. (e) What proportion of all Jiffy jars are under-volume? (f) Consider a simple random sample of 6 jars. Let x-bar be the sample mean weight for these 6 jars. Is x-bar a parameter or a statistic? (g) What are the mean and standard deviation of the (sampling) distribution for the sample mean? (h) Find the probability the sample mean fill of the 6 jars is less than 32.0 fl oz. (i) Now take a simple random sample of 24 jars. What are the mean and standard deviation of the distribution for x-bar? (j) Find the probability the sample mean fill of the 24 jars is less than 32.0 fl oz. ANS: (a) Response variable = weight of the jar (quantitative.) (b) Population = measurement of jars of Jiffy Peanut Butter. Parameter = 32 oz. (c) Average = 32.3 oz. No, its not a statistic. (d) P(X<32.0) =  EMBED Equation.3 or normalcdf(-1E99, 32, 32.3, 0.4) (e) 0.2266 (f) x-bar is a statistic. (g)  EMBED Equation.3  (h)  EMBED Equation.3 or normalcdf(-1E99, 32, 32.3, 1.63) (i)  EMBED Equation.3  (j)  EMBED Equation.3 essentially 0. 17) The weights, in pounds, of Portuguese Water Dogs has the N(45, 3) distribution. (a) If one dog is selected at random and X is its weight, find the probability X is within 1 lb. of the mean weight of 45. (b) Draw a simple random sample of 40 dogs. Find the probability the sample mean weight is within 1 lb. of the mean weight of 45. (c) Draw a simple random sample of 80 dogs. Find the probability the sample mean weight is within 1 lb. of the mean weight of 45. ANS: (a) P(44d"Xd"46) = normalcdf(44, 46, 45, 3) = 0.2611 (b)  EMBED Equation.3   EMBED Equation.3 normalcdf(44, 46, 45, 0.4743) = 0.9650 (c)  EMBED Equation.3   EMBED Equation.3 = normalcdf(44, 46, 45, 0.33541) = 0.9971 18) A bottling company uses a filling machine to fill plastic bottles with a popular cola. The bottles are supposed to contain 300 millilters (ml). In fact, the contents vary according to a normal distribution with mean m = 303 ml and standard deviation s = 3 ml. What is the probability that an individual bottle contains less than 300 ml? Now take a random sample of 10 bottles. What are the mean and standard deviation of the sample mean contents x-bar of these 10 bottles? What is the probability that the sample mean contents of the 10 bottles is less than 300 ml? ANS: (a) P(X<300) =  EMBED Equation.3 or normalcdf(-1E99, 300, 303, 3) (b)  EMBED Equation.3  (c)  EMBED Equation.3 or normalcdf(-1E99, 300, 303, 0.94868) 19) For 1998 as a whole, the mean return of all common stocks listed on the New York Stock Exchange (NYSE) was m = 16% and standard deviation s = 26%. Assume that the distribution of returns is roughly normal. What % of stocks lost money? Suppose we create a portfolio of 8 stocks by randomly selecting stocks from the NYSE and investing equal amounts of money in each stock. What are the mean and standard deviation of the sample mean returns x-bar for these 8 stocks? What is the probability the portfolio loses money? Explain the difference between this result and that of part (a). The probability is 0.05 that a portfolio constructed this way has a return of more than ________ ? (This would be the 95th percentile of portfolio returns; however, remember that these portfolios form a hypothetical population -- no one actually owns such a portfolio.) ANS: (a)  EMBED Equation.3 or normalcdf(-1E99, 0, 0.16, 0.26) (b)  EMBED Equation.3  (c)  EMBED Equation.3 or normalcdf(-1E99, 0, 0.16, 0.09192) This is the probability that the average of 8 randomly selected stocks loses money; the 0.2676 in part (a) is the probability a single stock loses money. (d) 0.9500 ( 1.645 EMBED Equation.3 or invNorm(0.95, 0.16, 0.09192) That is, 5% of these portfolios will make more than 31.12%. 20) The length of human pregnancies from conception to birth varies according to a distribution that is approximately normal with mean 264 days and standard deviation 16 days. Consider 15 pregnant women from a rural area. Assume they are equivalent to a random sample from all women. What are the mean and standard deviation of the sample mean length of pregnancy x-bar of these 15 pregnancies? If we want to predict, with 90% accuracy, the sample mean length of pregnancy for 15 randomly selected women, what values do we use? (That is, find value L AND U such that there's a 90% probability the sample mean x-bar lies between L and U.) What's the probability the sample mean length of pregnancy lasts less than 250 days? (Contrast this with the probability a single pregnant women is pregnant for less than 250 days, which is 0.1908.) Toxic waste is believed to have effected the health of residents of this area. Suppose the sample mean length of pregnancy is indeed 250 days; use the result of part (c) to argue that the waste has an effect of length of pregnancy. ANS: (a)  EMBED Equation.3  (b) 0.0500(-1.645 = EMBED Equation.3  or invNorm (0.05, 264, 4.13119) 0.9500 ( 1.645 =  EMBED Equation.3  or invNorm (0.95, 264, 4.13119) So, it will be between 258 and 271 days. (c)  EMBED Equation.3 or normalcdf(-1E99, 250, 264, 4.13119) (d) Assume the toxin has no effect on length of pregnancy -- the average length of pregnancy for all people (including people exposed to the toxin) is 264. The chance of an average length of pregnancy at least as low as the observed 250 is very remote -- it should occur in 1 in 3333 trials on average. This leads one to believe that perhaps the result isn't due to chance alone and, instead, that our assumption of 264 days on average is in question. (This result is "beyond a reasonable doubt.") 21) An airplane is only allowed a gross passenger weight of 8000 kg. If the weights of passengers traveling by air between Toronto and Vancouver have a mean of 78 kg and a standard deviation of 7 kg, the approximate probability that the combined weight of 100 paU V i j k l     J K ^ _ ` a BDFHdfqrＯwh[j hXMyh(MEHUjmzI h(MCJUVaJjhXMyh(MEHUj;zI h(MCJUVaJjh (h(MEHUjzI h(MCJUVaJjh!h(MEHUjzI h(MCJUVaJjh!h(MEHUjzI h(MCJUVaJjh(MUh(Mh*;@h(M5CJaJ" ) . m : j k C H X Y [`J qgd(M$a$gd(MJK^_Ȼ׬אtgXjMGK h(MCJUVaJjZh!h(MEHUjMzI h(MCJUVaJjh yh(MEHUjzI h(MCJUVaJjh!`h(MEHUjzI h(MCJUVaJjh!`h(MEHUjzI h(MCJUVaJh(Mjh(MUjh![h(MEHUj%zI h(MCJUVaJdUEJbZ[|otgd(M_`akm泦旊{n_jzI h(MCJUVaJj\)hh(MEHUjzI h(MCJUVaJj&hmh(MEHUjzI h(MCJUVaJj#hmh(MEHUjzI h(MCJUVaJj!hFGPh(MEHUjzI h(MCJUVaJhFGPh(MH*h(Mjh(MUjRhFGPh(MEHU  !!!!!!!! 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Find the probability that the time required to assemble all nine components (i.e. the total assembly time) is greater than 117 minutes. (a) 2514 (b) .2486 (c) .4772 (d) .0228 (e) .0013 Solution: d P (x > 117/9) = normalcdf (13, 1E99, 12,  EMBED Equation.3 ) = 0.02275 OR P (x > 117) = normalcdf (117, 1E99, 12 * 9,  EMBED Equation.3 ) = 0.02275 23) The sample mean is an unbiased estimator for the population mean. This means: (a) The sample mean always equals the population mean. (b) The average sample mean, over all possible samples, equals the population mean. (c) The sample mean is always very close to the population mean. (d) The sample mean will only vary a little from the population mean. (e) The sample mean has a normal distribution. Solution: b 24) The average monthly mortgage payment for recent home buyers in Winnipeg is  = $732, with.z8zBzLzVzbzz{T{{{@||*}B}D}t̃ 7$8$H$gd(Mzzzz{{{L{M{{{B}L}}}}~2^`thh(MmHsHU hh(Mh(Mjh(MUj}hQh(MEHUjIGK h(MCJUVaJ standard deviation of  = $421 A random sample of 125 recent home buyers is selected. The approximate probability that their average monthly mortgage payment will be more than $782 is: (a) 0.9082 (b) 0.4522 (c) 0.4082 (d) 0.0478 (e) 0.0918 Solution: e 21h:p(M/ =!"8#$% Dd b  c $A? ?3"`?2+2e~9$TKV1D `!2e~9$TKV1 dxڕ?,Q[9 Q!Dq9 T*$+$Iu$ JV B :J#QѐXoޟ d~&3 @O8I42#Qɨѳ2u7AD$W.w\hM#Lo,TH; z,qznY~jSԌ#6ZF˴(J{ûZ㨻V3R*= )n}n{LDpgàMeoj-_8^<^|R6 9{ 3'Md7h1M|傇Wى䁼qGǼkyw\Oss1Zpӽ2ӕpam`;md7,&7) \@Ū쟛V}{9l LD3_kyF8m.4HC윮6\f\YDd lb  c $A? ?3"`?2H,U{K"$% `!,U{K",xڕKTQ9oFS4]-0.eZ$#I1+-[G((6[BKis߻^;{ 8$!*p Х_aѮ^ 8C1OW7/E&BV89z*ht@CiIDDA.kZ+^6YvvuAL*zP4L"dr;!?n8'CQ y/? 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KA?H1 3zjx|K2B* Rj8 :@u!f010+ wfjQ9A $37X/\!(?71aǒXky`n 3q%0D ̌8"zb `?@ABDEFGHIJMPtQRTSUVWXY[Z\]^_`abcdfeghikjlmnopqrusvwxyz|{}~Root Entry F`oċOgData CҜWordDocument4ObjectPoolTnċ`oċ_1232793757Fnċ9nċOle CompObjfObjInfo  #$'*+,/2347:;<=>ADEHKLMPSTWZ]`adghilopqtwxy| FMicrosoft Equation 3.0 DS Equation Equation.39q?<R  x =11.25, x =1.75 90  =0.184Equation Native _1232793842 F9nċ9nċOle CompObj f FMicrosoft Equation 3.0 DS Equation Equation.39q?¶p 2x>11.5)=P(z>11.5"11.250.184=1.36)=0.0869ObjInfo Equation Native  _1232794085F9nċ9nċOle CompObjfObjInfoEquation Native _1232794939 F9nċ9nċ FMicrosoft Equation 3.0 DS Equation Equation.39q?­<R 2x<79)=P(z<79"82)6 32  )="2.83)=0.0023Ole CompObjfObjInfoEquation Native  FMicrosoft Equation 3.0 DS Equation Equation.39q?vjR  2p  = 0.2(0.8)100  =0.04 FMicrosoft Equation 3.0 DS Eq_1232794989F9nċ9nċOle CompObjfObjInfo!uation Equation.39q?ž@ j 2p >0.3)=P(z>0.3"0.20.04=2.5)=0.0062 FMicrosoft Equation 3.0 DS Equation Equation.39qEquation Native "_1232795173"F9nċ9nċOle %CompObj &fObjInfo!(Equation Native )_1232795592$F9nċ9nċOle -?µ<R P(2x<64)=P(z<64"65)6 35  )="0.99)=0.1611 FMicrosoft Equation 3.0 DS Equation Equation.39qCompObj#%.fObjInfo&0Equation Native 1_1232795651@)F9nċ9nċ?¬cR  2p  =0.10, 2p  = 0.10(0.90)50  =0.042 FMicrosoft Equation 3.0 DS Equation Equation.39qOle 5CompObj(*6fObjInfo+8Equation Native 9W?;$, P(0.11<2p <0.15)=P(0.11"0.100.042<z<0.15"0.100.042)=P(0.238<z<1.19)=0.289_1232795869.F9nċ9nċOle ?CompObj-/@fObjInfo0B FMicrosoft Equation 3.0 DS Equation Equation.39q?˜<R  x =40000, x =5000 160  =395.28 FMicrosoft Equation 3.0 DS EqEquation Native C_1232795981,;3F9nċ9nċOle FCompObj24Gfuation Equation.39q?<R P(2x<39000)=P(z<39000"40000395.28="2.53)=0.006 FMicrosoft Equation 3.0 DS EqObjInfo5IEquation Native J_12629641648F9nċ9nċOle NCompObj79OfObjInfo:QEquation Native R_1232796408=F9nċ9nċuation Equation.39qVxT  x =32, x =5 35  =0.338 FMicrosoft Equation 3.0 DS Equation Equation.39qOle UCompObj<>VfObjInfo?XEquation Native Y?d<R z x =1.28=2x"320.338 FMicrosoft Equation 3.0 DS Equation Equation.39q?€<R  2x =1_12327965831JBF9nċ9nċOle [CompObjAC\fObjInfoD^Equation Native __1232796630GF9nċ9nċOle bCompObjFHcf185, 2x =80 100  =8 FMicrosoft Equation 3.0 DS Equation Equation.39q?¦$ P(2x>1200)=P(z>1200"11ObjInfoIeEquation Native f_1232796892EOLF9nċ9nċOle j858=1.875)=0.03 FMicrosoft Equation 3.0 DS Equation Equation.39q?¨<R  2p  =0.70, 2p  = 0.7CompObjKMkfObjInfoNmEquation Native n_1232796945QF9nċ9nċ(0.3)600  =0.019 FMicrosoft Equation 3.0 DS Equation Equation.39q?º$ P(2p <0.65)=P(z<0.65"0.Ole rCompObjPRsfObjInfoSuEquation Native v700.019="2.63)=0.004 FMicrosoft Equation 3.0 DS Equation Equation.39q?E?R P(0.33d"2p d"0.37)_1232797316'VF9nċ9nċOle zCompObjUW{fObjInfoX}Equation Native ~a_1232797461[F9nċ9nċOle CompObjZ\f FMicrosoft Equation 3.0 DS Equation Equation.39q?¸<R  2p  =0.35, 2p  = 0.35(0.65)1500  =0.0123ObjInfo]Equation Native _1232797525Yc`F9nċ9nċOle  FMicrosoft Equation 3.0 DS Equation Equation.39q?G"$ P(0.33<2p <0.37)=P(0.33"0.350.0123<z<0.37"0.350.0123CompObj_afObjInfobEquation Native c_1232798022eF9nċ9nċ)=P("1.63<z<1.63)=0.8968 FMicrosoft Equation 3.0 DS Equation Equation.39q?;<R  2p  =p=0.21Ole CompObjdffObjInfogEquation Native W_1232798047^rjF9nċ9nċOle CompObjikfObjInfol FMicrosoft Equation 3.0 DS Equation Equation.39q?~$  2p  = 0.21(0.79)36  =0.033Equation Native _1232798256oF9nċ9nċOle CompObjnpf FMicrosoft Equation 3.0 DS Equation Equation.39q?¾<R P(2p <0.15)=P(z<0.15"0.210.033="1.80)=0.0359ObjInfoqEquation Native _1232798945mwtF9nċ9nċOle CompObjsufObjInfovEquation Native _1232799062yF9nċ9nċ FMicrosoft Equation 3.0 DS Equation Equation.39q?”<R  2x =2.35, 2x =1.06 36  =0.177 FMicrosoft Equation 3.0 DS EqOle CompObjxzfObjInfo{Equation Native uation Equation.39q?²<R P(2x>2.5)=P(z>2.5"2.350.177=0.85)=0.1977 FMicrosoft Equation 3.0 DS Eq_1232799203h~F9nċ9nċOle CompObj}fObjInfouation Equation.39q?b<R z="1.65=2x"2.350.177 FMicrosoft Equation 3.0 DS Equation Equation.39qEquation Native ~_1232799482F9nċ9nċOle CompObjfObjInfoEquation Native _1232799549F9nċ9nċOle ?<R  2x =2.8, 2x =0.5 50  =0.071. FMicrosoft Equation 3.0 DS Equation Equation.39q?®| P(2x>3.CompObjfObjInfoEquation Native _1232799625F9nċ9nċ0)=P(z>3.0"2.80.071=2.82)=0.0024 FMicrosoft Equation 3.0 DS Equation Equation.39q?'5 P(2.7<2xOle CompObjfObjInfoEquation Native C<3.0)=P(2.7"2.80.071<z<3.0"2.80.071)=P("1.41<z<2.82)=0.9183 FMicrosoft Equation 3.0 DS Equation Equation.39q_1232800080F9nċ9nċOle CompObjfObjInfo?š<R P(X>700)=P(z>700"61050=1.8)=0.0359 FMicrosoft Equation 3.0 DS Equation Equation.39qEquation Native _1232800193F9nċ9nċOle CompObjfObjInfoEquation Native L_1232800242F9nċ9nċOle ?0 # 50 40  ) FMicrosoft Equation 3.0 DS Equation Equation.39q?ª(2 P(2x>625)=P(z>625"610CompObjfObjInfoEquation Native _1233505899F9nċ9nċ7.91=1.90)=0.0287 FMicrosoft Equation 3.0 DS Equation Equation.39q<R P(z<32.0"32.30.40="0.Ole CompObjfObjInfoEquation Native 75)=0.2266 FMicrosoft Equation 3.0 DS Equation Equation.39q<R  2x ==32.3, 2x =0.4 6  =1.63_1233506212|F9nċ9nċOle CompObjfObjInfoEquation Native _1233506308F9nċ9nċOle CompObjf FMicrosoft Equation 3.0 DS Equation Equation.39qH. P(2x<32.0)=P(z<32.0"32.31.63="0.18)=0.4286ObjInfoEquation Native _1233506483F9nċ9nċOle     !$'*-.1458;<=@CDGJKNQRSVYZ]`adghknoruvwz} FMicrosoft Equation 3.0 DS Equation Equation.39q<R  2x =32.3, 2x =0.40 24  =0.08CompObjfObjInfoEquation Native _1233506545F9nċ9nċOle CompObjfObjInfo Equation Native   FMicrosoft Equation 3.0 DS Equation Equation.39q P(2x<32.0)=P(z<32.0"32.30.08="3.75)= FMicrosoft Equation 3.0 DS Eq_1295776346F9nċ9nċOle CompObjfObjInfouation Equation.39qX  2x =45, 2x =3 40  =0.4743 FMicrosoft Equation 3.0 DS Equation Equation.39qEquation Native _1233507170F9nċ9nċOle CompObjfObjInfoEquation Native U_1233507287F9nċ9nċOle 98L P(44d"2xd"46)= FMicrosoft Equation 3.0 DS Equation Equation.39q<R  2x =4CompObjfObjInfoEquation Native _1233507345F9nċ9nċ5, 2x =3 80  =0.33541 FMicrosoft Equation 3.0 DS Equation Equation.39q58 P(44d"2xd"46)Ole "CompObj#fObjInfo%Equation Native &Q_1233507455F9nċ9nċOle (CompObj)fObjInfo+ FMicrosoft Equation 3.0 DS Equation Equation.39qn<R P(z<300"3033="1)=0.1587Equation Native ,_1233507529F9nċ9nċOle /CompObj0f FMicrosoft Equation 3.0 DS Equation Equation.39q.  2x =303, 2x =3 10  =0.94868 FMicrosoft Equation 3.0 DS EqObjInfo2Equation Native 3_1233507580F9nċ9nċOle 6CompObj7fObjInfo9Equation Native :_1233507972F9nċ9nċuation Equation.39q5ĸ P(2x<300)=P(z<300"3030.94868="3.16)=0.0008 FMicrosoft Equation 3.0 DS EqOle >CompObj?fObjInfoAEquation Native Buation Equation.39q5ĸ P(X<0)=P(z<0"0.160.26="0.62)=0.2676 FMicrosoft Equation 3.0 DS Equation Equation.39q_1233507800F9nċ9nċOle ECompObjFfObjInfoH<R  2x =0.16, 2x =0.26 8  =0.09192 FMicrosoft Equation 3.0 DS Equation Equation.39qEquation Native I_1233507869F9nċ9nċOle LCompObjMfDAЮ ^80M jm-ѢU ‚{zR.\8{ދ~˝#k:Fd|yҌ5d$ֆdjpnvT{9Xì~']K=ZgdWZwI-Sheo iibXY}{v6MNdr @:4|OrA45HVI+ԘmƣK֭JUkz|Sp"{*J%k<J,6Z s㋛W;P\v<zĩnn-]ogv0 #@!Pk XU\\[ Cm)͙^ꕍxJklL֗k {YASs7qQm=o>aVJGp,-նkAXE7u3?&?B&J҉Kw[{ge&A0kHm͙ij%n|mTO0 lNxu?xFOL>#&ēoVIƱ* OO9Q U>G<( Js_$96H)6o ߜ7W\O<)#R̺~>{e/Lkq?EUoo{jγi;{l vv$MZq !(? 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