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- Q12) The distribution of the number of imperfections per 10 meters of synthetic fabric is given by the below table, If cost of this defect is given by the following equation g(x)=1/100 (x^2+2x+12), what is the * ?expected cost 1 0.37 3 4 0.41 0.16 0.05 0.01 0.1338 O 0.1538 O 0.1738 O 0.1938 OSuppose a sample Y1, ..., exponential distribution with E (Yi) = 0 and V (Yi) = 02. Let 0^ be an estimator of the mean, 8. Yn is selected from an 1. Explain what a parameter is. What is the parameter of interest in this case? 2. Name the properties of a good estimator and explain, shortly, what is meant by each. (a) (b) (c) 3. Given the properties you discussed in (2) and the parameter of interest identified in (1), what estimator would you propose? 4. What would the sampling distribution of this estimator be? Why?Calculate the test-statistic, t with the following information.n1=50n1=50, ¯x1=2.09x¯1=2.09, s1=0.96s1=0.96n2=40n2=40, ¯x2=2.4x¯2=2.4, s2=0.64s2=0.64Rounded to 2 decimal places.
- Two companies manufacture a rubber material in- tended for use in an automotive application. The part will be subjected to abrasive wear in the field application, so we decide to compare the material produced by each company in a test. Twenty-five samples of material from each company are tested in an abrasion test, and the amount of wear after 1000 cycles is observed. For company 1, the sample mean and standard de- viation of wear are = 20 milligrams/1000 cycles and s = 2 milligrams/1000 cycles, while for company 2 we obtain = 15 milligrams/1000 cycles and s, = 8 milligrams/1000 cycles. Find a 95% confidence interval on the difference in mean wear µ, - Hz. Assume each population is normally distributed but that their variances are not equal. What is the lower bound? Three decimal placesQ3/A/ Let X be a r.v with p.d.f f(x) = e-lkl - oQ2. Let f(x) = } 4(1 – x) (4x %3D 1/2 3/4) c) Find the probabilities given in part (b)Suppose the length of time a person takes to use an ATM machine is normally distributed with mean u = 110 seconds and standard deviation o = 6 seconds. There aren = 4 people ahead of Jackson in a line of people waiting to use the machine. He is concerned about T = total time the four people will take to use the machine. n USE SALT (a) What is the mean value, u, of T = total time for the four people ahead of Jackson? (b) Assuming that the times for the four people are independent of each other, determine the standard deviation of T. (c) Jackson hopes the total time T is less than or equal to 404 seconds. Find P(T < 404), the probability that the total waiting time is less than 404 seconds. (Round the answer to four decimal places.) P(T < 404) =Suppose that f (x) = e=* for 0 < x Determine the mean and variance of the random variable.Calculate the mean and variance when the probability variable X's moment-generating function is as follows f(x)= 2x-1/16 , x=1,2,3,4 my answer is mean = 25/8, var = 170/16 - (50/16)^2 but solution is mean = 2, var = 4/5 How do we solve the problem? Help meSuppose we want to test the hypothesis that mothers with low socioeconomic status (SES) deliver babies whose birth weights are different from normal. To test this hypothesis, a random sample of 100 birth weights is selected from a list of full-term babies of SES mothers. The mean birth weight is found to be 115 oz. Suppose the average birth weight of all babies (based on nationwide surveys of millions of deliveries) is known to be 120 oz with = 24 oz. Set = .05 Assume all conditions are met, what is the p-value of their test? Give your answer to 4 decimal places.Suppose the growth rate of a population is normal distribution with mean 50 and standard deviation 20. That is X~N(50,202)(1) Compute the probability of growth being larger than 70.(2) Compute the probability of growth being smaller than 40 (3) What is the rate of growth such that it touches the top 10% of growth level? (4) Find the range such that the growth is in the middle of 80% of growth level. please explain fully each stepThe mean weight of all Goliath grouper caught on the treasure coast is 435 pounds with a standard deviation of 51 pounds suppose a random sample of 77 Goliath groper from the tragicals are caught in Weighed.Determine the probability that the average weight of the sample of 77 Goliath Roper is within 6 pounds of the average weight of a goliath grouper caught on the treasure cost around the solution to four decimal places if necessary.P(x̅ within 6 pounds of μ)=SEE MORE QUESTIONS
