For the sampling distribution of means, mX sX s m 5.02 oz. n 1 500 100 A 500 1 2100 0.30 0.027

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N AN 2n

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(a) The combined weight will lie between 496 and 500 oz if the mean weight of the 100 ball bearings lies between 4.96 and 5.00 oz (Fig. 5-4). 4.96 in standard units 5.00 in standard units Required probability (area between z (area between z (area between z 0.4868 0.2704 2.22 and z 2.22 and z 0.74 and z 0.2164 0) 0) 0.74) 4.96 5.02 0.027 5.00 5.02 0.027 2.22 0.74

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Fig. 5-4

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CHAPTER 5 Sampling Theory

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Fig. 5-5

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(b) The combined weight will exceed 510 oz if the mean weight of the 100 bearings exceeds 5.10 oz (Fig. 5-5). 5.10 in standard units Required probability (area to right of z (area to right of z (area between z 0.5 0.4985 2.96) 0) 0 and z 2.96) 0.0015 5.10 5.02 0.027 2.96

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Therefore, there are only 3 chances in 2000 of picking a sample of 100 ball bearings with a combined weight exceeding 510 oz.

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5.6. Theorem 5-1, page 155.

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Since X1, X2, . . . , Xn are random variables having the same distribution as the population, which has mean , we have E(Xk) Then since the sample mean is defined as # X we have as required # E(X) 1 n [E(X1) c E(Xn)] 1 n (nm) m X1 c n Xn m k 1, 2, c, n

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5.7. Prove Theorem 5-2, page 156.

We have # X X1 n c X2 n c

Xn n 1 2 sb n2 s2 n

Then since X1, . . . , Xn are independent and have variance # Var (X) 1 Var (X1) n2

, we have by Theorems 3-5 and 3-7: na

1 Var (Xn) n2

Sampling distribution of proportions 5 5.8. Find the probability that in 120 tosses of a fair coin (a) between 40% and 60% will be heads, (b) 8 or more will be heads.

We consider the 120 tosses of the coin as a sample from the infinite population of all possible tosses of the coin. In this population the probability of heads is p 1, and the probability of tails is q 1 p 1. 2 2 (a) We require the probability that the number of heads in 120 tosses will be between 40% of 120, or 48, and 60% of 120, or 72. We proceed as in 4, using the normal approximation to the binomial distribution. Since the number of heads is a discrete variable, we ask for the probability that the number of heads lies between 47.5 and 72.5. (See Fig. 5-6.) m and expected number of heads s !npq B np 1 120a b 2 5.48 60

1 1 (120)a b a b 2 2

CHAPTER 5 Sampling Theory

Fig. 5-6

47.5 in standard units 72.5 in standard units Required probability

47.5 60 5.48 72.5 60 5.48

2.28 2.28

(area under normal curve between z 2(area between z 2(0.4887) 0.9774 2.28 and z 0 and z 2.28) 2.28)

Another method

mP p 1 2 0.50 sP pq An B 120

1 1 2 2

0.0456 2.19

40% in standard units 60% in standard units

0.40 0.50 0.0456 0.60 0.50 0.0456

2.19 and z 2.19, i.e., Therefore, the required probability is the area under the normal curve between z 2(0.4857) 0.9714. Although this result is accurate to two significant figures, it does not agree exactly since we have not used 1 1 the fact that the proportion is actually a discrete variable. To account for this, we subtract from 2n 2(120) 1 1 0.40 and add to 0.60. Therefore, the required proportions in standard units are, since 2n 2(120) 1>240 0.00417, 0.40 0.00417 0.0456 0.50 2.28 and 0.60 0.00417 0.0456 0.50 2.28

so that agreement with the first method is obtained. Note that (0.40 0.00417) and (0.60 0.00417) correspond to the proportions 47.5>120 and 72.5>120 in the first method above. (b) Using the second method of (a), we find that since 8 (0.6250

0.6250, 0.6250 0.00417 0.0456 0.50 2.65) 2.65

0.00417) in standard units

Required probability

(area under normal curve to right of z (area to right of z (area between z 0.5 0.4960 0) 0 and z 2.65) 0.0040

5.9. Each person of a group of 500 people tosses a fair coin 120 times. How many people should be expected to 5 report that (a) between 40% and 60% of their tosses resulted in heads, (b) 8 or more of their tosses resulted in heads