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Similar formulas are available for higher moments. The r th moments about the mean and the origin, respectively, are given by f1(x1 x)r c fk(xk x)r # # # a f (x x)r mr (29) n n mr r f1xr 1 c n fk xr k a fxr n (30)

Classify each of the following events as to whether the events are in the environment or in the system and whether the events are hidden or visible:.

(27) c2(u2 # u2) # (28)

14.20 In the coupled circuit p Fig. 14-35, nd the input admittance Y1 I1 =V1 and determine the of current i1 t for v1 2 2 cos t.

The two kinds of moments are related by m1 m2 m3 m4 etc. If we write Mr a f (u n u)r # Mr r a f ur n 0 mr 2 mr 3 mr 4

I1 2s2 s 1 3 V1 s s2 5s 1 p For s j, the input admittance is Y1 1 j =4 2=4 458. Therefore, i1 t cos t 458 . Y1

where u is given by (26), then the relations (31) also hold between the M s. But mr a f (x n x)r # a f [(a cu) n (a cu)]r # a fcr(u n u)r # crMr

(32) 3Mr4) 1

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14.21 Find the input impedance Z1 V1 =I1 in the coupled circuit of Fig. 14-36.

etc. The second equation of (32) is, of course, the same as (28). In a similar manner other statistics, such as skewness and kurtosis, can be found for grouped samples.

Fig. 14-36 Apply KVL around loops 1 and 2 in the s-domain. ( V1 sI1 1 sI2 2 I1 I2 3 1 0 1 sI1 1 sI2 2 I1 I2 12 sI2 3 4 ( or The result is I2 I1 and Z1 V1 2 s I1 3 The input impedance is V1 2 s I1 2 1 s I2 3 0 2 1 s I1 2 1 s I2 3 3

2. The door is unlocked by the system. System Visible 3. An employee lets a nonemployee through the door. Environment Hidden 4. An employee has an identical twin. Environment Hidden 5. An image has the minimal number of similarities for the matching algorithm. Hidden System

Sampling distribution of means 5.1. A population consists of the five numbers 2, 3, 6, 8, 11. Consider all possible samples of size two which can be drawn with replacement from this population. Find (a) the mean of the population, (b) the standard deviation of the population, (c) the mean of the sampling distribution of means, (d) the standard deviation of the sampling distribution of means, i.e., the standard error of means.

(a) (b) s2 and s (2 3.29. 6)2 (3 m 6)2 2 (6 5 3 6)2 6 5 (8 8 11 6)2 30 5 (11 6.0 6)2 16 9 0 5 4 25 10.8

The current through the resistor is I1 I2 0 and the resistor has no e ect on Z1 . purely inductive.

(c) There are 5(5) 25 samples of size two which can be drawn with replacement (since any one of five numbers on the first draw can be associated with any one of the five numbers on the second draw). These are (2, 2) (3, 2) (6, 2) (8, 2) (11, 2) (2, 3) (3, 3) (6, 3) (8, 3) (11, 3) (2, 6) (3, 6) (6, 6) (8, 6) (11, 6) (2, 8) (3, 8) (6, 8) (8, 8) (11, 8) (2, 11) (3, 11) (6, 11) (8, 11) (11, 11)

14.22 Two coupled coils, L1 0:8 H and L2 0:2 H, have a coe cient of coupling k 0:90. inductance M and the turns ratio N1 =N2 . Ans: 0:36 H, 2 Two coupled coils, N1 100 and N2 800, have a coupling coe cient k 0:85. current of 5.0 A in coil 2, the ux is 2 0:35 mWb. Find L1 , L2 , and M. Ans: 0:875 mH, 56 mH, 5.95 mH Find the mutual

The corresponding sample means are 2.0 2.5 (1) 4.0 5.0 6.5 2.5 3.0 4.5 5.5 7.0 4.0 4.5 6.0 7.0 8.5 5.0 5.5 7.0 8.0 9.5 6.5 7.0 8.5 9.5 11.0

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