Given the normally distributed variable X with mean 18 and standard deviation 2.5, find (a) P(X <10); (b) the value of k such that P(X k) = 0.1814; (d) P(15
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- When a population is finite, the formula that determines the standard error of the mean needs to be adjusted. If N is the size of the population and n is the size of the sample (where n ≥ 0.05N), the standard error of the mean is shown below. The finite population correction factor is given by the expression √(N-n)/(N-1). Determine the finite population correction factor for each of the following. (a) N = 1400 and n = 1050 (b) N = 1400 and n = 140 (d) N = 1400 and n = 70 (c) N = 1400 and n = 75 (e) What happens to the finite population correction factor as the sample size n decreases but the population size N remains the same? (a) The finite population correction factor is (Round to three decimal places as needed.) (b) The finite population correction factor is (Round to three decimal places as needed.) (c) The finite population correction factor is (Round to three decimal places as needed.) (d) The finite population correction factor is (Round to three decimal places as needed.) (e)…Is (∑i=2 Xi)/(n-1) a consistent and unbiased estimator of E(X)? Explain your result.Assume that the entire sample has 8 positive observations and 4 negatives observations. Variable X1: at the left branch has 9 positive observations and 3 negatives observations;at the right branch has 8 positive observations and 8 negative observations. What is the information gain or reduction in uncertainty of X1 using the Gini index? (round to two decimal spaces)
- Cardiologists use the short-range scaling exponent α1, which measures the randomness of heart rate patterns, as a tool to assess risk of heart attack. The article “Applying Fractal Analysis to Short Sets of Heart Rate Variability Data” compared values of α1 computed from long series of measurements (approximately 40,000 heartbeats) with those estimated from the first 300 beats to determine how well the long-term measurement (y) could be predicted the short-term one (x). Following are the data (obtained by digitizing a graph). Short Long 0.54 0.55 1.02 0.79 1.4 0.81 0.88 0.9 1.68 1.05 1.16 1.05 0.82 1.05 0.93 1.07 1.26 1.1 1.18 1.19 0.81 1.19 0.81 1.2 1.28 1.23 1.18 1.23 0.71 1.24 Note: This problem has a reduced data set for ease of performing the calculations required. This differs from the data set given for this problem in the text. Find a 95% confidence interval for the mean long-term measurement for those with short-term measurements…When a man observed a sobriety checkpoint conducted by a police department, he saw 656 drivers were screened and 5 were arrested for driving while intoxicated. Based on those results, we can estimate that P(W)equals=0.00762,where W denotes the event of screening a driver and getting someone who is intoxicated. What does P(W) denote, and what is its value? What does P(W) represent? P(W)=Cardiologists use the short-range scaling exponent α1, which measures the randomness of heart rate patterns, as a tool to assess risk of heart attack. The article “Applying Fractal Analysis to Short Sets of Heart Rate Variability Data” compared values of α1 computed from long series of measurements (approximately 40,000 heartbeats) with those estimated from the first 300 beats to determine how well the long-term measurement (y) could be predicted the short-term one (x). Following are the data (obtained by digitizing a graph). Short Long 0.54 0.55 1.02 0.79 1.4 0.81 0.88 0.9 1.68 1.05 1.16 1.05 0.82 1.05 0.93 1.07 1.26 1.1 1.18 1.19 0.81 1.19 0.81 1.2 1.28 1.23 1.18 1.23 0.71 1.24 Note: This problem has a reduced data set for ease of performing the calculations required. This differs from the data set given for this problem in the text. A. Compute the least-squares line for predicting the long-term measurement from the short-term measurement.…
- 1 (a) Discuss the two expressions -x;- )² and E(x;- )², both of which are used (n-1) i=1 to measure the spread of a set of observations x1, x2, . .., X. (b) A random sample of n observations is taken from a distribution; the sum of the observations is t, and the sum of the squares of the observations is 12. Explain how to estimate the mean and the variance of the distribution from which the random sample was taken. (c) Given the random sample described in part (b), write down expressions (based on 11 and t2) for estimates of the mean and variance of the mean of a further, independent, random sample of size m , from the original distribution. (d) Given that n = 25, 1, = 400 and t2 = 8800, construct a 99% confidence interval for the mean of the distribution, and use it to test whether or not this mean could be 20.Please solve within 20 minutesThe daily price of orange juice 30-day futures is normally distributed. InMarch through April 2007, the mean was 145.5 cents per pound, and standarddeviation = 25.0 cents per pound.4 Assuming the price is independent from day today, find P (x < 100) on the next day.
- please be clearThe 2010 U.S. Census found the chance of a household being a certain size. The data is in the pmf below ("Households by age," 2013). Let X be the number (size) in a household. E(X) = k·P(X = k) 7 (or more) P(X=k) 0.267 0.336 0.158 0.137 0.063 0.024 0.015 k 1 2 3 5 6 a) The probability of a household size being more than 5, P(X > 5) = % b) In the long run, we are expected to see a household size of, E(X)= on average. Round answer to three decimal places. c) The probability that the size of a household is equal to two is %. d) The probability of a household size being three OR six is %.A large automobile insurance company selected samples of single and married male policyholders and recorded the number who made an insurance claim over the preceding three-year period. Single Policyholders Married Policyholders 721 = 575 n2 = 740 Number making claims = 115 Number making claims = 111 a. Use α = 0.05. Test to determine whether the claim rates differ between single and married male policyholders. z-value p-value We can conclude (to 2 decimals) X (to 4 decimals) that there is the difference between claim rates. b. Provide a 95% confidence interval (to 4 decimals) for the difference between the proportions for the two populations. Enter negative answer as negative number.
