Problem #4: For a 60 year old, their force of mortality x years from now is1/440Problem #4(a):Problem #4(b):Problem #4(c):180+(a) Find xP60 and evaluate it when x is 15.(b) What is 960?(c) What is the force of mortality for a 75-year old person 10 years from now in this model.A probability of 35% is to be entered as 0.3500.answer correct to 4 decimalsanswer correct to 4 decimalsanswer correct to 4 decimals4

Given that The formula to calculate is, The formula to calculate

Answered: Problem #4: For a 60 year old, their…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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6. A home improvement store carries two models of power drills. Lifetime (i.e., time to failure) of Model 1 drills are exponentially distributed with a rate of of 0.15 failures per year, whereas Model 2 drills have a lifetime that is log-normally distributed with parameters u = 0.75 and o = 1.2. (a) Which model of power drill is more likely to fail within a year? (b) Which brand produces cables that are more likely to last at least 2 years? (c) Which model of power drill has longer expected lifetime?
QUESTION 1 The same scenario is used for all questions. New parents in a little town buy two major brands of infant formulas, labeled A and B. After some intensive study, a marketing researcher for one of the manufacturers found that the probability of parents purchasing A or B is based on the parents' most recent purchase. Suppose that the following transition probabilities are appropriate O Probabilities of Parents Purchasing A or B From A To A To B O A B B Which of the following is the correct two period tree diagram for a customer who last purchased Brand A? O A 0.75 A 0.20 0.25 0.75 0.75 0.25 0.25 0.75 A B A B A 0.25 B 0.80 0.75 0.25 0.20 0.80 0.75 0.25 0.20 0.80 0.25 0.75 0.20 0.80 A B A B A B A B A B A B (0.25)(0.75)-0.1875 (0.25)(0.25) 0.0625 (0.75)(0.20) 0.1500 (0.75)(0.80)-0.6000 (0.75)(0.75)=0.5625 (0.75)(0.25)-0.1875 (0.25)(0.20)-0.0500 (0.25)(0.80)-0.2000 (0.25)(0.25) 0.0625 (0.25)(0.75)-0.1875 (0.75)(0.20) 0.1500 (0.75)(0.80) 0.6000
q-14
Solve completely
A medical test has been designed to detect the presence of a certain disease. Among people who have the disease, the probability that the disease will be detected by the test is 0.94. However, the probability that the test will erroneously indicate the presence of the disease in those who do not actually have it is 0.03. It is estimated that 5% of the population who take this test have the disease. (Round your answers to three decimal places.) (a) If the test administered to an individual is positive, what is the probability that the person actually has the disease?(b) If an individual takes the test twice and the test is positive both times, what is the probability that the person actually has the disease? (Assume that the tests are independent.)
Problem #2: Death for (x) is distributed such that the probability that an individual dies in the next 3 months is always the same regardless of their age. The probability that (x) will survive for 5 years is 35.8486%. What is the APV (the EPV) of the payout of a 1.5 year term life insurance policy on (x) if it pays $100,000 at the end of a quarter in which (x) dies? Use an annual interest rate of 9% compounded monthly (that's not a typo; I am giving you the monthly rate). Problem #2: answer correct to the nearest dollar
An article in the European Physical Journal B ["Evidence for the exponential distribution of income in the USA" (2001, Vol. 20, pp. 585- 589)] shows that individual income in the US can be approximated with an exponential distribution. Suppose that the mean of individual income in a year is $25000. eTextbook and Media (b) What percentage of individuals in the US earn less than $20000? Round your answer to three decimal places (e.g. 0.987). P(X 50000) = (d) What is the income exceeded by 1% of individuals? Round your answer to the nearest integer (e.g. 9876). x =

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