HW-2_solutions

HW-2_solutions - PROBLEM 4.3 KNOWN = 121.6 psi Sx = 14 psi...

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4-5 PROBLEM 4.3 KNOWN: x = 121.6 psi S x = 14 psi N is very large FIND: P(x > 150 psi) ASSUMPTIONS: Normal distribution S x σ ; x x' SOLUTION The z variable is defined by z 1 = -(x 1 - x')/ σ = -(121.6 -150)/14 = 2.028 We look up P(2.028) from Table 4.3. Interpolation gives P(2.028) = 0.4786 This expresses the probability that 121.6 x 150 psi. Then, the probability that x > 150 psi is 0.5 - 0.4786 = 0.0214 or there is a 2.14% probability that any measurement will yield a value in excess of 150 psi.

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4-9 PROBLEM 4.5 SOLUTION In the matchbox game, the frequency distribution will more closely resemble a Gaussian (normal) distribution as N becomes larger. This is because each shot is independent of the other and each shot differs from the other by random variation. A 'better' player will have a mean distance in the outcome that is close to the target point and have a low variance. That is, the player will have a low systematic error (average distance from target point) and low random error (variations from the average point). This game and its interpretation are similar to the dart game discussed in Chapter 1 in the discussion of random and

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