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2 S1 I1;eff Z1 768 2 j3 1536 j2304 2 S2 I2;eff Z2 222 3 j6 666 j1332 ST S1 S2 2202 j3636

Then x !3 1.73 approx., which is the required mode. Note that this does give the maximum since the second derivative, 24x > 81, is negative for x !3. (b) The median is that value a for which P(X P(X a) a) 1>2. Now, for 0 x2) dx 4 9a2 81 2 0 a4 4 a 3,

that is, PT 2202 W, QT 3636 var (inductive), ST q 2202 2 3636 2 4251 VA and pf 2202 0:518 lagging 4251

4 a x(9 81 30

2648 4

CHAP. 10]

9 22 2

Since the phase angle of the voltage is unknown, only the magnitude of IT can be given. By current division, I2 and so

Fig. 7-1.

Therefore, the required median, which must lie between 0 and 3, is given by a2 from which a (c) 1.62 approx. E(X ) 4 3 2 x (9 81 30 x2) dx 9 9 22 2 4 3x3 81 x5 2 3 5 0

which is practically equal to the median. The mode, median, and mean are shown in Fig. 3-6.

106 222 1811 A2 13

Fig. 3-6

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10.16 Obtain the complete power triangle for the circuit shown in Fig. 10-20, if the total reactive power is 2500 var (inductive). Find the branch powers P1 and P2 .

3.34. A discrete random variable has probability function f(x) (b) the median, and (c) compare them with the mean.

1, 2, . . . . Find (a) the mode,

Fig. 10-20 The equivalent admittance allows the calculation of the total power triangle. Yeq Y1 Y2 0:2488 39:578 S Then, PT 2500 cot 39:578 3025 W ST 3025 j2500 3924 39:578 VA and pf PT =ST 0:771 lagging. The current ratio is I1 =I2 Y1 =Y2 0:177=0:0745. P1 I 2 4 21 1:88 P2 I2 12 from which P1 1975 W and P2 1050 W. and P1 P2 3025 W

Section 3 3.1 Organization 3.2 Tasks All documents will be reviewed. A configuration management tool will manage all documents and source code modules. All test plans will be done during the

(a) The mode is the value x having largest associated probability. In this case it is x probability is 1 > 2.

10.17 A load of 300 kW, with power factor 0.65 lagging, has the power factor improved to 0.90 lagging by parallel capacitors. How many kvar must these capacitors furnish, and what is the resulting percent reduction in apparent power

(b) If x is any value between 1 and 2, P(X x) 1 and P(X x) 1. Therefore, any number between 1 and 2 2 2 could represent the median. For convenience, we choose the midpoint of the interval, i.e., 3 > 2. (c) As found in Problem 3.3, m Problem 3.33. 2. Therefore, the ordering of the three measures is just the reverse of that in

Percentiles 3.35. Determine the (a) 10th, (b) 25th, (c) 75th percentile values for the distribution of Problem 3.33.

The angles corresponding to the power factors are rst obtained: cos 1 0:65 49:468 Then (see Fig. 10-21), Q 300 tan 49:468 350:7 kvar (inductive) Q Qc 300 tan 25:848 145:3 kvar (inductive) whence Qc 205:4 kvar (capacitive). Since S 300 461:5 kVA 0:65 S0 300 333:3 kVA 0:90 cos 1 0:90 25:848

(a) The 10th percentile is the value of a for which P(X a) 0.10, i.e., the solution of (18a2 Using the method of Problem 3.33, we find a 0.68 approx. (b) The 25th percentile is the value of a such that (18a2 (c) The 75th percentile is the value of a such that (18a2 a4) > 81 a4) > 81 0.25, and we find a 0.75, and we find a

Fig. 10-21

Other measures of dispersion 3.36. Determine, (a) the semi-interquartile range, (b) the mean deviation for the distribution of Problem 3.33.

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