Week_11_Probability_Distributions

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33) Mopeds (small motorcycles with an engine capacity below 50 cm3) are very popular in Europe because of their mobility, ease of operation, and low cost. The article “Procedure to Verify the Max- imum Speed of Automatic Transmission Mopeds in Periodic Motor Vehicle Inspections” (J. of Automobile Engr., 2008: 1615–1623) de- scribed a rolling bench test for determining maximum vehicle speed. A normal distribution with mean value 46.8 km/h and standard devi- ation 1.75 km/h is postulated. Consider randomly selecting a single such moped. a. What is the probability that maximum speed is at most 50 km/h? b. What is the probability that maximum speed is at least 48 km/h? c. What is the probability that maximum speed differs from the mean value by at most 1.5 standard deviations? Answer Given that, µ = 46 . 8 σ = 1 . 75 z = x µ σ a) maximum speed = 50 P ( X 50) = 50 46 . 8 1 . 75 = 3 . 2 1 . 75 = 1 . 83 According to z table 1 . 82 = 0 . 9664 Therefore,The probability that maximum speed at most 50 km/h is 0.9664 b) maximum speed = 48 P ( X 48) = 48 46 . 8 1 . 75 1
= 1 . 2 1 . 75 = 0 . 69 According to z table 0 . 68 = 0 . 2451 Therefore,The probability that maximum speed atleast 48 km/h is c) P ( 1 . 5 < z < 1 . 5) = P ( z < 1 . 5) P ( z < 1 . 5) = 0 . 9332 0 . 0668 = 0 . 8664 Therefore,The probability that maximum speed differs from the mean value by at most 1.5 is 0.8664 35) Suppose the diameter at breast height (in.) of trees of a certain type is normally distributed with m 5 8.8 and s 5 2.8, as suggested in the article “Simulating a Harvester-Forwarder Softwood Thinning” (Forest Products J., May 1997: 36–41). a. What is the probability that the diameter of a randomly selected tree will be at least 10 in.? Will exceed 10 in.? b. What is the probability that the diameter of a randomly selected tree will exceed 20 in.? c. What is the probability that the diameter of a randomly selected tree will be between 5 and 10 in.? d. What value c is such that the interval (8.8 2 c, 8.8 1 c) includes 98percent of all diameter values? e. If four trees are independently selected, what is the probabil- ity that at least one has a diameter exceeding 10 in.? Answer Given, µ = 8 . 8 σ = 2 . 8 z = x µ σ a. x=10 z = 10 8 . 8 2 . 8 z = 1 . 2 2 . 8 2
z = 0 . 428 P ( h > 10) = 1 P ( h < 10) = 1 P ( z < 0 . 43) According to z table = 1 0 . 6664 = 0 . 3346 Therefore,The probability that the diameter of a randomly selected tree will be at least 10 is 0.3346 b. x=20 z = 20 8 . 8 2 . 8 z = 11 . 2 2 . 8 z = 4 P ( h > 20) = 1 P ( h < 20) = 1 P ( z < 4) According to z table = 1 1 = 0 Therefore,The probability that the diameter of a randomly selected tree will exceed 20 is 0 c. P (5 < x < 10) = P ( h < 10) P ( h < 5) = P ( z < 10 8 . 8 2 . 8 ) P ( z < 5 8 . 8 2 . 8 ) = P ( z < 1 . 2 2 . 8 P ( z < 3 . 8 2 . 8 )) P ( z < 0 . 43) P ( z < 1 . 36) According to z table 0 . 6664 0 . 0869 = 0 . 5795 Therefore,The probability that the diameter of a randomly selected tree will be between 5 and 10 is 0.5795 3
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d. 98 100 = 0 . 98 P (8 . 8 c < h < 8 . 8 + c ) = 0 . 98 P ( h < 8 . 8 + c ) = 0 . 98 P ( z < c 2 . 8 ) = 0 . 98 c = 2 . 33 c = 2 . 33 × 2 . 8 c = 6 . 524 Therefore,The value of c is 6.524 e. z = 10 8 . 8 2 . 8 z = 1 . 2 2 . 8 z = 0 . 43 P ( x < 10) = P ( z < 0 . 43) According to z table = 0 . 6664 P ( Four trees < 10) = P ( x < 10) .P ( x < 10) .P ( x < 10) .P ( x < 10) = [ P ( x < 10)] 4 = (0 . 6664) 4 = 0 . 1972 P ( atleast one > 10) = 1 P ( Four trees < 10) = 1 0 . 1972 = 0 . 8028 Therefore,The probability that at least one has a diameter exceeding 10 is 0.8028 49) Consider babies born in the “normal” range of 37–43 weeks ges- tational age. Extensive data supports the assumption that for such babies born in the United States, birth weight is normally distributed with mean 3432 g and standard deviation 482 g. (The article “Are Ba- bies Normal?” (The American Statistician, 1999: 298–302) analyzed 4
data from a particular year; for a sensible choice of class intervals, a histogram did not look at all normal, but after further investiga- tions it was determined that this was due to some hospitals measuring weight in grams and others measuring to the nearest ounce and then converting to grams. A modified choice of class intervals that allowed for this gave a histogram that was well described by a normal distri- bution.) a. What is the probability that the birth weight of a randomly se- lected baby of this type exceeds 4000 g? Is between 3000 and 4000 g? b. What is the probability that the birth weight of a randomly se- lected baby of this type is either less than 2000 g or greater than 5000 g? c. What is the probability that the birth weight of a randomly se- lected baby of this type exceeds 7 lb? d. How would you characterize the most extreme .1percent of all birth weights? e. If X is a random variable with a normal distribution and a is a numerical constant (a is not equal to 0), then Y=aX also has a normal distribution. Use this to determine the distribution of birth weight expressed in pounds (shape, mean, and standard deviation), and then recalculate the probability from part (c). How does this compare to your previous answer? Given, µ = 3432 g σ = 482 g z = x µ σ a. P ( W < 4000) = P ( z > 4000 3432 482 ) = P ( z > 568 482 ) = P ( z > 1 . 18) = 1 P ( z < 1 . 18) According to z table = 1 0 . 8810 = 0 . 1190 5
P (3000 < W < 4000) = P ( W < 4000) P ( W < 3000) = P ( z < 4000 3432 482 ) P ( W < 3000 3432 482 ) = P ( z < 568 482 ) P ( W < 432 482 ) = P ( z < 1 . 18) P ( z < 0 . 90) According to z table = 0 . 8810 0 . 1841 = 0 . 6969 Therefore,The probability that the birth weight of a randomly selected baby of this type exceeds 4000 g is 0.1190 Therefore,The probability that the birth weight of a randomly selected baby of this type between 3000 and 4000 g is 0.6969 b. = P ( W < 2000) + P ( W > 5000) = P ( W < 2000) + 1 P ( W < 5000) = 1 P ( z < 5000 3432 482 ) + P ( z < 2000 3432 482 ) = 1 P ( z < 1568 482 ) + P ( z < 1432 482 ) = 1 P ( z < 3 . 25) P ( z < 2 . 97) According to z table = 1 0 . 9994 + (0 . 0015) = 0 . 0006 + 0 . 0015 = 0 . 0021 Therefore,The probability that the birth weight of a randomly selected baby of this type is either less than 2000 g or greater than 5000g is 0.0021 c. 1 lb = 453 . 59 453 . 59 7 7 lb = 3175 . 15 g P ( W > 3175 . 15) = 1 P ( W < 3175 . 15) 6
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= 1 P ( z < 3175 . 15 3432 482 ) 1 P ( z < 256 . 85 482 ) = 1 P ( z < 0 . 53) According to z table = 1 0 . 2981 = 0 . 7019 Therefore,The probability that the birth weight of a randomly selected baby of this type exceeds 7 lb is 0.7019 d. A = 0 . 1 100 A = 0 . 001 The Lower Bound Probability will A/2 P ( X low ) = 0 . 001 2 P ( X low ) = 0 . 0005 Z low = 3 . 295 3 . 295 = X low 3432 482 3 . 295 × 482 = X low 3432 1588 . 19 = X low 3432 X low = 3432 1588 . 19 X low = 1843 . 81 1 X low = X upp X upp = 1 0 . 0005 X upp = 0 . 9995 X upp = 3 . 30 3 . 30 = X upp 3432 482 X upp 3432 = 3 . 30 × 482 X upp 3432 = 1590 . 6 7
X upp = 1590 . 6 + 3432 X upp = 5022 . 6 Therefore,0.1 percent of babies weight less than 1843.81g or more than 5022.6g e. Given Y=ax Normal Distribution From Problem c; 1lb=453.59 Y = ax Y = 1 453 . 59 × x Y = 0 . 002204 × x So a = 0 . 002204 µ = 3432 g σ = 482 g µ.Y = 0 . 002204 × 3432 = 7 . 56 lb σ.Y = 0 . 002204 × 482 = 1 . 06 lb From Problem C P ( z > 7) = 1 P ( z < 7) = 1 P ( z < 7 7 . 56 1 . 06 ) = 1 P ( z < 0 . 56 1 . 06 ) = 1 P ( z < 0 . 56) = 1 0 . 2981 = 0 . 7019 Therefore,we get similar probability of problem c 8