240 Chapter 7: Sampling DistributionsCHAPTER 77.1 PHstat output:Common DataMean 100Standard Deviation 2Probability for a RangeProbability for X <= From X Value 95 X Value 95To X Value 97.5 Z Value -2.5Z Value for 95 -2.5 P(X<=95) 0.0062097Z Value for 97.5 -1.25P(X<=95) 0.0062 Probability for X > P(X<=97.5) 0.1056 X Value 102.2P(95<=X<=97.5) 0.0994 Z Value 1.1P(X>102.2) 0.1357Find X and Z Given Cum. Pctage.Cumulative Percentage 35.00%Z Value -0.38532X Value 99.22936(b) P(95 < X < 97.5) = P(– 2.50 < Z < – 1.25) = 0.1056 – 0.0062 = 0.0994(c) P(X > 102.2) = P(Z > 1.10) = 1.0 – 0.8643 = 0.1357(d) P(X > A) = P(Z > – 0.39) = 0.65X = 100 – 0.39(1025) = 99.227.2 PHStat output:Common DataMean 50Standard Deviation 0.5Probability for a RangeProbability for X <= From X Value 47 X Value 47To X Value 49.5 Z Value -6Z Value for 47 -6 P(X<=47) 9.866E-10Z Value for 49.5 -1P(X<=47) 0.0000 Probability for X > P(X<=49.5) 0.1587 X Value 51.5P(47<=X<=49.5) 0.1587 Z Value 3P(X>51.5) 0.0013Find X and Z Given Cum. Pctage.Cumulative Percentage 65.00% Probability for X<47 or X >51.5 Z Value 0.38532 P(X<47 or X >51.5) 0.0013X Value 50.19266(b) P(47 < X < 49.5) = P(– 6.00 < Z < – 1.00) = 0.1587 – 0.00 = 0.1587(c) P(X > 51.1) = P(Z > 2.20) = 1.0 – 0.9861 = 0.0139(d) P(X > A) = P(Z > 0.39) = 0.35 X = 50 + 0.39(0.5) = 50.195Solutions to End-of-Section and Chapter Review Problems 241 7.3 (a) For samples of 25 customer receipts for a supermarket for a year, the samplingdistribution of sample means is the distribution of means from all possible samples of 25customer receipts for a supermarket for that year.(b) For samples of 25 insurance payouts in a particular geographical area in a year, thesampling distribution of sample means is the distribution of means from all possiblesamples of 25 insurance payouts in that particular geographical area in that year.(c) For samples of 25 Call Center logs of inbound calls tracking handling time for a creditcard company during the year, the sampling distribution of sample means is thedistribution of means from all possible samples of 25 Call Center logs of inbound callstracking handling time for a credit card company during that year.7.4 (a) Sampling Distribution of the Mean for n = 2 (without replacement)iX1 1, 31X = 22 1, 62X = 3.53 1, 73X = 44 1, 94X = 55 1, 105X = 5.56 3, 66X = 4.57 3, 77X = 58 3, 98X = 69 3, 109X = 6.510 6, 710X = 6.511 6, 911X = 7.512 6, 1012X = 813 7, 913X = 814 7, 1014X = 8.515 9, 1015X = 9.5(a) Mean of All Possible Sample Means: Mean of All Population Elements:μX =9015=6136791066μ+++++==Both means are equal to 6. This property is called unbiasedness.242 Chapter 7: Sampling Distributions7.4 (b) Sampling Distribution of the Mean for n = 3 (without replacement) cont.X1 1, 3, 61X = 3 1/32 1, 3, 72X = 3 2/33 1, 3, 93X = 4 1/34 1, 3, 104X = 4 2/35 1, 6, 75X = 4 2/36 1, 6, 96X = 5 1/37 1, 6, 107X = 5 2/38 3, 6, 78X = 5 1/39 3, 6, 99X = 610 3, 6, 1010X = 6 1/311 6, 7, 911X = 7 1/312 6, 7, 1012X = 7 2/313 6, 9, 1013X = 8 1/314 7, 9, 1014X = 8 2/315 1, 7, 915X = 5 2/316 1, 7, 1016X = 617 1, 9, 1017X = 6 2/318 3, 7, 918X = 6 1/319 3, 7, 1019X = 6 2/320 3, 9, 1020X = 7 1/3μX =12020=6This is equal toμ, the population mean.(c) The distribution for n = 3 has less variability. The larger sample size has resulted in samplemeans being closer toμ.(d) (a) Sampling Distribution of the Mean for n = 2 (with replacement)Solutions to End-of-Section and Chapter Review Problems 243 7.4cont.XiX1X = 22 1, 32X = 3.53 1, 63X = 44 1, 74X = 55 1, 95X = 5.56 1, 106X = 27 3, 17X = 38 3, 38X = 4.59 3, 69X = 510 3, 710X = 611 3, 911X = 6.512 3, 1012X = 3.513 6, 113X = 4.514 6, 314X = 615 6, 615X = 6.516 6, 716X = 7.517 6, 917X = 818 6, 1018X= 419 7, 119X = 520 7, 320X = 6.521 7, 621X = 722 7, 722X = 823 7, 923X = 8.524 7, 1024X = 525 9, 125X = 626 9, 326X = 7.527 9, 627X = 828 9, 728X = 929 9, 929X = 9.530 9, 1030X = 5.531 10, 131X = 6.532 10, 332X = 833 10, 633X = 8.534 10, 734X = 9.535 10, 935X = 1036 10, 1036244 Chapter 7: Sampling Distributions 7.4 (d) (a) Mean of All Possible Mean of All cont. Sample Means: Population Elements:216636X μ== μ=1+3+6+7+7+126=6Both means are equal to 6. This property is called unbiasedness. (b) Repeat the same process for the sampling distribution of the mean for n = 3 (withreplacement). There will be 36216= different samples.6X μ= This is equal to μ, the population mean.(c) The distribution for n = 3 has less variability. The larger sample size has resulted inmore sample means being close to μ. 7.5(a)Because the population diameter of tennis balls is approximately normally distributed, the sampling distribution of samples of 9 will also be approximately normal with a mean ofX μμ== 2.63 and X σ== 0.01.(b)X(c)Upper bound: X = 2.6384Solutions to End-of-Section and Chapter Review Problems 2457.6 (a)When n = 4 , the shape of the sampling distribution of X should closely resemble the shape of the distribution of the population from which the sample is selected. Because the mean is larger than the median, the distribution of the sales price of new houses is skewed to the right, and so is the sampling distribution of X although it will be less skewed than the population.(b)If you select samples of n = 100, the shape of the sampling distribution of the sample mean will be very close to a normal distribution with a mean of $322,100 and a standarddeviation of X σ== $9,000.(c) X σ =900010090000==nσ(d)7.7 (a) X σ1==246 Chapter 7: Sampling Distributions 7.7 (b) PHStat output: cont.(c) X σ0.5==(d) With the sample size increasing from n = 25 to n = 100, more sample means will becloser to the distribution mean. The standard error of the sampling distribution of size 100 is much smaller than that of size 25, so the likelihood that the sample mean will fall within ±0.5 minutes of the mean is much higher for samples of size 100 (probability = 0.6827) than for samples of size 25 (probability = 0. 3829).7.8(b) P (X < A ) = P (Z < 1.0364) = 0.85 X = 27 + 1.0364 (1) = 28.0364(c) To be able to use the standard normal distribution as an approximation for the area underthe curve, we must assume that the population is symmetrically distributed such that the central limit theorem will likely hold for samples of n = 16.(d)X = 27 + 1.0364 (0.5) = 27.5182Solutions to End-of-Section and Chapter Review Problems 2477.9 (a)p = 48/64 = 0.75(b)p σ =64)30.0(70.0 = 0.05737.10 (a)p = 20/50 = 0.40(b)p σ= 7.11 (a) p = 14/40 = 0.35(b)p σ =40)70.0(30.0 = 0.07257.12 (a) 0.501p μπ==, 0.05p σ===Partial PHstat output:Probability for X > X Value 0.55Z Value 0.98P(X>0.55) 0.1635P (p > 0.55) = P (Z > 0.98) = 1 – 0.8365 = 0.1635(b) 0.60p μπ==, 0.04899p σ===Partial PHstat output:Probability for X > X Value 0.55Z Value -1.020621P(X>0.55) 0.8463P (p > 0.55) = P (Z > – 1.021) = 1 – 0.1539 = 0.8461(c) 0.49p μπ==, 0.05p σ===Partial PHstat output:Probability for X > X Value 0.55Z Value 1.2002401P(X>0.55) 0.1150P (p > 0.55) = P (Z > 1.20) = 1 – 0.8849 = 0.1151(d) Increasing the sample size by a factor of 4 decreases the standard error by a factor of 2.248 Chapter 7: Sampling Distributions7.12 (d) (a) Partial PHstat output:cont.Probability for X >X Value 0.55Z Value 1.9600039P(X>0.55) 0.0250P(p > 0.55) = P (Z > 1.96) = 1 – 0.9750 = 0.0250(b) Partial PHstat output:Probability for X >X Value 0.55Z Value -2.041241P(X>0.55) 0.9794P(p > 0.55) = P (Z > – 2.04) = 1 – 0.0207 = 0.9793(c) Partial PHstat output:Probability for X >X Value 0.55Z Value 2.4004801P(X>0.55) 0.0082P(p > 0.55) = P (Z > 2.40) = 1 – 0.9918 = 0.0082If the sample size is increased to 400, the probably in (a), (b) and (c) is smaller,larger, and smaller, respectively because the standard error of the samplingdistribution of the sample proportion becomes smaller and, hence, the samplingdistribution is more concentrated around the true population proportion.7.13 (a) Partial PHstat output:Probability for a RangeFrom X Value 0.5To X Value 0.6Z Value for 0.5 0Z Value for 0.6 2.828427P(X<=0.5) 0.5000P(X<=0.6) 0.9977P(0.5<=X<=0.6) 0.4977P(0.50 < p < 0.60) = P(0 < Z < 2.83) = 0.4977(b) Partial PHstat output:Find X and Z Given Cum. Pctage.Cumulative Percentage 95.00%Z Value 1.644854X Value 0.558154P(– 1.645 < Z < 1.645) = 0.90p = .50 – 1.645(0.0354) = 0.4418 p = .50 + 1.645(0.0354) = 0.5582(c) Partial PHstat output:Probability for X >X Value 0.65Z Value 4.2426407P(X>0.65) 0.0000P(p > 0.65) = P (Z > 4.24) = virtually zeroSolutions to End-of-Section and Chapter Review Problems 2497.13 (d)Partial PHstat output:cont.Probability for X > X Value 0.6Z Value 2.8284271P(X>0.6) 0.0023If n = 200, P (p > 0.60) = P (Z > 2.83) = 1.0 – 0.9977 = 0.0023Probability for X > X Value 0.55Z Value 3.1622777P(X>0.55) 0.00078If n = 1000, P (p > 0.55) = P (Z > 3.16) = 1.0 – 0.99921 = 0.00079More than 60% correct in a sample of 200 is more likely than more than 55% correct in asample of 1000.7.14 (a) ==πμp 0.80, ()np ππσ−=1(b)(c)250 Chapter 7: Sampling Distributions 7.14 (d) ==πμp 0.80, ()n p ππσ−=1cont.(b)(c)7.15 (a) ==πμp 0.57, ()np ππσ−=1 = 0.0495(b)Solutions to End-of-Section and Chapter Review Problems 2517.15 (c)(d) ==πμp 0.57, ()np ππσ−=1 = 0.0248(a)(b)(c)7.16 (a)population proportion and, hence,π=0.15. Also the sampling distribution of the sample proportion will be close to a normal distribution according to the central limit theorem.252 Chapter 7: Sampling Distributions7.16p μπ==0.15,p σ=== 0.0252cont. P (0.12 < p < 0.18) = P (-1.1882< Z < 1.1882) = 0.7652A = 0.1085B = 0.1915A = 0.1005B = 0.1995 7.17 (a) p μπ==0.49, p σ=== 0.0500A = 0.4078B = 0.5722Solutions to End-of-Section and Chapter Review Problems 2537.17 (c)Partial PHStat output:cont.A = 0.3920B = 0.5880(d) (a) p μπ==0.49, p σ=== 0.02500.6007A = 0.4489B = 0.5311 (c)A = 0.4410B = 0.5390254 Chapter 7: Sampling Distributions 7.18 (a) ==πμp 0.36, ()n p ππσ−=1= 0.0480(b)= 0.0240(c)Increasing the sample size by a factor of 4 decreases the standard error by a factor of . The sampling distribution of the proportion becomes more concentrated around the true proportion of 0.36 and, hence, the probability in (b) becomes smaller than that in (a).7.19 Because the average of all the possible sample means of size n is equal to the population mean.7.20 The variation of the sample means becomes smaller as larger sample sizes are taken. This is dueto the fact that an extreme observation will have a smaller effect on the mean in a larger sample than in a small sample. Thus, the sample means will tend to be closer to the population mean as the sample size increases.7.21 As larger sample sizes are taken, the effect of extreme values on the sample mean becomessmaller and smaller. With large enough samples, even though the population is not normally distributed, the sampling distribution of the mean will be approximately normally distributed.7.22 The population distribution is the distribution of a particular variable of interest, while thesampling distribution represents the distribution of a statistic.7.23 When the items of interest and the items not of interest are at least 5, the normal distribution canbe used to approximate the binomial distribution.Solutions to End-of-Section and Chapter Review Problems 2557.24 0.753X μ=5004.0==nX σσ = 0.0008PHStat output:Common DataMean 0.753 Standard Deviation 0.0008Probability for a RangeProbability for X <=From X Value 0.75 X Value 0.74 To X Value 0.753 Z Value -16.25 Z Value for 0.75 -3.75 P(X<=0.74) 1.117E-59 Z Value for 0.753 0P(X<=0.75) 0.0001 Probability for X >P(X<=0.753) 0.5000 X Value 0.76 P(0.75<=X<=0.753) 0.4999 Z Value 8.75P(X>0.76) 0.0000 Find X and Z Given Cum. Pctage.Cumulative Percentage 7.00% Probability for X<0.74 or X >0.76 Z Value -1.475791 P(X<0.74 or X >0.76) 0.0000X Value 0.751819Probability for a Range From X Value 0.74 To X Value 0.75 Z Value for 0.74 -16.25 Z Value for 0.75 -3.75 P(X<=0.74) 0.0000 P(X<=0.75) 0.0001 P(0.74<=X<=0.75) 0.00009(a) P (0.75 < X < 0.753) = P (– 3.75 < Z < 0) = 0.5 – 0.00009 = 0.4999 (b) P (0.74 < X < 0.75) = P (– 16.25 < Z < – 3.75) = 0.00009 (c) P (X > 0.76) = P (Z > 8.75) = virtually zero (d) P (X < 0.74) = P (Z < – 16.25) = virtually zero (e) P (X < A ) = P (Z < – 1.48) = 0.07 X = 0.753 – 1.48(0.0008) = 0.7518256 Chapter 7: Sampling Distributions 7.25 2.0X μ=505.0==nX σσ = 0.01PHStat output:Common DataMean 2 Standard Deviation 0.01Probability for a RangeProbability for X <= From X Value 1.99X Value 1.98 To X Value 2Z Value -2 Z Value for 1.99 -1P(X<=1.98) 0.0227501 Z Value for 2 0P(X<=1.99) 0.1587Probability for X >P(X<=2) 0.5000X Value 2.01 P(1.99<=X<=2) 0.3413Z Value 1P(X>2.01) 0.1587 Find X and Z Given Cum. Pctage.Cumulative Percentage 1.00%Probability for X<1.98 or X>2.01Z Value-2.326348P(X<1.98 or X >2.01) 0.1814 X Value 1.976737Find X and Z Given Cum. Pctage.Cumulative Percentage 99.50% Z Value 2.575829X Value2.025758(a) P (1.99 < X < 2.00) = P (– 1.00 < Z < 0) = 0.5 – 0.1587 = 0.3413 (b) P (X < 1.98) = P (Z < – 2.00) = 0.0228(c) P (X > 2.01) = P (Z > 1.00) = 1.0 – 0.8413 = 0.1587(d) P (X > A ) = P ( Z > – 2.33) = 0.99 A= 2.00 – 2.33(0.01) = 1.9767 (e) P (A < X < B ) = P (– 2.58 < Z < 2.58) = 0.99A = 2.00 – 2.58(0.01) = 1.9742B = 2.00 + 2.58(0.01) = 2.0258Solutions to End-of-Section and Chapter Review Problems 2577.26 4.7X μ=0.400.085X σ===PHstat output:Common DataMean 4.7 Standard Deviation 0.08Probability for X >Find X and Z Given Cum. Pctage. X Value 4.6 Cumulative Percentage 23.00% Z Value -1.25 Z Value -0.738847 P(X>4.6) 0.8944 X Value 4.640892Find X and Z Given Cum. Pctage. Find X and Z Given Cum. Pctage. Cumulative Percentage 15.00% Cumulative Percentage 85.00% Z Value -1.036433 Z Value 1.036433 X Value 4.6170853X Value 4.782915(a) P (4.60 < X ) = P (– 1.25 < Z ) = 1 – 0.1056 = 0.8944 (b) P (A < X < B ) = P (– 1.04 < Z < 1.04) = 0.70A = 4.70 – 1.04(0.08) = 4.6168 ounces X = 4.70 + 1.04(0.08) = 4.7832 ounces(c) P (X > A ) = P (Z > – 0.74) = 0.77A = 4.70 – 0.74(0.08) = 4.6408 7.27X μ=5.00.400.085X σ===(a)Partial PHStat output:Probability for X > X Value 4.6 Z Value -5 P(X>4.6)1.0000P (4.60 < X ) = P (– 5 < Z ) = essentially 1.0 (b) Partial PHStat output:Find X and Z Given Cum. Pctage. Cumulative Percentage 15.00% Z Value -1.036433 X Value 4.917085A = 5.0 – 1.0364(0.08) = 4.9171 ounces X = 5.0 + 1.0364(0.08) = 5.0829 ounces258 Chapter 7: Sampling Distributions7.27 (c) Partial PHStat output: cont.Find X and Z Given Cum. Pctage. Cumulative Percentage 23.00% Z Value -0.738847 X Value4.940892P (X > A ) = P (Z > – 0.7388) = 0.77 A = 5.0 – 0.7388(0.08) = 4.94097.28 μμ=X = 15.23, X σ=(a)(b)(c)7.29 (a) μ=3.17 σ = 10Solutions to End-of-Section and Chapter Review Problems 2597.29 (b)Partial PHStat output:cont.(c)(d) μμ=X= 3.17, X σ== 5(e)(f)(g)Since the sample mean of returns of a sample of stocks is distributed closer to the population mean than the return of a single stock, the probabilities in (a) and (b) are higher than those in (d) and (e) while the probability in (c) is lower than that in (f).。
