chap04 - PROBLEM 4.1 KNOWN: N > 1000; x= 9.2 units ; Sx...

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4-1 PROBLEM 4.1 KNOWN: N > 1000; x = 9.2 units ; S x = 1.1 units FIND: Range of x in which 50% of all measurements should fall. ASSUMPTIONS: Measurand follows a normal density function Data set sufficiently large such that x x’ and S x σ SOLUTION By assuming that the data is sufficiently large such that its population behaves as an infinite population, we can find the interval defined by x' - z 1 σ x i x' + z 1 σ as follows. We can find P(x' + z 1 σ ) from the one-sided integral solution to This solution is given in Table 4.3 for p(z 1 ) = 0.25 (one half of the 50% probability sought) as z 1 = 0.674. Then, we d β e ] [2 1 ) P(z 1 2 Z 0 /2 β 1/2 1 π =
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4-2 should expect that 50% of the x i values lie in the interval given by 9.2 – 0.7425 x i 9.2 + 0.7425 (50%) COMMENT We can see from Table 4.4 that as N becomes large the value for t approaches a value given by z 1 .
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4-3 PROBLEM 4.2 KNOWN: N > 10 000; x = 204 units ; S x = 18 units FIND: x' - z 1 σ x x' + z 1 σ at P = 90% ASSUMPTIONS Measurand follows a normal density function Data set sufficiently large such that x x’ and S x σ SOLUTION Using the definition z 1 = (x 1 - x')/ σ we find the z 1 value corresponding to the one-sided probability integral p(z 1 ) = 0.45 from Table 4.3. This gives, z 1 = 1.65 Then, 1.65 = (x 1 - x')/ σ or 1.65 σ = x 1 - x' or
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4-4 x 1 = x' + 1.65 σ But a normal (Gaussian) distribution is symmetric about the mean value. Hence, for 90% probability (2)(0.45), x' - 1.65 σ x x' + 1.65 σ (90%) or 174.3 x 233.7 units (90%)
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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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4-6
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4-7 PROBLEM 4.4 KNOWN: Toss of four coins FIND: Develop the histogram for the outcome of any toss. State the probability of obtaining three heads on one toss. SOLUTION A coin has two distinct sides, so each toss has two possible outcomes. With four coins, there will be 2 4 = 16 possible outcomes of any one toss of the four coins. The probability of three heads is 4 in 16 or 25%. The possible outcomes are: Number n j of Heads 4 1 3 4 2 6 1 4 0 1 The histogram is shown below. Because of the few number of Problem 4.4 0 2 4 6 8 01234 Number of Heads in a Toss Possible Occurrences, n
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4-8 tosses, the development of the histogram is primitive. But the symmetry is obvious. This type of distribution is best described as a Binomial distribution (see Table 4.2). COMMENT The binomial distribution shape is similar to the Gaussian (normal) distribution, except that it can lack the extended "tails" found with the Gaussian shape. As the number of possible outcomes (number of coins tossed) becomes large (say 30 or more), the two distributions become nearly identical over a wide interval about the mean and the Gaussian distribution can be used for ease.
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4-9 PROBLEM 4.5 SOLUTION In the matchbox game, the frequency distribution will more
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chap04 - PROBLEM 4.1 KNOWN: N > 1000; x= 9.2 units ; Sx...

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