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Normal approximation to binomial distribution 4.17. Find the probability of getting between 3 and 6 heads inclusive in 10 tosses of a fair coin by using (a) the binomial distribution, (b) the normal approximation to the binomial distribution.

Fig. 12-3

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(a) Let X be the random variable giving the number of heads that will turn up in 10 tosses (Fig. 4-15). Then P(X 3) a a 10 1 3 1 7 ba b a b 2 2 3 10 1 5 1 5 ba b a b 2 2 5 15 128 63 256 P(X 4) a a 10 1 4 1 6 ba b a b 2 2 4 10 1 6 1 4 ba b a b 2 2 6 105 512 105 512

CHAP. 12]

Then the required probability is P(3 X 6) 15 128 105 512 63 256 105 512 99 128 0.7734

Fig. 4-15

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frequency increases, the performance is called high-frequency roll-o and the circuit is a low-pass network. On the contrary, a high-pass network will have low-frequency roll-o , with jHv1 j decreasing as the frequency decreases. Four two-element circuits are shown in Fig. 12-4, two high-pass and two lowpass.

Fig. 4-16

Fig. 12-4

(b) The probability distribution for the number of heads that will turn up in 10 tosses of the coin is shown graphically in Figures 4-15 and 4-16, where Fig. 4-16 treats the data as if they were continuous. The required probability is the sum of the areas of the shaded rectangles in Fig. 4-16 and can be approximated by the area under the corresponding normal curve, shown dashed. Treating the data as continuous, it follows that 3 to 6 heads can be considered as 2.5 to 6.5 heads. Also, the mean and variance for the binomial distribution are given by m np 10 A 1 B 5 and s # A1 B A1 B (10) 2 2 1.58. !npq 2 Now 2.5 in standard units 6.5 in standard units 2.5 5 1.58 6.5 5 1.58 1.58 0.95

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Fig. 4-17

The RL high-pass circuit shown in Fig. 12-5 is open-circuited or under no-load. The input impedance frequency response is determined by plotting the magnitude and phase angle of Hz1 ! R1 j!L2 jHz j H

Required probability (Fig. 4-17)

Fig. 12-5

0.95) 0) 0.95)

which compares very well with the true value 0.7734 obtained in part (a). The accuracy is even better for larger values of n.

or, normalizing and writing !x R1 =L2 , Hz1 ! 1 j !=!x R1 q 1 !=!x 2 tan 1 !=!x

4.18. A fair coin is tossed 500 times. Find the probability that the number of heads will not differ from 250 by (a) more than 10, (b) more than 30.

Five values of ! provide su cient data to plot jHz j=R1 and H , as shown in Fig. 12-6. The magnitude approaches in nity with increasing frequency, and so, at very high frequencies, the network current I1 will be zero. In a similar manner, the frequency response of the output-to-input voltage ratio can be obtained. Voltage division under no-load gives Hv1 ! so that j!L2 1 R1 j!L2 1 j !x =! and H tan 1 !x =!

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(a) We require the probability that the number of heads will lie between 240 and 260, or considering the data as continuous, between 239.5 and 260.5. 239.5 in standard units Required probability 239.5 250 11.18 0.94 260.5 in standard units 0.94 and z 2(0.3264) 0.94 0.94) 0.6528

(area under normal curve between z (twice area between z 0 and z 0.94)

1 jHv j q 1 !x =! 2

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