Chapter 9Chapter 99.1a. 1/6b. 1/69.2 a )1X(P =P(1,1)= 1/36b )6X(P = P(6,6) = 1/369.3a P(X = 1) = (1/6)5= .0001286b P(X = 6) = (1/6)5= .00012869.4 The variance of Xis smaller than the variance of X.9.5 The sampling distribution of the mean is normal with a mean of 40 and a standard deviation of 12/100= 1.2.9.6 No, because the sample mean is approximately normally distributed.9.7 a )1050X(P = 16/20010001050n/XP = P(Z > 1.00) = 1 – P(Z < 1.00) = 1 – .8413 = .1587b  )960X(P16/2001000960n/XP = P(Z < –.80) = .2119c  )1100X(P16/20010001100n/XP = P(Z > 2.00) = 1 – P(Z < 2.00) = 1 – .9772 = .02289.8 a )1050X(P =25/20010001050n/XP = P(Z > 1.25) = 1 – P(Z < 1.25) = 1 – .8944 = .1056b  )960X(P25/2001000960n/XP = P(Z < –1.00) = .1587c  )1100X(P25/20010001100n/XP = P(Z > 2.50) = 1 – P(Z < 2.50) = 1 – .9938 = .00629.9 a )1050X(P =100/20010001050n/XP = P(Z > 2.50) = 1 – P(Z < 2.50) = 1 – .9938 = .0062213b  )960X(P100/2001000960n/XP = P(Z < –2.00) = .0228c  )1100X(P100/20010001100n/XP = P(Z > 5.00) = 09.10 a  )52X49(P4/55052n/X4/55049P = P(–.40 < Z < .80) = P(Z < .80) − P(Z < −.40) = .7881 −.3446 = .4435b  )52X49(P16/55052n/X16/55049P = P(–.80 < Z < 1.60) = P(Z < 1.60) − P(Z < −.80) = .9452 −.2119 = .7333c  )52X49(P25/55052n/X25/55049P = P(–1.00 < Z < 2.00) = P(Z < 2.00) − P(Z < −1.00) = .9772 −.1587 = .81859.11 a  )52X49(P 4/105052n/X4/105049P = P(–.20 < Z < .40) = P(Z < .40) − P(Z < −.20) = .6554 −.4207= .2347b  )52X49(P16/105052n/X16/105049P = P(–.40 < Z < .80) = P(Z < .80) − P(Z < −.40) = .7881 −.3446= .4435c  )52X49(P25/105052n/X25/105049P = P(–.50 < Z < 1.00) = P(Z < 1.00) − P(Z < −.50) = .8413 −.3085= .53289.12 a  )52X49(P 4/205052/4/205049nXP = P(–.10 < Z < .20) = P(Z < .20) − P(Z < −.10) = .5793 −.4602= .1191b  )52X49(P16/205052n/X16/205049P = P(–.20 < Z < .40) = P(Z < .40) − P(Z < −.20) = .6554 −.4207= .2347c  )52X49(P25/205052n/X25/205049P = P(–.25 < Z < .50) = P(Z < .50) − P(Z < −.25) = .6915 −.4013= .2902 2149.13 a1NnN= 1000,1100000,1= .9492b1NnN= 1000,3100000,3= .9834c1NnN= 1000,5100000,5= .9900d. The finite population correction factor is approximately 1.9.14 a x=1NnNn= 1000,10000,1000,10000,1500= 15.00b x=1NnNn= 1000,10500000,10500500= 21.80c x=1NnNn= 1000,10100000,10100500= 49.759.15 a P(X > 66) = 26466XP= P(Z > 1.00) = 1 – P(Z < 1.00) = 1 – .8413 = .1587b  )66X(P4/26466n/XP = P(Z > 2.00) = 1 – P(Z < 2.00) = 1 – .9772 = .0228c  )66X(P100/26466n/XP = P(Z > 10.00) = 09.16 We can answer part (c) and possibly part (b) depending on how nonnormal the population is.9.17 a P(X > 120) = 2.5117120XP= P(Z > 0.58) = 1 – P(Z < .58) = 1 – .7190 = .2810b  )120X(P4/2.5117120n/XP = P(Z > 1.15) = 1 – P(Z < 1.15) = 1 – .8749 = .1251c [P(X >120)]4=[.2810]4 = .006239.18 a P(X > 60) = 65260XP= P(Z > 1.33) = 1 – P(Z < 1.33) = 1 – .9082 = .0918b  )60X(P3/65260n/XP = P(Z > 2.31) = 1 – P(Z < 2.31) = 1 – .9896 = .0104c [P(X >60)]3=[.0918]3= .000772159.19 a P(X > 12) = 31012XP= P(Z > .67) = 1 – P(Z < .67) = 1 – .7486 = .2514b  )25/275X(P )11X(P25/31011n/XP = P(Z > 1.67) = 1 – P(Z < 1.67) = 1 – .9525 = 04759.20 a P(X < 75) = 67875XP= P(Z < –.50) = .3085b  )75X(P50/67875n/XP = P(Z < –3.54) = 1 – P(Z < 3.54) = 1 – 1 = 09.21 a P(X > 7) = 5.167XP= P(Z > .67) = 1 – P(Z < .67) = 1 – .7486 = .2514b  )7X(P5/5.167n/XP = P(Z > 1.49) = 1 – P(Z < 1.49) = 1 – .9319 = .0681c [P(X >7)]5=[.2514]5= .001009.22 a  )97.5X(P36/18.05.697.5n/XP = P(Z < –2.67) =.0038b It appears to be false.9.23  )16/000,10X(P )625X(P16/200600625n/XP = P(Z > .50) = 1 – P(Z < .50) = 1 – .6915 = .30859.24 The professor needs to know the mean and standard deviation of the population of the weights of elevator usersand that the distribution is not extremely nonnormal.9.25  )16/140,1X(P )25.71X(P16/107525.71n/XP = P(Z > –1.50) = 1 − P(Z < −1.50) = 1 − 0668 = .93329.26 P(Total time > 300) =  )60/300X(P )5(XP60/3.18.45n/XP = P(Z > 1.19) 216= 1 – P(Z < 1.19) = 1 – .8830 = .11709.27 No because the central limit theorem says that the sample mean is approximately normally distributed.9.28 P(Total number of cups > 240) =  )125/240X(P )92.1X(P125/6.0.292.1n/XP= P(Z > –1.49) = 1 − P(Z < −1.49) = 1 − .0681 = .93199.29 P(Total number of faxes > 1500) =  )5/1500X(P )300X(P5/75275300n/XP= P(Z > .75) = 1 – P(Z < .75) = 1 – .7734 = .22669.30a P(Pˆ> .60) = 300/)5.1)(5(.5.60.n/)p1(ppPˆP= P(Z > 3.46) = 0b. P(Pˆ> .60) = 300/)55.1)(55(.55.60.n/)p1(ppPˆP= P(Z > 1.74) = 1 – P(Z < 1.74) = 1 – .9591 =