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Unformatted text preview: 1 Homework 5 (6.5 pts) due Sept. 30 (1 pt.) #10: Only 5% of a large population of 50ohm resistors have resistances that exceed 55 ohms. a) For samples of size 200 from this population, describe the sampling distribution of the sample proportion of resisters that have resistances in excess of 55 ohms. Since nπ = (200)(0.05) = 10 ≥ 5 and n(1 – π) = (200)(0.95) = 190 ≥ 5 This is a normal distribution with μ p = π = 0.05 and = (1 − ) = ¡ 0.05 ¢ (0.95) 200 = £ 0.0002375 = 0.01541 b) What is the probability that the proportion of resistors with resistances exceeding 55 ohms in a random sample of 200 will be more than 6%? ¡ > 0.06 ¢ = ¤ > 0.06 − 0.05 0.01541 ¥ = ¡ > 0.65 ¢ = 1 − ¡ ≤ 0.65 ¢ = 1 − 0.7422 = 0.2578 (1 pt.) 7.2 (p.293). A random sample of ten homes in a particular area, each heated with natural gas, is selected, and the amount of gas (therms) used during January is determined for each home. The resulting observations are 103 156 118 89 125 147 122 109 138 99 a) Use an unbiased estimator to compute a point estimate of μ, the average amount of gas used by all houses in the area. An unbiased estimator of μ is x ̅ . ¦ = § = 1206 10 = 120.6 b) Use an unbiased estimator to computer a point estimate of π, the proportion of all homes that use over 100 therms....
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 Spring '08
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 Statistics, Normal Distribution, Standard Deviation, unbiased estimator, Pogonias cromis, Pogonias cromis falls

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