# chap07.doc - 140 Chapter 7 Sampling Distributions CHAPTER 7...

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140 Chapter 7: Sampling DistributionsCHAPTER 77.1(a)P(X < 95) = P(Z< – 2.50) = 0.0062(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(99 < X < 101) = P(– 0.50 < Z< 0.50) = 0.6915 – 0.3085 = 0.3830(e)P(X > A) = P(Z> – 0.39) = 0.65X = 100 – 0.39(1025) = 99.22(f)(a)P(X < 95) = P(Z< – 2.00) = 0.0228(b)P(95 < X < 97.5) = P(– 2.00 < Z< – 1.00)= 0.1587 – 0.0228 = 0.1359(c)P(X > 102.2) = P(Z> 0.88) = 1.0 – 0.8106 = 0.1894(d)P(99 < X < 101) = P(– 0.40 < Z< 0.40)= 0.6554 – 0.3446 = 0.3108(e)P(X > A) = P(Z> – 0.39) = 0.65X = 100 – 0.39(1016) = 99.0257.2(a)P(X < 47) = P(Z< – 6.00) = virtually zero(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(49 < X < 51) = P(– 2.00 < Z< + 2.00) = 0.9772 – 0.0228 = 0.9544(e)P(X > A) = P(Z> 0.39) = 0.35X = 50 + 0.39(0.5) = 50.195(f)(a)P(X < 47) = P(Z< – 3.00) = 0.00135(b)P(47 < X < 49.5) = P(– 3.00 < Z< – 0.50)= 0.3085 – 0.00135 = 0.30715(c)P(X > 51.1) = P(Z> 1.10) = 1.0 – 0.8643 = 0.1357(d)P(49 < X < 51) = P(– 1.00 < Z< + 1.00)= 0.8413 – 0.1587 = 0.6826(e)P(X > A) = P(Z> 0.39) = 0.35X = 50 + 0.39(1) = 50.397.3(a)For samples of 25 travel expense vouchers for a university in an academic year, the sampling distribution of sample means is the distribution of means from all possible samples of 25 vouchers that could occur.(b)For samples of 25 absentee records in 1997 for employees of a large manufacturing company, the sampling distribution of sample means is the distribution of means from all possible samples of 25 records that could occur.(c)For samples of 25 sales of unleaded gasoline at service stations located in a particularcounty, the sampling distribution of sample means is the distribution of means from all possible samples of 25 sales that could occur.
141 Chapter 7: Sampling Distributions7.4 (a)Sampling Distribution of the Mean for n= 2 (without replacement)Sample Number Outcomes Sample Means X i11, 3X 1= 221, 6X 2= 3.531, 7X 3= 441, 7X 4= 451, 12X 5= 6.563, 6X 6= 4.573, 7X 7= 583, 7X 8= 593, 12X 9= 7.5106, 7X 10= 6.5116, 7X 11= 6.5126, 12X 12= 9137, 7X 13= 7147, 12X 14= 9.5157, 12X 15= 9.5Mean of All PossibleMean of AllSample Means:Population Elements:X 90156136771266Both means are equal to 6. This property is called unbiasedness.
Solutions to End-of-Section and Chapter Review Problems 1427.4(b)Sampling Distribution of the Mean for n= 3 (without replacement)cont.Sample Number Outcomes Sample Means X i11, 3, 6X 1= 3 1/321, 3, 7X 2= 3 2/331, 3, 7X 3= 3 2/341, 3, 12X 4= 5 1/351, 6, 7X 5= 4 2/361, 6, 7X 6= 4 2/371, 6, 12X 7= 6 1/383, 6, 7X 8= 5 1/393, 6, 7X 9= 5 1/3103, 6, 12X 10= 7116, 7, 7X 11= 6 2/3126, 7, 12X 12= 8 1/3136, 7, 12X 13= 8 1/3147, 7, 12X 14= 8 2/3151, 7, 7X 15= 5161, 7, 12X 16= 6 2/3171, 7, 12X 17= 6 2/3183, 7, 7X 18= 5 2/3193, 7, 12X 19= 7 1/3203, 7, 12X 20= 7 1/3X 120206This is equal to , the population mean.(c)The distribution for n= 3 has less variability. The larger sample size has resulted in more sample means being close to .(d)The sampling distributions in part (a) have the smaller variability because fewer samples are possible and there is less chance of obtaining an extremely small or large sample mean.7.5
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