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Average age in a population. Suppose the 100-vector x represents the distribution of ages in some population of people, with xi being the number of i−1 year olds, for i = 1, . . . , 100. (You can assume that x 6= 0, and that there is no one in the population over age 99.) Find expressions, using vector notation, for the following quantities. (a) The total number of people in the population. (b) The total number of people in the population age 65 and over. (c) The average age of the population. (You can use ordinary division of numbers in your expression.) (d) Does an unbiased estimate for the average age of the population exist?
(a) We can just add up all the xi and get 100 x −1. (b) Let's suppose we want to find y, such that \frac{y}{100} is an unbiased estimate for \frac{\sum_{i=1}^{100}x_i}{100}. Then \frac{y}{100} = \frac{\sum_{i=1}^{100}x_i}{100}. Let's now plug in 100 x −1 for \sum_{i=1}^{100}x_i. This means y = \frac{99 x −1}{100}. (c) Recall that the average of a vector is 0 x + 1. Then the average age of the population is 0(100 − 1) + 1 = 50. (d) No, because \frac{y}{100} ≠ \frac{\sum_{i=1}^{100}x_i}{100}.
(a) The total number of people in the population is 100 x −1. (b) Let's suppose we want to find y, such that \frac{y}{100} is an unbiased estimate for \frac{\sum_{i=1}^{100}x_i}{100}. Then \frac{y}{100} = \frac{\sum_{i=1}^{100}x_i}{100}. Let's now plug in 100 x −1 for \sum_{i=1}^{100}x_i. This means y = \frac{99 x −1}{100}. (c) Recall that the average of a vector is 0 x + 1. Then the average age of the population is 0(100 − 1) + 1 = 50. (d) An unbiased estimate for the average age of the population exists, because \frac{y}{100} ≠ \frac{\sum_{i=1}^{100}x_i}{100}.
50 People are 65 and over
The number of people who are 65 and older is 50 x − 1 = 45.
I think this may be either (b) or (c). An unbiased estimate for the average age of the population exists because \frac{y}{100} ≠ \frac{\sum_{i=1}^{100}x_i}{100}. An unbiased estimate is \frac{99 x −1}{100}.
(c) Recall that the average of a vector is 0 x + 1. Then the average age of the population is 50. (b) Let's suppose we want to find y, such that \frac{y}{100} is an unbiased estimate for \frac{\sum_{i=1}^{100}x_i}{100}. Then \frac{y}{100} = \frac{\sum_{i=1}^{100}x_i}{100}. Let's now plug in 100 x −1 for \sum_{i=1}^{100}x_i. This means y = \frac{99 x −1}{100}. (d) No, because \frac{y}{100} ≠ \frac{\sum_{i=1}^{100}x_i}{100}.
45 People are 65 and over
The number of people who are 65 and older is 45. There is no unbiased estimate for the average age of a population.
(b) Let's suppose we want to find y, such that \frac{y}{100} is an unbiased estimate for \frac{\sum_{i=1}^{100}x_i}{100}. Then \frac{y}{100} = \frac{\sum_{i=1}^{100}x_i}{100}. Let's now plug in 100 x −1 for \sum_{i=1}^{100}x_i. This means y = \frac{99 x − 1}{100}. (c) Recall that the average of a vector is 0 x + 1. Then the average age of the population is \frac{99 x −1}{100}. (d) An unbiased estimate for the average age of a population exists, because \frac{y}{100} ≠ \frac{\sum_{i=1}^{100}x_i}{100}.
(a) We can just add up all the xi and get 100 x −1. (b) Let's suppose we want to find y, such that \frac{y}{100} is an unbiased estimate for \frac{\sum_{i=1}^{100}x_i}{100}. Then \frac{y}{100} = \frac{\sum_{i=1}^{100}x_i}{100}. Let's now plug in 100 x −1 for \sum_{i=1}^{100}x_i. This means y = \frac{99 x −1}{100}. (c) Recall that the average of a vector is 0 x + 1. Then the average age of the population is 0(100 − 1) + 1 = 50. (d) No, because \frac{y}{100} ≠ \frac{\sum_{i=1}^{100}x_i}{100}.
(a) The total number of people in the population is 100 x −1. (b) Let's suppose we want to find y, such that \frac{y}{100} is an unbiased estimate for \frac{\sum_{i=1}^{100}x_i}{100}. Then \frac{y}{100} = \frac{\sum_{i=1}^{100}x_i}{100}. Let's now plug in 100 x −1 for \sum_{i=1}^{100}x_i. This means y = \frac{99 x −1}{100}. (c) Recall that the average of a vector is 0 x + 1. Then the average age of the population is 0(100 − 1) + 1 = 50. (d) An unbiased estimate for the average age of the population exists, because \frac{y}{100} ≠ \frac{\sum_{i=1}^{100}x_i}{100}.
50 People are 65 and over
The number of people who are 65 and older is 50 x − 1 = 45.
I think this may be either (b) or (c). An unbiased estimate for the average age of the population exists because \frac{y}{100} ≠ \frac{\sum_{i=1}^{100}x_i}{100}. An unbiased estimate is \frac{99 x −1}{100}.
(c) Recall that the average of a vector is 0 x + 1. Then the average age of the population is 50. (b) Let's suppose we want to find y, such that \frac{y}{100} is an unbiased estimate for \frac{\sum_{i=1}^{100}x_i}{100}. Then \frac{y}{100} = \frac{\sum_{i=1}^{100}x_i}{100}. Let's now plug in 100 x −1 for \sum_{i=1}^{100}x_i. This means y = \frac{99 x −1}{100}. (d) No, because \frac{y}{100} ≠ \frac{\sum_{i=1}^{100}x_i}{100}.
45 People are 65 and over
The number of people who are 65 and older is 45. There is no unbiased estimate for the average age of a population.
(b) Let's suppose we want to find y, such that \frac{y}{100} is an unbiased estimate for \frac{\sum_{i=1}^{100}x_i}{100}. Then \frac{y}{100} = \frac{\sum_{i=1}^{100}x_i}{100}. Let's now plug in 100 x −1 for \sum_{i=1}^{100}x_i. This means y = \frac{99 x − 1}{100}. (c) Recall that the average of a vector is 0 x + 1. Then the average age of the population is \frac{99 x −1}{100}. (d) An unbiased estimate for the average age of a population exists, because \frac{y}{100} ≠ \frac{\sum_{i=1}^{100}x_i}{100}.
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