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(c) The theoretical mean of the sampling distribution of means, given by mX, equals the population mean , which is 67.45 inches (see Problem 5.33), in close agreement with the value 67.58 inches of part (b). The theoretical standard deviation (standard error) of the sampling distribution of means, given by sX, equals s> !n, where the population standard deviation s 2.92 inches (see Problem 5.40) and the sample size n 4. Since s> !n 2.92> !4 1.46 inches, we have close agreement with the value 1.41 inches of part (b). Discrepancies are due to the fact that only 30 samples were selected and the sample size was small.

74 16.4 103.2 28.7 105 V(3, 4)

5.44. The standard deviation of the weights of a very large population of students is 10.0 lb. Samples of 200 students each are drawn from this population, and the standard deviations of the weights in each sample are computed. Find (a) the mean, (b) the standard deviation of the sampling distribution of standard deviations.

The output le contains the following: DC TRANSFER CURVES Vs V(1) 1:050E 02 1:139E 01 Therefore, V 3; 4 73:07 V. V(3, 4) 7:307E 01

We can consider that sampling is either from an infinite population or with replacement from a finite population. From Table 5-1, page 160, we have: (a) (b) sS mS s 22n s 10.0 lb 10 2400 0.50 lb

5.45. What percentage of the samples in Problem 5.44 would have standard deviations (a) greater than 11.0 lb, (b) less than 8.8 lb

Fig. 15-24

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The sampling distribution of standard deviations is approximately normal with mean 10.0 lb and standard deviation 0.50 lb. (a) 11.0 lb in standard units (11.0 10.0)>0.50 2.0. Area under normal curve to right of z (0.5 0.4772) 0.0228; hence, the required percentage is 2.3%. (b) 8.8 lb in standard units (8.8 10.0)>0.50 2.4. Area under normal curve to left of z (0.5 0.4918) 0.0082; hence, the required percentage is 0.8%. 2.0 is 2.4 is

Subjective comments on quality Problem: Initial req ambiguity Problem: Problem: Problem:

The source le is DC analysis, VS Is R1 R2 R3 R4 .DC .PRINT .END Fig. 15-24 2 0 0 3 0 1 1 2 1 3 3 0 Vs 200 DC I(R4) DC DC 27 47 4 23 200 200 V 20 A

5.46. A sample of 6 observations is drawn at random from a continuous population. What is the probability that the last 2 observations are less than the first 4

Assume that the population has density function f (x). The probability that 3 of the first 4 observations are greater than u while the 4th observation lies between u and u du is given by

The output le contains the following results: DC TRANSFER CURVE Vs I(R4) 2:000E 02 1:123E 01 Current I R4 11:23 A ows from node 3 to node 0 according to the order of nodes in the data statement for R4.

The probability that the last 2 observations are less than u (and thus less than the first 4) is given by

Find the three loop currents in the circuit of Fig. 15-25 using PSpice and compare your solution with the analytical approach.

Then the probability that the first 4 are greater than u and the last 2 are less than u is the product of (1) and (2), i.e., (3)

Since u can take on values between ` and `, the total probability of the last 2 observations being less than the first 4 is the integral of (3) from ` to `, i.e.,

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CHAP. 15]

4C3 3

Fig. 15-25

1 4C3 3 0

0. Therefore, (4) becomes v)3dv (3) (4) (7) 1 15

The source le is DC analysis, V1 V2 R1 R2 R3 R4 R5 .DC .PRINT .END Fig. 15-25 2 0 0 4 0 1 1 2 1 3 3 0 3 4 V1 25 DC I(R1) DC DC 2 5 10 4 2 25 I(R3) 25 50

v2(1

1 I(R5)

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