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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

Fig. 15-1

0.6250, 0.6250 0.00417 0.0456 0.50 2.65) 2.65

(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

CHAP. 15]

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

We rst label the nodes by the numbers 0, 1, 2 and the elements by the symbols R1, R2, C, and Vs [Fig. 15-1(b)]. We then create in ASCII the source le shown below and give it a name, for instance, EXMP1.CIR. DC analysis, Fig. Vs R1 R2 C .END 15-1 1 0 1 2 0 2 0 2 DC 3k 6k 5 uF 9V

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This problem is closely related to Problem 5.8. Here we consider 500 samples, of size 120 each, from the infinite population of all possible tosses of a coin. (a) Part (a) of Problem 5.8 states that of all possible samples, each consisting of 120 tosses of a coin, we can expect to find 97.74% with a percentage of heads between 40% and 60%. In 500 samples we can expect to find about 97.74% of 500, or 489, samples with this property. It follows that about 489 people would be expected to report that their experiment resulted in between 40% and 60% heads. It is interesting to note that 500 489 11 people would be expected to report that the percentage of heads was not between 40% and 60%. These people might reasonably conclude that their coins were loaded, even though they were fair. This type of error is a risk that is always present whenever we deal with probability. (b) By reasoning as in (a), we conclude that about (500)(0.0040) their tosses resulted in heads. 2 persons would report that 8 or more of

A test coverage criterion is a rule about how to select tests and when to stop testing. One basic issue in testing research is how to compare the e ectiveness of di erent test coverage criteria. The standard approach is to use the subsumes relationship.

5.10. It has been found that 2% of the tools produced by a certain machine are defective. What is the probability that in a shipment of 400 such tools, (a) 3% or more, (b) 2% or less will prove defective

Executing the command PSPICE EXMP1, the computer solves the circuit and writes the following results in the le EXMP1.OUT. NODE VOLTAGE NODE (1) 9.0000 (2) VOLTAGE SOURCE CURRENTS NAME CURRENT Vs 1:000E 03 TOTAL POWER DISSIPATION VOLTAGE 6.0000

mP p 0.02 and sP pq An A 1>800 0.03 0.02(0.98) 400 0.14 20 0.007

The printed output speci es that the voltage at node 2 with reference to node 0 is 6 V, the current entering the voltage source Vs is 10 3 A, and the total power dissipated in the circuit is 9 10 3 W.

0.00125, we have 0.02 1.25) 1.25 0.1056

If we had not used the correction we would have obtained 0.0764. Another method (3% of 400) 12 defective tools. On a continuous basis, 12 or more tools means 11.5 or more. m (2% of 400) 8 (11.5 s !npq !(400)(0.02)(0.98) 2.8

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