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CHAPTER 5 Sampling Theory
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This problem is closely related to Problem 5.8. Here we consider 500 samples, of size 120 each, from the infinite population of all possible tosses of a coin. (a) Part (a) of Problem 5.8 states that of all possible samples, each consisting of 120 tosses of a coin, we can expect to find 97.74% with a percentage of heads between 40% and 60%. In 500 samples we can expect to find about 97.74% of 500, or 489, samples with this property. It follows that about 489 people would be expected to report that their experiment resulted in between 40% and 60% heads. It is interesting to note that 500 489 11 people would be expected to report that the percentage of heads was not between 40% and 60%. These people might reasonably conclude that their coins were loaded, even though they were fair. This type of error is a risk that is always present whenever we deal with probability. (b) By reasoning as in (a), we conclude that about (500)(0.0040) their tosses resulted in heads. 2 persons would report that 8 or more of
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5.10. It has been found that 2% of the tools produced by a certain machine are defective. What is the probability that in a shipment of 400 such tools, (a) 3% or more, (b) 2% or less will prove defective
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mP p 0.02 and sP pq An A 1>800 0.03 0.02(0.98) 400 0.14 20 0.007
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(a) Using the correction for discrete variables, 1>2n (0.03 0.00125) in standard units
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0.00125, we have 0.02 1.25) 1.25 0.1056
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Required probability
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0.00125 0.007 (area under normal curve to right z
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If we had not used the correction we would have obtained 0.0764. Another method (3% of 400) 12 defective tools. On a continuous basis, 12 or more tools means 11.5 or more. m (2% of 400) 8 (11.5 s !npq !(400)(0.02)(0.98) 2.8
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Then, 11.5 in standard units (b) (0.02
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8)>2.8
1.25, and as before the required probability is 0.1056. 0.02 0.00125 0.007 0.02 0.18) 0.18
0.00125) in standard units
Required probability
(area under normal curve to left of z 0.5000 0.0714 0.5714
If we had not used the correction, we would have obtained 0.5000. The second method of part (a) can also be used.
5.11. The election returns showed that a certain candidate received 46% of the votes. Determine the probability that a poll of (a) 200, (b) 1000 people selected at random from the voting population would have shown a majority of votes in favor of the candidate.
(a) mP p 0.46 and sP pq An 0.46(0.54) B 200 0.0352
Since 1>2n 1>400 0.0025, a majority is indicated in the sample if the proportion in favor of the candidate is 0.50 0.0025 0.5025 or more. (This proportion can also be obtained by realizing that 101 or more indicates a majority, but this as a continuous variable is 100.5; and so the proportion is 100.5>200 0.5025.) Then, 0.5025 in standard units (0.5025 0.46)>0.0352 1.21 and Required probability (area under normal curve to right of z 0.5000 0.3869 0.1131 1.21)
CHAPTER 5 Sampling Theory
(b) mP 0.46, sP !pq>n !0.46(0.54)1000
0.0158, and 2.69 2.69)
0.5025 in standard units Required probability
0.5025 0.46 0.0158
(area under normal curve to right of z 0.5000 0.4964 0.0036
Sampling distributions of differences and sums 5.12. Let U1 be a variable that stands for any of the elements of the population 3, 7, 8 and U2 a variable that stands for any of the elements of the population 2, 4. Compute (a) mU1, (b) mU2, (c) mU1 U2, (d) sU1, (e) sU2, (f) sU1 U2.
(a) mU1 mean of population U1 1 (3 3 1 (2 2 7 8) 6
mean of population U2
(c) The population consisting of the differences of any member of U1 and any member of U2 is 3 3 Then mU1 2 4 7 7 2 4 8 8 U2)
2 4 1
or 5
1 1 6
6 4 ( 1) 6 3 4 3
mean of (U1
which illustrates the general result mU1
as is seen from (a) and (b). 6)2 (7 3 6)2 (8 6)2 14 3
(d) or sU1 (e) or sU2 (f) s2 1 U 1.
s2 1 U 14 . A3
variance of population U1
s2 2 U
variance of population U2
3)2 2
variance of population (U1 (1 3)2 (5 3)2 (6
U2) 3)2 6 ( 1 3)2 (3 3)2 (4 3)2 17 3
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