Consider an auto insurance portfolio where the number of accidents follows a Poisson distributionwith parameter λ= 1000. Suppose the damage sizes for separate accidents are i.i.d. (independentidentically distributed) r.v.'s having an exponential distribution with a mean of $2500.Each policy involves a deductible of $500. Let N₁ be the number of accidents that result in claims,and N₂ be the number of accidents that do not result in claims.Answer the following questions 1-5.Q1Are N₁, N₂ dependent?Q2Depends on a situation,NoYesWhat is the name of the distributions of N₁, №₂?O Marked PoissonGammaExponentialCompound Poisson Q3What is the mean and variance of N₁? The numbers below are rounded.Q4181.27, 32858.811000, 10001000, 1000000818.73, 818.73818.73, 670318.81What is the mean and variance for N₂? The numbers below are rounded.181.27, 181.271000, 10000001000, 1000818.73, 818.73818.73, 670318.81

Yes, . has Poisson distribution and has exponential distribution.

Answered: Consider an auto insurance portfolio…

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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1. ABC inc. stock is currently selling for $30, one year from today the stock price can either increase by 20% or decrease by 15%. The probability of an increase in the stock price is equal to 0.3. The one-year risk-free rate is 5% What is the value of a European put that expires in one year with an exercise price of $24. 2. Graphically, show the value and the profit and loss of the following butterfly position: Long in a call with an exercise price of $30, short in 2 calls with an exercise price of $45, and long in a call with an exercise price of 60. All calls are written on the same stock and have the same maturity. 3. "Early exercise of an American option on a stock that does not pay any dividend is not optimal regardless of whether the option is a Call or a Put". True, False, or Uncertain. Explain.
Period People (Y) Ocupation percentage (X) 2009 1T 46,977,904 59.7 2009 2T 47,453,163 59.3 2009 3T 48,738,589 60.2 2009 4T 48,903,792 60.4 2010 1T 48,069,274 59.9 2010 2T 49,133,132 60.2 2010 3T 49,190,032 58.7 2010 4T 48,478,718 58.8 2011 1T 48,505,168 59.3 2011 2T 49,482,112 59.5 2011 3T 50,127,032 59.7 2011 4T 50,772,496 60.5 2012 1T 50,192,842 60.3 2012 2T 51,477,178 60.7 2012 3T 51,927,050 60.7 2012 4T 51,317,999 60 2013 1T 50,847,242 60 2013 2T 51,895,865 60.4 2013 3T 52,034,353 60.2 2013 4T 52,370,886 60.7 2014 1T 51,559,018 60.3 2014 2T 51,836,752 59.8 2014 3T 52,192,043 59.6 2014 4T 52,108,400 59.4 2015 1T 52,007,842 59.7 2015 2T 52,623,721 59.6 2015 3T 53,179,919 59.7 2015 4T 53,809,017 60.3 2016 1T 52,918,649 59.7 2016 2T 53,539,565 59.6 2016 3T 54,226,803 59.9 2016 4T 54,034,800 59.6 2017 1T 53,681,720 59.7 2017 2T 54,068,791 59.3 2017 3T 54,369,915 59.1 2017 4T 54,696,638 59.3 2018 1T 54,590,773…
The economic dynamism, which is the index of productive growth (in dollars) for countries that are designated by the World Bank as middle-income are in following table ("SOCR data 2008," 2013): Table: Economic Dynamism ($) of Middle Income Countries 25.8057 37.4511 51.915 43.6952 47.8506 43.7178 58.0767 41.1648 38.0793 37.7251 39.6553 42.0265 48.6159 43.8555 49.1361 61.9281 41.9543 44.9346 46.0521 48.3652 43.6252 50.9866 59.1724 39.6282 33.6074 21.6643 Compute a 95% confidence interval for the mean economic dynamism of middle-income countries by answering the following questions (iv) Determine t score corresponding to desired confidence level Enter value in decimal form rounded to nearest hundredth. Examples of correctly entered answers: 2.01 0.37 0.60 0.00 (v) Determine error bound of the mean: Enter value in decimal form rounded to nearest ten-thousandth. Examples of correctly entered answers: 0.0000 0.0003 3.0060 2.0700…
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If the iron cutting machine in a factory works continuously for a certain period of time under the same conditions, it goes into malfunction. In this type of failure, the machine takes 24 hours to cool down. After the machine has cooled down, it continues to work the next day. Assume that the running time (minutes) of this machine until it malfunctions has a normal distribution. The working time of the iron cutting machine until the failure was measured for 9 different days, which were chosen at random. Continuous working time (minutes): 14-26-18-29-12-19-22-30-19 a) Establish the 95% confidence interval for the average uninterrupted working time of this iron cutting machine and interpret the result.b) The factory chief claims that the average continuous operating time of the machine is less than 23 minutes. Test the claim that the average continuous working time of this iron cutting machine is less than 23 minutes at the 0.05 significance level and the result isPlease comment.

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