You are keen on tracking down a 95% certainty span for the mean number of visits for exercise based recuperation patients. The information beneath show the quantity of visits for 14 haphazardly chose exercise based recuperation patients. Round responses to 3 decimal spots where conceivable.

9 6 10 15 19 6 23 26 19 16 11 25 16 11

a. To figure the certainty span utilize a t or z dissemination.

b. With 95% certainty the populace mean number of visits per active recuperation patient is among ___ and ___ visits.

c. On the off chance that numerous gatherings of 14 arbitrarily chose non-intrusive treatment patients are examined, an alternate certainty stretch would be created from each gathering. About ___ percent of these certainty stretches will contain the genuine populace mean number of visits per patient and about ___ percent won't contain the genuine populace mean number of visits per patient.

Educator Smith loves dim chocolate, and she might want to discover the extent of the populace that shares her inclination. She haphazardly examined 700 individuals, and 630 individuals picked dull chocolate. To develop a 95% certainty stretch, she should utilize:

TInterval

1-PropZInt

2-SampZInt

@23@

2-PropZInt

ZInterval

2-SampTInt

Overall, $27,230 each year in educational cost and expenses. The standard deviation is $6,522. Accept the dispersion is ordinary. Leave X alone the expense for an arbitrarily chosen school. Round all responses to 4 decimal spots where conceivable.

a. What is the circulation of X? X ~ N(,)

b. Discover the likelihood that a haphazardly chosen Private charitable four-year school will cost under 29,094 every year.

c. Track down the 61th percentile for this circulation. $ (Round to the closest dollar.)

An editorial manager has a pile of k reports to audit. The request in which the reports are surveyed is arbitrary with each requesting being similarly likely. Of the k records to survey, two are named "Unwinding Through Mathematics" and "The Joy of Calculus." Give an articulation for every one of the probabilities underneath as a component of k. Work on your last articulation however much as could reasonably be expected so your answer does exclude any articulations in the structure (an over b).

(a) What is the likelihood that "Unwinding Through Mathematics" is first to audit?

(b) What is the likelihood that "Unwinding Through Mathematics" and "The Joy of Calculus" are close to one another in the stack?

Let X be an exponential random variable with parameter A. (a) Apply the Markov inequality to bound Pr[X 2 2/A]. (b) Use the Chebyshev inequality to compute the bound. (0) Use the one-sided Chebyshev inequality to compute the bound. (d) Compute the actual probability of this event. bound using Chebyshev Inequality. (b) From part (a), we see that the lower bound provided by Chebyshev Inequality is not very accurate. However, the inequality is very useful when applied to the sample mean from a large random sample. Let X = (X1 + ... + X400)/400 be the sample mean for a random sample from a normal population with mean 100 and standard deviation 15. Find the value of P( X - 100) O. Let X1, X2. ...; Xn be a random sample from the log-normal distribution where 0 = (#, o' ) is unknown. (a) Show that the log-normal distribution belongs to the exponential family. [2 marks] (b) Find the MVUE for a?. [5 marks] 0 (c) The Fisher information matrix is Is = ra in this case. Show that the MVUE for of is not efficient. [3 marks]Problem 5 A river has the following observed annual flood peaks at a gaging station for the period from 1982 to 1995. Determine the 2-year, 10-year, and 100-year flood discharges based on the log- normal distribution. The frequency factors for different exceedance probabilities for the log- normal distribution are shown in the following table. Annual Flood Peak Discharge Year (cfs ) Year (cfs) 1982 8000 1989 9400 1983 8800 1990 14200 1984 7400 1991 7600 1985 6700 1992 5800 1986 11100 1993 14300 1987 12200 1994 11600 1988 5700 1995 10400 Frequency Factors versus Exceedance Probabilities for the Log-normal Distribution P K P K P K 0.0001 3.719 0.100 1.282 0.600 -0.253 0.0005 3.291 0.150 1.036 0.650 -0.385 0.001 3.090 0.200 0.842 0.700 -0.524 0.002 2.880 0.250 0.674 0.750 -0.674 0.003 2.760 0.300 0.524 0.800 -0.842 0.004 2.650 0.350 0.385 0.850 -1.036 0.005 2.576 0.400 0.253 0.900 -1.282 0.010 2.326 0.450 0.126 0.950 -1.465 0.025 1.960 0.500 0.000 0.975 -1.960 0.050 1.845 0.550 -0.126 0.990 -2.326